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<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $rx$ is irrational.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
If $r x$ were rational, then $x=\frac{r x}{r}$ would also be rational.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $E$ be a nonempty subset of an ordered set; suppose $\alpha$ is a lower bound of $E$ and $\beta$ is an upper bound of $E$. Prove that $\alpha \leq \beta$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Since $E$ is nonempty, there exists $x \in E$. Then by definition of lower and upper bounds we have $\alpha \leq x \leq \beta$, and hence by property $i i$ in the definition of an ordering, we have $\alpha<\beta$ unless $\alpha=x=\beta$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Prove that no order can be defined in the complex field that turns it into an ordered field.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
By Part (a) of Proposition $1.18$, either $i$ or $-i$ must be positive. Hence $-1=i^2=(-i)^2$ must be positive. But then $1=(-1)^2$, must also be positive, and this contradicts Part $(a)$ of Proposition 1.18, since 1 and $-1$ cannot both be positive.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
If $z_1, \ldots, z_n$ are complex, prove that $|z_1 + z_2 + \ldots + z_n| \leq |z_1| + |z_2| + \cdots + |z_n|$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
We can apply the case $n=2$ and induction on $n$ to get
$$
\begin{aligned}
\left|z_1+z_2+\cdots z_n\right| &=\left|\left(z_1+z_2+\cdots+z_{n-1}\right)+z_n\right| \\
& \leq\left|z_1+z_2+\cdots+z_{n-1}\right|+\left|z_n\right| \\
& \leq\left|z_1\right|+\left|z_2\right|+\cdots+\left|z_{n-1}\right|+\left|z_n\right|
\end{aligned}
$$
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
If $z$ is a complex number such that $|z|=1$, that is, such that $z \bar{z}=1$, compute $|1+z|^{2}+|1-z|^{2}$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
$|1+z|^2=(1+z)(1+\bar{z})=1+\bar{z}+z+z \bar{z}=2+z+\bar{z}$. Similarly $|1-z|^2=(1-z)(1-\bar{z})=1-z-\bar{z}+z \bar{z}=2-z-\bar{z}$. Hence
$$
|1+z|^2+|1-z|^2=4 \text {. }
$$
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Prove that $|\mathbf{x}+\mathbf{y}|^{2}+|\mathbf{x}-\mathbf{y}|^{2}=2|\mathbf{x}|^{2}+2|\mathbf{y}|^{2}$ if $\mathbf{x} \in R^{k}$ and $\mathbf{y} \in R^{k}$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
The proof is a routine computation, using the relation
$$
|x \pm y|^2=(x \pm y) \cdot(x \pm y)=|x|^2 \pm 2 x \cdot y+|y|^2 .
$$
If $\mathrm{x}$ and $\mathrm{y}$ are the sides of a parallelogram, then $\mathrm{x}+\mathrm{y}$ and $\mathbf{x}-\mathrm{y}$ are its diagonals. Hence this result says that the sum of the squares on the diagonals of a parallelogram equals the sum of the squares on the sides.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
If $k = 1$ and $\mathbf{x} \in R^{k}$, prove that there does not exist $\mathbf{y} \in R^{k}$ such that $\mathbf{y} \neq 0$ but $\mathbf{x} \cdot \mathbf{y}=0$
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Not true when $k=1$, since the product of two nonzero real numbers is nonzero.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
If $A$ and $B$ are disjoint closed sets in some metric space $X$, prove that they are separated.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
We are given that $A \cap B=\varnothing$. Since $A$ and $B$ are closed, this means $A \cap \bar{B}=\varnothing=\bar{A} \cap B$, which says that $A$ and $B$ are separated.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Prove that every compact metric space $K$ has a countable base.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
$K$ can be covered by a finite union of neighborhoods of radius $1 / n$, and this shows that this implies that $K$ is separable.
It is not entirely obvious that a metric space with a countable base is separable. To prove this, let $\left\{V_n\right\}_{n=1}^{\infty}$ be a countable base, and let $x_n \in V_n$. The points $V_n$ must be dense in $X$. For if $G$ is any non-empty open set, then $G$ contains $V_n$ for some $n$, and hence $x_n \in G$. (Thus for a metric space, having a countable base and being separable are equivalent.)
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $E\subset\mathbb{R}^k$ is uncountable, and let $P$ be the set of condensation points of $E$. Prove that at most countably many points of $E$ are not in $P$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
If $x \in W^c$, and $O$ is any neighborhood of $x$, then $x \in V_n \subseteq O$ for some n. Since $x \notin W, V_n \cap E$ is uncountable. Hence $O$ contains uncountably many points of $E$, and so $x$ is a condensation point of $E$. Thus $x \in P$, i.e., $W^c \subseteq P$.
Conversely if $x \in W$, then $x \in V_n$ for some $V_n$ such that $V_n \cap E$ is countable. Hence $x$ has a neighborhood (any neighborhood contained in $V_n$ ) containing at most a countable set of points of $E$, and so $x \notin P$, i.e., $W \subseteq P^c$. Hence $P=W^c$.
It is clear that $P$ is closed (since its complement $W$ is open), so that we need only show that $P \subseteq P^{\prime}$. Hence suppose $x \in P$, and $O$ is any neighborhood of $x$. (By definition of $P$ this means $O \cap E$ is uncountable.) We need to show that there is a point $y \in P \cap(O \backslash\{x\})$. If this is not the case, i.e., if every point $y$ in $O \backslash\{x\}$ is in $P^c$, then for each such point $y$ there is a set $V_n$ containing $y$ such that $V_n \cap E$ is at most countable. That would mean that $y \in W$, i.e., that $O \backslash\{x\}$ is contained in $W$. It would follow that $O \cap E \subseteq\{x\} \cup(W \cap E)$, and so $O \cap E$ contains at most a countable set of points, contrary to the hypothesis that $x \in P$. Hence $O$ contains a point of $P$ different from $x$, and so $P \subseteq P^{\prime}$. Thus $P$ is perfect.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Prove that every open set in $\mathbb{R}$ is the union of an at most countable collection of disjoint segments.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $O$ be open. For each pair of points $x \in O, y \in O$, we define an equivalence relation $x \sim y$ by saying $x \sim y$ if and only if $[\min (x, y), \max (x, y)] \subset$ 0 . This is an equivalence relation, since $x \sim x([x, x] \subset O$ if $x \in O)$; if $x \sim y$, then $y \sim x$ (since $\min (x, y)=\min (y, x)$ and $\max (x, y)=\max (y, x))$; and if $x \sim y$ and $y \sim z$, then $x \sim z([\min (x, z), \max (x, z)] \subseteq[\min (x, y), \max (x, y)] \cup$ $[\min (y, z), \max (y, z)] \subseteq O)$. In fact it is easy to prove that
$$
\min (x, z) \geq \min (\min (x, y), \min (y, z))
$$
and
$$
\max (x, z) \leq \max (\max (x, y), \max (y, z))
$$
It follows that $O$ can be written as a disjoint union of pairwise disjoint equivalence classes. We claim that each equivalence class is an open interval.
To show this, for each $x \in O$; let $A=\{z:[z, x] \subseteq O\}$ and $B=\{z:[x, z] \subseteq$ $O\}$, and let $a=\inf A, b=\sup B$. We claim that $(a, b) \subset O$. Indeed if $a<z<b$, there exists $c \in A$ with $c<z$ and $d \in B$ with $d>z$. Then $z \in[c, x] \cup[x, d] \subseteq O$. We now claim that $(a, b)$ is the equivalence class containing $x$. It is clear that each element of $(a, b)$ is equivalent to $x$ by the way in which $a$ and $b$ were chosen. We need to show that if $z \notin(a, b)$, then $z$ is not equivalent to $x$. Suppose that $z<a$. If $z$ were equivalent to $x$, then $[z, x]$ would be contained in $O$, and so we would have $z \in A$. Hence $a$ would not be a lower bound for $A$. Similarly if $z>b$ and $z \sim x$, then $b$ could not be an upper bound for $B$.
We have now established that $O$ is a union of pairwise disjoint open intervals. Such a union must be at most countable, since each open interval contains a rational number not in any other interval.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Prove that $\lim_{n \rightarrow \infty}\sqrt{n^2 + n} -n = 1/2$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Multiplying and dividing by $\sqrt{n^2+n}+n$ yields
$$
\sqrt{n^2+n}-n=\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1} .
$$
It follows that the limit is $\frac{1}{2}$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
For any two real sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$, prove that $\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right) \leq \limsup _{n \rightarrow \infty} a_{n}+\limsup _{n \rightarrow \infty} b_{n},$ provided the sum on the right is not of the form $\infty-\infty$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Since the case when $\limsup _{n \rightarrow \infty} a_n=+\infty$ and $\limsup _{n \rightarrow \infty} b_n=-\infty$ has been excluded from consideration, we note that the inequality is obvious if $\limsup _{n \rightarrow \infty} a_n=+\infty$. Hence we shall assume that $\left\{a_n\right\}$ is bounded above.
Let $\left\{n_k\right\}$ be a subsequence of the positive integers such that $\lim _{k \rightarrow \infty}\left(a_{n_k}+\right.$ $\left.b_{n_k}\right)=\limsup _{n \rightarrow \infty}\left(a_n+b_n\right)$. Then choose a subsequence of the positive integers $\left\{k_m\right\}$ such that
$$
\lim _{m \rightarrow \infty} a_{n_{k_m}}=\limsup _{k \rightarrow \infty} a_{n_k} .
$$
The subsequence $a_{n_{k_m}}+b_{n_{k_m}}$ still converges to the same limit as $a_{n_k}+b_{n_k}$, i.e., to $\limsup _{n \rightarrow \infty}\left(a_n+b_n\right)$. Hence, since $a_{n_k}$ is bounded above (so that $\limsup _{k \rightarrow \infty} a_{n_k}$ is finite), it follows that $b_{n_{k_m}}$ converges to the difference
$$
\lim _{m \rightarrow \infty} b_{n_{k_m}}=\lim _{m \rightarrow \infty}\left(a_{n_{k_m}}+b_{n_{k_m}}\right)-\lim _{m \rightarrow \infty} a_{n_{k_m}} .
$$
Thus we have proved that there exist subsequences $\left\{a_{n_{k_m}}\right\}$ and $\left\{b_{n_{k_m}}\right\}$ which converge to limits $a$ and $b$ respectively such that $a+b=\limsup _{n \rightarrow \infty}\left(a_n+b_n^*\right)$. Since $a$ is the limit of a subsequence of $\left\{a_n\right\}$ and $b$ is the limit of a subsequence of $\left\{b_n\right\}$, it follows that $a \leq \limsup _{n \rightarrow \infty} a_n$ and $b \leq \limsup _{n \rightarrow \infty} b_n$, from which the desired inequality follows.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Prove that the convergence of $\Sigma a_{n}$ implies the convergence of $\sum \frac{\sqrt{a_{n}}}{n}$ if $a_n\geq 0$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Since $\left(\sqrt{a_n}-\frac{1}{n}\right)^2 \geq 0$, it follows that
$$
\frac{\sqrt{a_n}}{n} \leq \frac{1}{2}\left(a_n^2+\frac{1}{n^2}\right) .
$$
Now $\Sigma a_n^2$ converges by comparison with $\Sigma a_n$ (since $\Sigma a_n$ converges, we have $a_n<1$ for large $n$, and hence $\left.a_n^2<a_n\right)$. Since $\Sigma \frac{1}{n^2}$ also converges ($p$ series, $p=2$ ), it follows that $\Sigma \frac{\sqrt{a_n}}{n}$ converges.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Prove that the Cauchy product of two absolutely convergent series converges absolutely.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Since both the hypothesis and conclusion refer to absolute convergence, we may assume both series consist of nonnegative terms. We let $S_n=\sum_{k=0}^n a_n, T_n=\sum_{k=0}^n b_n$, and $U_n=\sum_{k=0}^n \sum_{l=0}^k a_l b_{k-l}$. We need to show that $U_n$ remains bounded, given that $S_n$ and $T_n$ are bounded. To do this we make the convention that $a_{-1}=T_{-1}=0$, in order to save ourselves from having to separate off the first and last terms when we sum by parts. We then have
$$
\begin{aligned}
U_n &=\sum_{k=0}^n \sum_{l=0}^k a_l b_{k-l} \\
&=\sum_{k=0}^n \sum_{l=0}^k a_l\left(T_{k-l}-T_{k-l-1}\right) \\
&=\sum_{k=0}^n \sum_{j=0}^k a_{k-j}\left(T_j-T_{j-1}\right) \\
&=\sum_{k=0}^n \sum_{j=0}^k\left(a_{k-j}-a_{k-j-1}\right) T_j \\
&=\sum_{j=0}^n \sum_{k=j}^n\left(a_{k-j}-a_{k-j-1}\right) T_j
&=\sum_{j=0}^n a_{n-j} T_j \\
&\leq T \sum_{m=0}^n a_m \\
&=T S_n \\
&\leq S T .
\end{aligned}
$$
Thus $U_n$ is bounded, and hence approaches a finite limit.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
If $\left\{E_{n}\right\}$ is a sequence of closed nonempty and bounded sets in a complete metric space $X$, if $E_{n} \supset E_{n+1}$, and if $\lim _{n \rightarrow \infty} \operatorname{diam} E_{n}=0,$ then $\bigcap_{1}^{\infty} E_{n}$ consists of exactly one point.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Choose $x_n \in E_n$. (We use the axiom of choice here.) The sequence $\left\{x_n\right\}$ is a Cauchy sequence, since the diameter of $E_n$ tends to zero as $n$ tends to infinity and $E_n$ contains $E_{n+1}$. Since the metric space $X$ is complete, the sequence $x_n$ converges to a point $x$, which must belong to $E_n$ for all $n$, since $E_n$ is closed and contains $x_m$ for all $m \geq n$. There cannot be a second point $y$ in all of the $E_n$, since for any point $y \neq x$ the diameter of $E_n$ is less $\operatorname{than} d(x, y)$ for large $n$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $f$ is a real function defined on $\mathbb{R}$ which satisfies $\lim_{h \rightarrow 0} f(x + h) - f(x - h) = 0$ for every $x \in \mathbb{R}$. Show that $f$ does not need to be continuous.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
$$
f(x)= \begin{cases}1 & \text { if } x \text { is an integer } \\ 0 & \text { if } x \text { is not an integer. }\end{cases}
$$
(If $x$ is an integer, then $f(x+h)-f(x-h) \equiv 0$ for all $h$; while if $x$ is not an integer, $f(x+h)-f(x-h)=0$ for $|h|<\min (x-[x], 1+[x]-x)$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $f$ be a continuous real function on a metric space $X$. Let $Z(f)$ (the zero set of $f$ ) be the set of all $p \in X$ at which $f(p)=0$. Prove that $Z(f)$ is closed.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
$Z(f)=f^{-1}(\{0\})$, which is the inverse image of a closed set. Hence $Z(f)$ is closed.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $E$ be a dense subset of $X$. Prove that if $g(p) = f(p)$ for all $p \in P$ then $g(p) = f(p)$ for all $p \in X$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
The function $\varphi: X \rightarrow R^1$ given by
$$
\varphi(p)=d_Y(f(p), g(p))
$$
is continuous, since
$$
\left|d_Y(f(p), g(p))-d_Y(f(q), g(q))\right| \leq d_Y(f(p), f(q))+d_Y(g(p), g(q))
$$
(This inequality follows from the triangle inequality, since
$$
d_Y(f(p), g(p)) \leq d_Y(f(p), f(q))+d_Y(f(q), g(q))+d_Y(g(q), g(p)),
$$
and the same inequality holds with $p$ and $q$ interchanged. The absolute value $\left|d_Y(f(p), g(p))-d_Y(f(q), g(q))\right|$ must be either $d_Y(f(p), g(p))-d_Y(f(q), g(q))$ or $d_Y(f(q), g(q))-d_Y(f(p), g(p))$, and the triangle inequality shows that both of these numbers are at most $d_Y(f(p), f(q))+d_Y(g(p), g(q))$.)
By the previous problem, the zero set of $\varphi$ is closed. But by definition
$$
Z(\varphi)=\{p: f(p)=g(p)\} .
$$
Hence the set of $p$ for which $f(p)=g(p)$ is closed. Since by hypothesis it is dense, it must be $X$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that there exist a set $E \subset \mathbb{R}$ and a real continuous function $f$ defined on $E$, such that there does not exist a continuous real function $g$ on $\mathbb{R}$ such that $g(x)=f(x)$ for all $x \in E$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $E:=(0,1)$, and define $f(x):=1 / x$ for all $x \in E$. If $f$ has a continuous extension $g$ to $\mathbb{R}$, then $g$ is continuous on $[-1,1]$, and therefore bounded by the intermediate value theorem. However, $f$ is not bounded in any neighborhood of $x=0$, so therefore $g$ is not bounded either, a contradiction.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $f$ be a real uniformly continuous function on the bounded set $E$ in $R^{1}$. Prove that $f$ is bounded on $E$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $a=\inf E$ and $b=\sup E$, and let $\delta>0$ be such that $|f(x)-f(y)|<1$ if $x, y \in E$ and $|x-y|<\delta$. Now choose a positive integer $N$ larger than $(b-a) / \delta$, and consider the $N$ intervals $I_k=\left[a+\frac{k-1}{b-a}, a+\frac{k}{b-a}\right], k=1,2, \ldots, N$. For each $k$ such that $I_k \cap E \neq \varnothing$ let $x_k \in E \cap I_k$. Then let $M=1+\max \left\{\left|f\left(x_k\right)\right|\right\}$. If $x \in E$, we have $\left|x-x_k\right|<\delta$ for some $k$, and hence $|f(x)|<M$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $f$ is a uniformly continuous mapping of a metric space $X$ into a metric space $Y$ and prove that $\left\{f\left(x_{n}\right)\right\}$ is a Cauchy sequence in $Y$ for every Cauchy sequence $\{x_n\}$ in $X$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Suppose $\left\{x_n\right\}$ is a Cauchy sequence in $X$. Let $\varepsilon>0$ be given. Let $\delta>0$ be such that $d_Y(f(x), f(u))<\varepsilon$ if $d_X(x, u)<\delta$. Then choose $N$ so that $d_X\left(x_n, x_m\right)<\delta$ if $n, m>N$. Obviously $d_Y\left(f\left(x_n\right), f\left(x_m\right)\right)<\varepsilon$ if $m, n>N$, showing that $\left\{f\left(x_n\right)\right\}$ is a Cauchy sequence.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Prove that every continuous open mapping of $R^{1}$ into $R^{1}$ is monotonic.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Suppose $f$ is continuous and not monotonic, say there exist points $a<b<c$ with $f(a)<f(b)$, and $f(c)<f(b)$. Then the maximum value of $f$ on the closed interval $[a, c]$ is assumed at a point $u$ in the open interval $(a, c)$. If there is also a point $v$ in the open interval $(a, c)$ where $f$ assumes its minimum value on $[a, c]$, then $f(a, c)=[f(v), f(u)]$. If no such point $v$ exists, then $f(a, c)=(d, f(u)]$, where $d=\min (f(a), f(c))$. In either case, the image of $(a, c)$ is not open.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $K$ and $F$ are disjoint sets in a metric space $X, K$ is compact, $F$ is closed. Prove that there exists $\delta>0$ such that $d(p, q)>\delta$ if $p \in K, q \in F$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Following the hint, we observe that $\rho_F(x)$ must attain its minimum value on $K$, i.e., there is some point $r \in K$ such that
$$
\rho_F(r)=\min _{q \in K} \rho_F(q) .
$$
Since $F$ is closed and $r \notin F$, it follows from Exercise $4.20$ that $\rho_F(r)>0$. Let $\delta$ be any positive number smaller than $\rho_F(r)$. Then for any $p \in F, q \in K$, we have
$$
d(p, q) \geq \rho_F(q) \geq \rho_F(r)>\delta .
$$
This proves the positive assertion.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $f$ be defined for all real $x$, and suppose that $|f(x)-f(y)| \leq (x-y)^{2}$ for all real $x$ and $y$. Prove that $f$ is constant.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Dividing by $x-y$, and letting $x \rightarrow y$, we find that $f^{\prime}(y)=0$ for all $y$. Hence $f$ is constant.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $g$ is a real function on $R^{1}$, with bounded derivative (say $\left|g^{\prime}\right| \leq M$ ). Fix $\varepsilon>0$, and define $f(x)=x+\varepsilon g(x)$. Prove that $f$ is one-to-one if $\varepsilon$ is small enough.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
If $0<\varepsilon<\frac{1}{M}$, we certainly have
$$
f^{\prime}(x) \geq 1-\varepsilon M>0,
$$
and this implies that $f(x)$ is one-to-one, by the preceding problem.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $f$ is defined and differentiable for every $x>0$, and $f^{\prime}(x) \rightarrow 0$ as $x \rightarrow+\infty$. Put $g(x)=f(x+1)-f(x)$. Prove that $g(x) \rightarrow 0$ as $x \rightarrow+\infty$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $\varepsilon>0$. Choose $x_0$ such that $\left|f^{\prime}(x)\right|<\varepsilon$ if $x>x_0$. Then for any $x \geq x_0$ there exists $x_1 \in(x, x+1)$ such that
$$
f(x+1)-f(x)=f^{\prime}\left(x_1\right) .
$$
Since $\left|f^{\prime}\left(x_1\right)\right|<\varepsilon$, it follows that $|f(x+1)-f(x)|<\varepsilon$, as required.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $f^{\prime}(x), g^{\prime}(x)$ exist, $g^{\prime}(x) \neq 0$, and $f(x)=g(x)=0$. Prove that $\lim _{t \rightarrow x} \frac{f(t)}{g(t)}=\frac{f^{\prime}(x)}{g^{\prime}(x)}.$
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Since $f(x)=g(x)=0$, we have
$$
\begin{aligned}
\lim _{t \rightarrow x} \frac{f(t)}{g(t)} &=\lim _{t \rightarrow x} \frac{\frac{f(t)-f(x)}{t-x}}{\frac{g(t)-g(x)}{t-x}} \\
&=\frac{\lim _{t \rightarrow x} \frac{f(t)-f(x)}{t-x}}{\lim _{t \rightarrow x} \frac{g(t)-g(x)}{t-x}} \\
&=\frac{f^{\prime}(x)}{g^{\prime}(x)}
\end{aligned}
$$
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $f$ is a real, three times differentiable function on $[-1,1]$, such that $f(-1)=0, \quad f(0)=0, \quad f(1)=1, \quad f^{\prime}(0)=0 .$ Prove that $f^{(3)}(x) \geq 3$ for some $x \in(-1,1)$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Following the hint, we observe that Theorem $5.15$ (Taylor's formula with remainder) implies that
$$
\begin{aligned}
f(1) &=f(0)+f^{\prime}(0)+\frac{1}{2} f^{\prime \prime}(0)+\frac{1}{6} f^{(3)}(s) \\
f(-1) &=f(0)-f^{\prime}(0)+\frac{1}{2} f^{\prime \prime}(0)-\frac{1}{6} f^{(3)}(t)
\end{aligned}
$$
for some $s \in(0,1), t \in(-1,0)$. By subtracting the second equation from the first and using the given values of $f(1), f(-1)$, and $f^{\prime}(0)$, we obtain
$$
1=\frac{1}{6}\left(f^{(3)}(s)+f^{(3)}(t)\right),
$$
which is the desired result. Note that we made no use of the hypothesis $f(0)=0$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that the collection $$\mathcal{T}_\infty = \{U | X - U \text{ is infinite or empty or all of X}\}$$ does not need to be a topology on the set $X$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $X=\mathbb{R}, U_1=(-\infty, 0)$ and $U_2=(0, \infty)$. Then $U_1$ and $U_2$ are in $\mathcal{T}_{\infty}$ but $U_1 \cup U_2=\mathbb{R} \backslash\{0\}$ is not.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
If $\mathcal{T}_\alpha$ is a family of topologies on $X$, show that $\bigcup \mathcal{T}_\alpha$ does not need to be a topology on $X$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
On the other hand, the union $\bigcup_\alpha \mathcal{T}_\alpha$ is in general not a topology on $X$. For instance, let $X=\{a, b, c\}$. Then $\mathcal{T}_1=\{\emptyset, X,\{a\}\}$ and $\mathcal{T}_2=\{\emptyset, X,\{b\}\}$ are topologies on $X$ but $\mathcal{T}_1 \cup \mathcal{T}_2=$ $\{\emptyset, X,\{a\},\{b\}\}$ is not.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $\mathcal{T}_\alpha$ be a family of topologies on $X$. Show that there is a unique largest topology on $X$ contained in all the collections $\mathcal{T}_\alpha$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Now we prove that there exists a unique largest topology contained in all $\mathcal{T}_\alpha$. Uniqueness of such topology is clear. Consider $\mathcal{T}=\bigcap_\alpha \mathcal{T}_\alpha$. We already know that $\mathcal{T}$ is a topology by, and clearly $\mathcal{T} \subset \mathcal{T}_\alpha$ for all $\alpha$. If $\mathcal{O}$ is another topology contained in all $\mathcal{T}_\alpha$, it must be contained in their intersection, so $\mathcal{O} \subset \mathcal{T}$. I follows that $\mathcal{T}$ is the unique largest topology contained in all $\mathcal{T}_\alpha$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that if $\mathcal{A}$ is a subbasis for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\mathcal{A}$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
If we now considered $\mathcal{A}$ as a subbasis, then the elements of $\mathcal{T}$ are union of finite intersections of elements of $\mathcal{A}$. The inclusion $\mathcal{O} \subset \mathcal{T}$ is again clear and $\mathcal{T} \subset \mathcal{O}$ holds since every union of finite intersections of elements of $\mathcal{A}$ belongs to $\mathcal{O}$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that the collection $\{(a,b) \mid a < b, a \text{ and } b \text{ rational}\}$ is a basis that generates the standard topology on $\mathbb{R}$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Exercise 13.8. (a) First note that $\mathcal{B}$ is a basis for a topology on $\mathbb{R}$. This follows from the fact that the union of its elements is all of $\mathbb{R}$ and the intersection of two elements of $\mathcal{B}$ is either empty or another element of $\mathcal{B}$. Let $\mathcal{T}$ be the standard topology on $\mathbb{R}$. Clearly the topology generated by $\mathcal{B}$ is coarser than $\mathcal{T}$. Let $U \in \mathcal{T}$ and $x \in U$. Then $U$ contains an open interval with centre $x$. Since the rationals are dense in $\mathbb{R}$ with the standard topology, there exists $q \in \mathbb{Q}$ such that $x \in(x-q, x+q) \subset U$. This proves that $\mathcal{T}$ is coarser than the topology generated by $\mathcal{B}$. We conclude that $\mathcal{B}$ generates the standard topology on $\mathbb{R}$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that if $Y$ is a subspace of $X$, and $A$ is a subset of $Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Exercise 16.1. Let $\mathcal{T}$ be the topology $A$ inherits as a subspace of $Y$, and $\mathcal{O}$ be the topology it inherits as a subspace of $X$. A (standard) basis element for $\mathcal{T}$ has the form $U \cap A$ where $U$ is open in $Y$, so is of the form $(Y \cap V) \cap A=V \cap A$ where $V$ is open in $X$. Therefore every basis element for $\mathcal{T}$ is also a basis element for $\mathcal{O}$. Conversely, a (standard) basis element for $\mathcal{O}$ have the form $W \cap A=W \cap Y \cap A$ where $W$ is open in $X$. Since $W \cap Y$ is open in $Y$, this is a basis element for $\mathcal{T}$, so every basis element for $\mathcal{O}$ is a basis element for $\mathcal{T}$. It follows that $\mathcal{T}=\mathcal{O}$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that the countable collection \[\{(a, b) \times (c, d) \mid a < b \text{ and } c < d, \text{ and } a, b, c, d \text{ are rational}\}\] is a basis for $\mathbb{R}^2$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
We know that $\mathcal{B}=\{(a,b)|a<b, a \text{ and } b \text rational\}$ is a basis for $\mathcal{R}$, therefore the set we are concerned with in the above question is a basis for $\mathcal{R}^2$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $Y$ be an ordered set in the order topology. Let $f, g: X \rightarrow Y$ be continuous. Show that the set $\{x \mid f(x) \leq g(x)\}$ is closed in $X$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
We prove that $U=\{x \mid g(x)<f(x)\}$ is open in $X$. Let $a \in U$, so that $g(a)<f(a)$. If there is an element $c$ between $g(a)$ and $f(a)$, then $a \in g^{-1}((-\infty, c)) \cap f^{-1}((c,+\infty))$. If there are no elements between $g(a)$ and $f(a)$, then $a \in g^{-1}\left((-\infty, f(a)) \cap f^{-1}((g(a),+\infty))\right.$. Note that all these preimages are open since $f$ and $g$ are continuous. Thus $U=V \cup W$, where
$$
V=\bigcup_{c \in X} g^{-1}((-\infty, c)) \cap f^{-1}((c,+\infty)) \quad \text { and } \quad W=\bigcup_{\substack{g(a)<f(a) \\(g(a), f(a))=\emptyset}} g^{-1}\left((-\infty, f(a)) \cap f^{-1}((g(a),+\infty))\right.
$$
are open in $X$. So $U$ is open in $X$ and therefore $X \backslash U=\{x \mid f(x) \leq g(x)\}$ is closed in $X$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $A \subset X$; let $f: A \rightarrow Y$ be continuous; let $Y$ be Hausdorff. Show that if $f$ may be extended to a continuous function $g: \bar{A} \rightarrow Y$, then $g$ is uniquely determined by $f$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $h, g: \bar{A} \rightarrow Y$ be continuous extensions of $f$. Suppose that there is a point $x \in \bar{A}$ such that $h(x) \neq g(x)$. Since $h=g$ on $A$, we must have $x \in A^{\prime}$. Since $Y$ is Hausdorff, there is a neighbourhood $U$ of $h(x)$ and a neighbourhood $V$ of $g(x)$ such that $U \cap V=\emptyset$. Since $h$ and $g$ are continuous, $h^{-1}(U) \cap g^{-1}(V)$ is a neighbourhood of $x$. Since $x \in A^{\prime}$, there is a point $y \in h^{-1}(U) \cap g^{-1}(V) \cap A$ different from $x$. But $h=g$ on $A$, so $g^{-1}(V) \cap A=h^{-1}(V) \cap A$ and hence $y \in h^{-1}(U) \cap h^{-1}(V)=h^{-1}(U \cap V)=\emptyset$, a contradiction. It follows that $h=g$ on $\bar{A}$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that $\mathbb{R} \times \mathbb{R}$ in the dictionary order topology is metrizable.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
The dictionary order topology on $\mathbb{R} \times \mathbb{R}$ is the same as the product topology $\mathbb{R}_d \times \mathbb{R}$, where $\mathbb{R}_d$ denotes $\mathbb{R}$ with the discrete topology. We know that $\mathbb{R}_d$ and $\mathbb{R}$ are metrisable. Thus, it suffices to show that the product of two metrisable spaces is metrisable.
So let $X$ and $Y$ be metrisable spaces, with metrics $d$ and $d^{\prime}$ respectively. On $X \times Y$, define
$$
\rho(x \times y, w \times z)=\max \left\{d(x, w), d^{\prime}(y, z)\right\} .
$$
Then $\rho$ is a metric on $X \times Y$; it remains to prove that it induces the product topology on $X \times Y$. If $B_d\left(x, r_1\right) \times B_d\left(y, r_2\right)$ is a basis element for the product space $X \times Y$, and $r=\min \left\{r_1, r_2\right\}$, then $x \times y \in B_\rho(x \times y, r) \subset B_d\left(x, r_1\right) \times B_d\left(y, r_2\right)$, so the product topology is coarser than the $\rho$-topology. Conversely, if $B_\rho(x \times y, \delta)$ is a basis element for the $\rho$-topology, then $x \times y \in B_d(x, \delta) \times B_{d^{\prime}}(y, \delta) \subset$ $B_\rho(x \times y, \delta)$, so the product topology is finer than the $\rho$-topology. It follows that both topologies are equal, so the product space $X \times Y$ is metrisable.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Define $f_{n}:[0,1] \rightarrow \mathbb{R}$ by the equation $f_{n}(x)=x^{n}$. Show that the sequence $\left(f_{n}\right)$ does not converge uniformly.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
The sequence $\left(f_n\right)_n$ does not converge uniformly, since given $0<\varepsilon<1$ and $N \in \mathbb{Z}_{+}$, for $x=\varepsilon^{1 / N}$ we have $d\left(f_N(x), f(x)\right)=\varepsilon$. We can also apply Theorem 21.6: the convergence is not uniform since $f$ is not continuous.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $p: X \rightarrow Y$ be a continuous map. Show that if there is a continuous map $f: Y \rightarrow X$ such that $p \circ f$ equals the identity map of $Y$, then $p$ is a quotient map.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $1_Y: Y \rightarrow Y$ be the identity map in $Y$. If $U$ is a subset of $Y$ and $p^{-1}(U)$ is open in $X$, then $f^{-1}\left(p^{-1}(U)\right)=1_Y^{-1}(U)=U$ is open in $Y$ by continuity of $f$. Thus $p$ is a quotient map.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $p \colon X \rightarrow Y$ be an open map. Show that if $A$ is open in $X$, then the map $q \colon A \rightarrow p(A)$ obtained by restricting $p$ is an open map.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $U$ be open in $A$. Since $A$ is open in $X, U$ is open in $X$ as well, so $p(U)$ is open in $Y$. Since $q(U)=p(U)=p(U) \cap p(A)$, the set $q(U)$ is open in $p(A)$. Thus $q$ is an open map.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $\left\{A_{\alpha}\right\}$ be a collection of connected subspaces of $X$; let $A$ be a connected subset of $X$. Show that if $A \cap A_{\alpha} \neq \varnothing$ for all $\alpha$, then $A \cup\left(\bigcup A_{\alpha}\right)$ is connected.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
For each $\alpha$ we have $A \cap A_\alpha \neq \emptyset$, so each $A \cup A_\alpha$ is connected by Theorem 23.3. In turn $\left\{A \cup A_\alpha\right\}_\alpha$ is a collection of connected spaces that have a point in common (namely any point in $A)$, so $\bigcup_\alpha\left(A \cup A_\alpha\right)=A \cup\left(\bigcup_\alpha A_\alpha\right)$ is connected.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $A \subset X$. Show that if $C$ is a connected subspace of $X$ that intersects both $A$ and $X-A$, then $C$ intersects $\operatorname{Bd} A$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Suppose that $C \cap B d A=C \cap \bar{A} \cap \overline{X-A}=\emptyset$. Then $C \cap A$ and $C \cap(X \backslash A)$ are a pair of disjoint non-empty sets whose union is all of $C$, neither of which contains a limit point of the other. Indeed, if $C \cap(X-A)$ contains a limit point $x$ of $C \cap A$, then $x \in C \cap(X-A) \cap A^{\prime} \subset C \cap \bar{A} \cap \overline{X-A}=\emptyset$, a contradiction, and similarly $C \cap A$ does not contain a limit point of $C \cap(X-A)$. Then $C \cap A$ and $C \cap(X-A)$ constitute a separation of $C$, contradicting the fact that $C$ is connected (Lemma 23.1).
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $p: X \rightarrow Y$ be a quotient map. Show that if each set $p^{-1}(\{y\})$ is connected, and if $Y$ is connected, then $X$ is connected.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Suppose that $U$ and $V$ constitute a separation of $X$. If $y \in p(U)$, then $y=p(x)$ for some $x \in U$, so that $x \in p^{-1}(\{y\})$. Since $p^{-1}(\{y\})$ is connected and $x \in U \cap p^{-1}(\{y\})$, we have $p^{-1}(\{y\}) \subset U$. Thus $p^{-1}(\{y\}) \subset U$ for all $y \in p(U)$, so that $p^{-1}(p(U)) \subset U$. The inclusion $U \subset p^{-1}(p(U))$ if true for any subset and function, so we have the equality $U=p^{-1}(p(U))$ and therefore $U$ is saturated. Similarly, $V$ is saturated. Since $p$ is a quotient map, $p(U)$ and $p(V)$ are disjoint non-empty open sets in $Y$. But $p(U) \cup p(V)=Y$ as $p$ is surjective, so $p(U)$ and $p(V)$ constitute a separation of $Y$, contradicting the fact that $Y$ is connected. We conclude that $X$ is connected.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $f \colon X \rightarrow X$ be continuous. Show that if $X = [0, 1]$, there is a point $x$ such that $f(x) = x$. (The point $x$ is called a fixed point of $f$.)
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
If $f(0)=0$ or $f(1)=1$ we are done, so suppose $f(0)>0$ and $f(1)<1$. Let $g:[0,1] \rightarrow[0,1]$ be given by $g(x)=f(x)-x$. Then $g$ is continuous, $g(0)>0$ and $g(1)<0$. Since $[0,1]$ is connected and $g(1)<0<g(0)$, by the intermediate value theorem there exists $x \in(0,1)$ such that $g(x)=0$, that is, such that $f(x)=x$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $G$ be a topological group; let $C$ be the component of $G$ containing the identity element $e$. Show that $C$ is a normal subgroup of $G$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Given $x \in G$, the maps $y \mapsto x y$ and $y \mapsto y x$ are homeomorphisms of $G$ onto itself. Since $C$ is a component, $x C$ and $C x$ are both components that contain $x$, so they are equal. Hence $x C=C x$ for all $x \in G$, so $C$ is a normal subgroup of $G$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $p: X \rightarrow Y$ be a closed continuous surjective map such that $p^{-1}(\{y\})$ is compact, for each $y \in Y$. (Such a map is called a perfect map.) Show that if $Y$ is compact, then $X$ is compact.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
We first show that if $U$ is an open set containing $p^{-1}(\{y\})$, then there is a neighbourhood $W$ of $y$ such that $p^{-1}(W)$ is contained in $U$. Since $X-U$ is closed in $X$, $p(X-U)$ is closed in $Y$ and does not contain $y$, so $W=Y \backslash p(X \backslash U)$ is a neighbourhood of $y$. Moreover, since $X \backslash U \subset p^{-1}(p(X \backslash U)$ ) (by elementary set theory), we have
$$
p^{-1}(W)=p^{-1}(Y \backslash p(X \backslash U))=p^{-1}(Y) \backslash p^{-1}(p(X \backslash U)) \subset X \backslash(X \backslash U)=U .
$$
Now let $\mathcal{A}$ be an open covering of $X$. For each $y \in Y$, let $\mathcal{A}_y$ be a subcollection of $\mathcal{A}$ such that
$$
p^{-1}(\{y\}) \subset \bigcup_{A \in \mathcal{A}_y} A .
$$
Since $p^{-1}(\{y\})$ is compact, there exists a finite subcollection of $\mathcal{A}_y$ that also covers $p^{-1}(\{y\})$, say $\left\{A_y^1, \ldots, A_y^{n_y}\right\}$. Thus $\bigcup_{i=1}^{n_y} A_y^i$ is open and contains $p^{-1}(\{y\})$, so there exists a neighbourhood $W_y$ of $y$ such that $p^{-1}\left(W_y\right)$ is contained in $\bigcup_{i=1}^{n_y} A_y^i$. Then $\left\{W_y\right\}_{y \in Y}$ is an open covering of $Y$, so there exist $y_1, \ldots, y_k \in Y$ such that $\left\{W_{y_j}\right\}_{j=1}^k$ also covers $Y$. Then
$$
X=p^{-1}(Y) \subset p^{-1}\left(\bigcup_{j=1}^k W_{y_j}\right)=\bigcup_{j=1}^k p^{-1}\left(W_{y_j}\right) \subset \bigcup_{j=1}^k\left(\bigcup_{i=1}^{n_{y_j}} A_{y_j}^i\right)
$$
so
$$
\left\{A_{y_j}^i\right\}_{\substack{j=1, \ldots, k . \\ i=1, \ldots, n_{y_j}}}
$$
is a finite subcollection of $\mathcal{A}$ that also covers $X$. Therefore, $X$ is compact.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
A space $X$ is said to be countably compact if every countable open covering of $X$ contains a finite subcollection that covers $X$. Show that for a $T_1$ space $X$, countable compactness is equivalent to limit point compactness.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
First let $X$ be a countable compact space. Note that if $Y$ is a closed subset of $X$, then $Y$ is countable compact as well, for if $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$is a countable open covering of $Y$, then $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}} \cup(X \backslash Y)$ is a countable open covering of $X$; there is a finite subcovering of $X$, hence a finite subcovering of $Y$. Now let $A$ be an infinite subset. We show that $A$ has a limit point. Let $B$ be a countable infinite subset of $A$. Suppose that $B$ has no limit point, so that $B$ is closed in $X$. Then $B$ is countable compact. Since $B$ has no limit point, for each $b \in B$ there is a neighbourhood $U_b$ of $b$ that intersects $B$ in the point $b$ alone. Then $\left\{U_b\right\}_{b \in B}$ is an open covering of $B$ with no finite subcovering, contradicting the fact that $B$ is countable compact. Hence $B$ has a limit point, so that $A$ has a limit point as well. Since $A$ was arbitrary, we deduce that $X$ is limit point compact. (Note that the $T_1$ property is not necessary in this direction.)
Now assume that $X$ is a limit point compact $T_1$ space. We show that $X$ is countable compact. Suppose, on the contrary, that $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$is a countable open covering of $X$ with no finite subcovering. For each $n$, take a point $x_n$ in $X$ not in $U_1 \cup \cdots \cup U_n$. By assumption, the infinite set $A=\left\{x_n \mid n \in \mathbb{Z}_{+}\right\}$has a limit point $y \in X$. Since $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$covers $X$, there exists $N \in \mathbb{Z}_{+}$such that $y \in U_1 \cup \cdots \cup U_N$. Now $X$ is $T_1$, so for each $i=1, \ldots, N$ there exists a neighbourhood $V_i$ of $y$ that does not contain $x_i$. Then
$$
V=\left(V_1 \cap \cdots \cap V_N\right) \cap\left(U_1 \cup \cdots \cup U_N\right)
$$
is a neighbourhood of $y$ that does not contain any of the points $x_i$, contradicting the fact that $y$ is a limit point of $A$. It follows that every countable open covering of $X$ must have a finite subcovering, so $X$ is countable compact.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $(X, d)$ be a metric space. If $f: X \rightarrow X$ satisfies the condition $d(f(x), f(y))=d(x, y)$ for all $x, y \in X$, then $f$ is called an isometry of $X$. Show that if $f$ is an isometry and $X$ is compact, then $f$ is bijective and hence a homeomorphism.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Note that $f$ is an imbedding. It remains to prove that $f$ is surjective. Suppose it is not, and let $a \in f(X)$. Since $X$ is compact, $f(X)$ is compact and hence closed (every metric space is Hausdorff). Thus, there exists $\varepsilon>0$ such that the $\varepsilon^{-}$ neighbourhood of $a$ is contained in $X \backslash f(X)$. Set $x_1=a$, and inductively $x_{n+1}=f\left(x_n\right)$ for $n \in \mathbb{Z}_{+}$. We show that $d\left(x_n, x_m\right) \geq \varepsilon$ for $n \neq m$. Indeed, we may assume $n<m$. If $n \geq 1$, then $d\left(x_n, x_m\right)=$ $d\left(f^{-1}\left(x_n\right), f^{-1}\left(x_m\right)\right)=d\left(x_{n-1}, x_{m-1}\right)$. By induction it follows that $d\left(x_n, x_m\right)=d\left(x_{n-i}, x_{m-i}\right)$ for all $i \geq 1$, and hence $d\left(x_n, x_m\right)=d\left(a, x_{m-n}\right)=d\left(a, f\left(x_{m-n-1}\right)\right)$. Since $f\left(x_{m-n-1}\right) \in f(X)$ and $B(a, \varepsilon) \cap f(X)=\emptyset$, we have $d\left(x_n, x_m\right) \geq \varepsilon$, as claimed. Thus $\left\{x_n\right\}_{n \in \mathbb{Z}_{+}}$is a sequence with no convergent subsequence, so $X$ is not sequentially compact. This contradicts the fact that $X$ is compact. Therefore $f$ is surjective and hence a homeomorphism.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that $[0, 1]^\omega$ is not locally compact in the uniform topology.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Consider $\mathbf{0} \in[0,1]^\omega$ and suppose that $[0,1]^\omega$ is locally compact at $\mathbf{0}$. Then there exists a compact $C$ containing an open ball $B=B_\rho(\mathbf{0}, \varepsilon) \subset[0,1]^\omega$. Note that $\bar{B}=[0, \varepsilon]^\omega$. Then $[0, \varepsilon]^\omega$ is closed and contained in the compact $C$, so it is compact. But $[0, \varepsilon]^\omega$ is homeomorphic to $[0,1]^\omega$, which is not compact by Exercise 28.1. This contradiction proves that $[0,1]^\omega$ is not locally compact in the uniform topology.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that if $X$ is a countable product of spaces having countable dense subsets, then $X$ has a countable dense subset.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $\left(X_n\right)$ be spaces having countable dense subsets $\left(A_n\right)$. For each $n$, fix an arbitrary $x_n \in X_n$. Consider the subset $A$ of $X$ defined by
$$
A=\bigcup\left\{\prod U_n: U_n=A_n \text { for finitely many } n \text { and is }\left\{x_n\right\} \text { otherwise }\right\} .
$$
This set is countable because the set of finite subsets of $\mathbb{N}$ is countable and each of the inner sets is countable. Now, let $x \in X$ and $V=\prod V_n$ be a basis element containing $x$ such that each $V_n$ is open in $X_n$ and $V_n=X_n$ for all but finitely many $n$. For each $n$, if $V_n \neq X_n$, choose a $y_n \in\left(A_n \cap V_n\right)$ (such a $y_n$ exists since $A_n$ is dense in $\left.X_n\right)$. Otherwise, let $y_n=x_n$. Then $\left(y_n\right) \in(A \cap V)$, proving that $A$ is dense in $X$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that if $X$ is regular, every pair of points of $X$ have neighborhoods whose closures are disjoint.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $x, y \in X$ be two points such that $x \neq y$. Since $X$ is regular (and thus Hausdorff), there exist disjoint open sets $U, V \subseteq X$ such that $x \in U$ and $y \in V$.
Note that $y \notin \bar{U}$. Otherwise $V$ must intersect $U$ in a point different from $y$ since $V$ is an open neighborhood of $y$, which is a contradiction since $U$ and $V$ are disjoint.
Since $X$ is regular and $\bar{U}$ is closed, there exist disjoint open sets $U^{\prime}, V^{\prime} \subseteq X$ such that $\bar{U} \subseteq U^{\prime}$ and $y \in V^{\prime}$.
And now $U$ and $V^{\prime}$ are neighborhoods of $x$ and $y$ whose closures are disjoint. If $\bar{U} \cap \overline{V^{\prime}} \neq \emptyset$, then it follows that $U^{\prime} \supseteq U$ intersects $\overline{V^{\prime}}$. Since $U^{\prime}$ is open, it follows that $U^{\prime}$ intersects $V^{\prime}$, which is a contradiction.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that every order topology is regular.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $X$ be an ordered set.
First we show that $X$ is a $T_1$-space. For $x \in X$ we have that
$$
X \backslash\{x\}=\langle-\infty, x\rangle \cup\langle x,+\infty\rangle
$$
which is an open set as an union of two open intervals. Therefore, the set $\{x\}$ is closed.
Step 2
2 of 3
Now to prove that $X$ is regular we use Lemma $\mathbf{3 1 . 1 .}$
Let $x \in X$ be any point and $U \subseteq X$ any open neighborhood of $x$. Then there exist $a, b \in X$ such that $x \in\langle a, b\rangle \subseteq U$. Now we have four possibilities.
1. If there exist $x_1, x_2 \in U$ such that $a<x_1<x<x_2<b$, then
$$
x \in\left\langle x_1, x_2\right\rangle \subseteq \overline{\left\langle x_1, x_2\right\rangle} \subseteq\left[x_1, x_2\right] \subseteq\langle a, b\rangle \subseteq U
$$
2. If there exists $x_1 \in U$ such that $a<x_1<x$, but there's no $x_2 \in U$ such that $x<x_2<b$, then
$$
x \in\left\langle x_1, b\right\rangle=\left(x_1, x\right] \subseteq \overline{\left(x_1, x\right]} \subseteq\left[x_1, x\right] \subseteq\langle a, b\rangle \subseteq U
$$
3. If there exists $x_2 \in U$ such that $x<x_2<b$, but there's no $x_1 \in U$ such that $a<x_1<x$, then
$$
x \in\left\langle a, x_2\right\rangle=\left[x, x_2\right) \subseteq \overline{\left[x, x_2\right)} \subseteq\left[x, x_2\right] \subseteq\langle a, b\rangle \subseteq U
$$
4. If there's no $x_1 \in U$ such that $a<x_1<x$ and no $x_2 \in U$ such that $x<x_2<b$, then
$$
x \in\langle a, b\rangle=\{x\}=\overline{\{x\}}=\{x\} \subseteq U
$$
We have that $\overline{\{x\}}=\{x\}$ because $X$ is a $T_1$-space.
In all four cases we proved that there exists an open interval $V$ such that $x \in V \subseteq \bar{V} \subseteq U$, so $X$ is regular.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that if $\prod X_\alpha$ is Hausdorff, then so is $X_\alpha$. Assume that each $X_\alpha$ is nonempty.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Suppose that $X=\prod_\beta X_\beta$ is Hausdorff and let $\alpha$ be any index.
Let $x, y \in X_\alpha$ be any points such that $x \neq y$. Since all $X_\beta$ are nonempty, there exist points $\mathbf{x}, \mathbf{y} \in X$ such that $x_\beta=y_\beta$ for every $\beta \neq \alpha$ and $x_\alpha=x, y_\alpha=y$.
Since $x \neq y$, it follows that $\mathbf{x} \neq \mathbf{y}$. Since $X$ is Hausdorff, there exist open disjoint sets $U, V \subseteq X$ such that $\mathbf{x} \in U$ and $\mathbf{y} \in V$.
For $\beta \neq \alpha$ we have that $x_\beta=y_\beta \in \pi_\beta(U) \cap \pi_\beta(V)$, hence $\pi_\beta(U)$ and $\pi_\beta(V)$ are not disjoint.
Since $U$ and $V$ are disjoint, it follows that $\pi_\alpha(U) \cap \pi_\alpha(V)=\emptyset$.
We also have that $x \in \pi_\alpha(U)$ and $y \in \pi_\alpha(V)$ and since the projections are open maps, it follows that the sets $\pi_\alpha(U)$ and $\pi_\alpha(V)$ are open.
This proves that $x$ and $y$ can be separated by open sets, so $X_\alpha$ is Hausdorff.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that if $\prod X_\alpha$ is normal, then so is $X_\alpha$. Assume that each $X_\alpha$ is nonempty.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Suppose that $X=\prod_\beta X_\beta$ is normal and let $\alpha$ be any index.
Since $X$ is normal, it follows that $X$ is Hausdorff (or regular), which then implies that $X_\alpha$ is Hausdorff (or regular). This imples that $X_\alpha$ satisfies the $T_1$ axiom.
Now the proof that $X_\alpha$ satisfies the $T_4$ axiom is the same as for regular spaces.
If $F, G \subseteq X_\alpha$ are disjoint closed sets, then $\prod_\beta F_\beta$ and $\prod_\beta G_\beta$, where $F_\alpha=F, G_\alpha=G$ and $F_\beta=G_\beta=X_\beta$ for $\beta \neq \alpha$, are disjoint closed sets in $X$.
Since $X$ is normal (and therefore satisfies the $T_4$ axiom), there exist disjoint open sets $U, V \subseteq X$ such that $\prod_\beta F_\beta \subseteq U$ and $\prod_\beta G_\beta \subseteq V$
Then $\pi_\alpha(U)$ and $\pi_\alpha(V)$ are disjoint open sets in $X_\alpha$ such that $F \subseteq \pi_\alpha(U)$ and $G \subseteq \pi_\alpha(V)$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that every locally compact Hausdorff space is completely regular.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
$X$ is a subspace of a compact Hausdorff space $Y$, its one-point compactification. $Y$ is normal, and so by the Urysohn lemma $Y$ is completely regular. Therefore by corollary $X$ is completely regular.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $X$ be a compact Hausdorff space that is the union of the closed subspaces $X_1$ and $X_2$. If $X_1$ and $X_2$ are metrizable, show that $X$ is metrizable.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Both $X_1$ and $X_2$ are compact, Hausdorff and metrizable, so by exercise 3 they are second countable, i.e. there are countable bases $\left\{U_{i, n} \subset X_i \mid n \in \mathbb{N}\right\}$ for $i \in\{1,2\}$. By the same exercise it is enough to show that $X$ is second countable. If $X_1 \cap X_2=\emptyset$ both $X_1$ and $X_2$ are open and the union $\left\{U_{i, n} \mid i \in\{1,2\} ; n \in \mathbb{N}\right\}$ of their countable bases form a countable base for $X$.
Suppose now $X_1 \cap X_2 \neq \emptyset$. Let $x \in X$ and $U \subset X$ be an open neighborhood of $x$. If $x \in X_i-X_j=X-$ $X_j$ then $U \cap X_i$ is open in $X_i$ and there is a basis neighborhood $U_{i, n}$ of $x$ such that $x \in U_{i, n} \cap X-X_j$ is an open neighborhood of $x$ in the open subset $X-X_j$, so $U_{i, n} \cap X-X_j$ is also open in $X$.
Suppose now that $x \in X_1 \cap X_2$. We have that $U \cap X_i$ is open in $X_i$ so there is a basis neighborhood $U_{i, n_i}$ contained in $U \cap X_i$. By definition of sub-space topology there is some open subset $V_{i, n_i} \subset X$ such that $U_{i, n_i}=$ $X_i \cap V_{i, n_i}$. Then
$$
x \in V_{1, n_1} \cap V_{2, n_2}=\left(V_{1, n_1} \cap V_{2, n_2} \cap X_1\right) \cup\left(V_{1, n_1} \cap V_{2, n_2} \cap X_2\right)=\left(U_{1, n_1} \cap V_{2, n_2}\right) \cup\left(V_{1, n_1} \cap U_{2, n_2}\right) \subset U
$$
Therefore the open subsets $U_{i, n} \cap X-X_j$ and $V_{1, n_1} \cap V_{2, n_2}$ form a countable base for $X$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces; let $Y$ be complete. Let $A \subset X$. Show that if $f \colon A \rightarrow Y$ is uniformly continuous, then $f$ can be uniquely extended to a continuous function $g \colon \bar{A} \rightarrow Y$, and $g$ is uniformly continuous.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
Let $\left(X, d_X\right)$ and $\left(Y, d_Y\right)$ be metric spaces; let $Y$ be complete. Let $A \subset X$. It is also given that $f: A \longrightarrow X$ is a uniformly continuous function. Then we have to show that $f$ can be uniquely extended to a continuous function $g: \bar{A} \longrightarrow Y$, and $g$ is uniformly coninuous.
We define the function $g$ as follows:
$$
g(x)= \begin{cases}f(x) & \text { if } x \in A \\ \lim _{n \rightarrow \infty} f\left(x_n\right) & \text { if } x \in \bar{A},\end{cases}
$$
where $\left\{x_n\right\} \subset A$ is some sequence in $A$ such that $\lim _{n \rightarrow \infty} x_n=x$. Now we have to check that the above definition is well defined. We first note that since $\left\{x_n\right\} \subset A$ is convergent in $X,\left\{x_n\right\}$ is Cauchy in $A$ and since $f$ is uniformly continuous on $A,\left\{f\left(x_n\right)\right\}$ is Cauchy in $Y$. Further, since $Y$ is complete $\left\{f\left(x_n\right)\right\}$ is convergent. So let us now consider two sequences $\left\{x_n\right\}$ and $\left\{y_n\right\}$ such that $\lim _{n \rightarrow \infty} x_n=\lim _{n \rightarrow \infty} y_n=x$. Then we need to prove that $\lim _{n \rightarrow \infty} f\left(x_n\right)=\lim _{n \rightarrow \infty} f\left(y_n\right)=f(x)$. Let
$$
\lim _{n \rightarrow \infty} f\left(x_n\right)=a \text { and } b=\lim _{n \rightarrow \infty} f\left(y_n\right)
$$
Now since $f$ is uniformly continuous for any given $\epsilon>0$ there exists $\delta>0$ such that
$$
d_Y(f(x), f(y))<\epsilon \text { whenever } d_X(x, y)<\delta \text { and } x, y \in A
$$
So for this $\delta>0$ there exists $N \in \mathbb{N}$ such that
$$
d_X\left(x_n, x\right)<\frac{\delta}{2} \text { and } d_X\left(y_n, x\right)<\frac{\delta}{2}, \text { foe all } n \geq N .
$$
Therefore, we have that for all $n \geq N$,
$$
d_X\left(x_n, y_n\right)<\delta
$$
Thus the equation (1) yields us that
$$
d_Y\left(f\left(x_n\right), f\left(y_n\right)\right)<\epsilon \text { for all } n \geq N .
$$
Now since $\lim _{n \rightarrow \infty} f\left(x_n\right)=a$ and $b=\lim _{n \rightarrow \infty} f\left(y_n\right)$, so for the above $\epsilon>0$ we have a natural number $K \geq N$ such that
$$
\begin{gathered}
d_Y\left(f\left(x_n\right), a\right)<\epsilon \text { for all } n \geq K \text { and } \\
d_Y\left(f\left(y_n\right), b\right)<\epsilon \text { for all } n \geq K .
\end{gathered}
$$
Moreover, since $K \geq N$, from $(2)$ we get
$$
d_Y\left(f\left(x_n\right), f\left(y_n\right)\right)<\epsilon \text { for all } n \geq K .
$$
Now we calculate the following, for $n \geq K$,
$$
\begin{array}{rlr}
d_Y(a, b) & \leq & d_Y\left(a, f\left(x_n\right)\right)+d_Y\left(f\left(x_n\right), f\left(y_n\right)\right)+d\left(f\left(y_n\right), b\right) \\
& < & \epsilon+\epsilon+\epsilon \text { by }(3),(4) \text { and }(5) \\
& = & 3 \epsilon
\end{array}
$$
where the first inequality holds because of triangular inequality. Since $\epsilon>0$ is arbitrary the above calculation shows that $d_Y(a, b)=0$. Thus, the above definition is independent of the choice of the sequence $\left\{x_n\right\}$ and hence the map $g$ is well defined. Moreover, from the construction it follows that $g$ is continuous on $\bar{A}$.
Moreover, we observe that $g$ is unique extension of $f$ by the construction.
So it remains to show that $g$ is uniformly continuous. In order to that we take a Cauchy sequence $\left\{a_n\right\} \subset \bar{A}$. Then since $\bar{A}$ is a closed set so the sequence $\left\{a_n\right\}$ is convergent and hence $\left\{g\left(a_n\right)\right\}$ is also a convergent sequence as $g$ is continuous on $\bar{A}$. So $\left\{g\left(a_n\right)\right\}$ is a Cauchy sequence in $Y$. Since a function is uniformly continuous if and only if it sends Cauchy sequences to Cauchy sequences, we conclude that $g$ is uniformly continuous.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Show that $\frac{-1 + \sqrt{3}i}{2}$ is a cube root of 1 (meaning that its cube equals 1).
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
$$
\left(\frac{-1+\sqrt{3} i}{2}\right)^2=\frac{-1-\sqrt{3} i}{2},
$$
hence
$$
\left(\frac{-1+\sqrt{3} i}{2}\right)^3=\frac{-1-\sqrt{3} i}{2} \cdot \frac{-1+\sqrt{3} i}{2}=1
$$
This means $\frac{-1+\sqrt{3} i}{2}$ is a cube root of 1.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Prove that if $a \in \mathbf{F}$, $v \in V$, and $av = 0$, then $a = 0$ or $v = 0$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
If $a=0$, then we immediately have our result. So suppose $a \neq 0$. Then, because $a$ is some nonzero real or complex number, it has a multiplicative inverse $\frac{1}{a}$. Now suppose that $v$ is some vector such that
$$
a v=0
$$
Multiply by $\frac{1}{a}$ on both sides of this equation to get
$$
\begin{aligned}
\frac{1}{a}(a v) & =\frac{1}{a} 0 & & \\
\frac{1}{a}(a v) & =0 & & \\
\left(\frac{1}{a} \cdot a\right) v & =0 & & \text { (associativity) } \\
1 v & =0 & & \text { (definition of } 1/a) \\
v & =0 & & \text { (multiplicative identity) }
\end{aligned}
$$
Hence either $a=0$ or, if $a \neq 0$, then $v=0$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Give an example of a nonempty subset $U$ of $\mathbf{R}^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $\mathbf{R}^2$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
$$
U=\left\{(x, y) \in \mathbf{R}^2:|x|=|y|\right\}
$$
For $(x, y) \in U$ and $\lambda \in \mathbb{R}$, it follows $\lambda(x, y)=$ $(\lambda x, \lambda y)$, so $|\lambda x|=|\lambda||x|=|\lambda||y|=|\lambda y|$. Therefore, $\lambda(x, y) \in U$.
On the other hand, consider $a=(1,-1), b=$ $(1,1) \in U$. Then, $a+b=(1,-1)+(1,1)=$ $(2,0) \notin U$. So, $U$ is not a subspace of $\mathbb{R}^2$.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Prove that the union of two subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
To prove this one way, suppose for purposes of contradiction that for $U_1$ and $U_2$, which are subspaces of $V$, that $U_1 \cup U_2$ is a subspace and neither is completely contained within the other. In other words, $U_1 \nsubseteq U_2$ and $U_2 \nsubseteq U_1$. We will show that you can pick a vector $v \in U_1$ and a vector $u \in U_2$ such that $v+u \notin U_1 \cup U_2$, proving that if $U_1 \cup U_2$ is a subspace, one must be completely contained inside the other.
If $U_1 \nsubseteq U_2$, we can pick a $v \in U_1$ such that $v \notin U_2$. Since $v$ is in the subspace $U_1$, then $(-v)$ must also be, by definition. Similarly, if $U_2 \nsubseteq U_1$, then we can pick a $u \in U_2$ such that $u \notin U_1$. Since $u$ is in the subspace $U_2$, then $(-u)$ must also be, by definition.
If $v+u \in U_1 \cup U_2$, then $v+u$ must be in $U_1$ or $U_2$. But, $v+u \in U_1 \Rightarrow v+u+(-v) \in U_1 \Rightarrow u \in U_1$
Similarly,
$$
v+u \in U_2 \Rightarrow v+u+(-u) \in U_2 \Rightarrow v \in U_2
$$
This is clearly a contradiction, as each element was defined to not be in these subspaces. Thus our initial assumption must have been wrong, and $U_1 \subseteq U_2$ or $U_2 \subseteq U_1$
To prove the other way, Let $U_1 \subseteq U_2$ (WLOG). $U_1 \subseteq U_2 \Rightarrow U_1 \cup U_2=U_2$. Since $U_2$ is a subspace, $U_1 \cup U_2$ is as well. QED.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose that $V$ is finite dimensional and that $T \in \mathcal{L}(V, W)$. Prove that there exists a subspace $U$ of $V$ such that $U \cap \operatorname{null} T=\{0\}$ and range $T=\{T u: u \in U\}$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
The point here is to note that every subspace of a vector space has a complementary subspace.
In this example, $U$ will precisely turn out to be the complementary subspace of null $T$. That is, $V=U \oplus \operatorname{null} T$
How should we characterize $U$ ? This can be achieved by extending a basis $B_1=\left\{v_1, v_2, \ldots, v_m\right\}$ of null $T$ to a basis of $V$. Let $B_2=\left\{u_1, u_2, \ldots, u_n\right\}$ be such that $B=B_1 \cup B_2$ is a basis of $V$.
Define $U=\operatorname{span}\left(B_2\right)$. Now, since $B_1$ and $B_2$ are complementary subsets of the basis $B$ of $V$, their spans will turn out to be complementary subspaces of $V$. Let's prove that $V=U \oplus$ null $T$.
Let $v \in V$. Then, $v$ can be expressed as a linear combination of the vectors in $B$.
Let $v=a_1 u_1+\cdots+a_n u_n+c_1 v_1+\cdots+c_m v_m$. However, since $\left\{u_1, u_2, \ldots, u_n\right\}$ is a basis of $U, a_1 u_1+$ $\cdots+a_n u_n=u \in U$ and since $\left\{v_1, v_2, \ldots, v_m\right\}$ is a basis of null $T, c_1 v_1+\cdots+c_m v_m=w \in$ null $T$.
Hence, $v=u+w \in U+\operatorname{null} T$. This shows that
$$
V=U+\operatorname{null} T
$$
Now, let $v \in U \cap \operatorname{null} T$.
Since $v \in U, u$ can be expressed as a linear combination of basis vectors of $U$. Let
$$
v=a_1 u_1+\cdots+a_n u_n
$$
Similarly, since $v \in \operatorname{null} T$, it can also be expressed as a tinear combination of the basis vectors of null $T$. Let
$$
v=c_1 v_1+\cdots+c_m v_m
$$
The left hand sides of the above two equations are equal. Therefore, we can equate the right hand sides.
$$
\begin{aligned}
& a_1 u_1+\cdots+a_n u_n=v=c_1 v_1+\cdots+c_m v_m \\
& a_1 u_1+\cdots+a_n u_n-c_1 v_1-\cdots-c_m v_m=0
\end{aligned}
$$
We have found a linear combination of $u_i^{\prime}$ 's and $v_i$ 's which is equal to zero. However, they are basis vectors of $V$. Hence, all the multipliers $c_i$ 's and $a_i$ 's must be zero implying that $v=0$.
Therefore, if $v \in U \cap$ null $T$, then $v=0$. this means that
$$
U \cap \operatorname{null} T=\{0\}
$$
The above shows that $U$ satisfies the first of the required conditions.
Now let $w \in$ range $T$. Then, there exists $v \in V$ such that $T v=w$. This allows us to write $v=u+w$ where $u \in U$ and $w \in$ null $T$. This implies
$$
\begin{aligned}
w & =T v \\
& =T(u+w) \\
& =T u+T w \\
& =T u+0 \quad \quad(\text { since } w \in \operatorname{null} T) \\
& =T u
\end{aligned}
$$
This shows that if $w \in$ range $T$ then $w=T u$ for some $u \in U$. Therefore, range $T \subseteq\{T u \mid u \in U\}$.
Since $U$ is a subspace of $V$, it follows that $T u \in$ range $T$ for all $u \in U$. Thus, $\{T u \mid u \in U\} \subseteq$ range $T$.
Therefore, range $T=\{T u \mid u \in U\}$.
This shows that $U$ satisfies the second required condition as well.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $T \in \mathcal{L}(V)$. Prove that if $U_{1}, \ldots, U_{m}$ are subspaces of $V$ invariant under $T$, then $U_{1}+\cdots+U_{m}$ is invariant under $T$.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
First off, assume that $U_1, \ldots, U_m$ are subspaces of $V$ invariant under $T$. Now, consider a vector $u \in$ $U_1+\ldots+U_m$. There does exist $u_1 \in U_1, \ldots, u_m \in U_m$ such that $u=u_1+\ldots+u_m$.
Once you apply $T$ towards both sides of the previous equation, we would then get $T u=T u_1+\ldots+$ $T u_m$.
Since each $U_j$ is invariant under $T$, then we would have $T u_1 \in U_1+\ldots+T u_m$. This would then make the equation shows that $T u \in U_1+\ldots+T u_m$, which does imply that $U_1+. .+U_m$ is invariant under $T$
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $S, T \in \mathcal{L}(V)$. Prove that $S T$ and $T S$ have the same eigenvalues.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
To start, let $\lambda \in F$ be an eigenvalue of $S T$. Now, we would want $\lambda$ to be an eigenvalue of $T S$. Since $\lambda$, by itself, is an eigenvalue of $S T$, then there has to be a nonzero vector $v \in V$ such that $(S T) v=\lambda v$.
Now, With a given reference that $(S T) v=\lambda v$, you will then have the following: $(T S)(T v)=$ $T(S T v)=T(\lambda v)=\lambda T v$
If $T v \neq 0$, then the listed equation above shows that $\lambda$ is an eigenvalue of $T S$.
If $T v=0$, then $\lambda=0$, since $S(T v)=\lambda T v$. This also means that $T$ isn't invertible, which would imply that $T S$ isn't invertible, which can also be implied that $\lambda$, which equals 0 , is an eigenvalue of $T S$.
Step 3
3 of 3
Now, regardless of whether $T v=0$ or not, we would have shown that $\lambda$ is an eigenvalue of $T S$. Since $\lambda$ (was) an arbitrary eigenvalue of $S T$, we have shown that every single eigenvalue of $S T$ is an eigenvalue of $T S$. When you do reverse the roles of both $S$ and $T$, then we can conclude that that every single eigenvalue of $T S$ is also an eigenvalue of $S T$. Therefore, both $S T$ and $T S$ have the exact same eigenvalues.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $T \in \mathcal{L}(V)$ is such that every subspace of $V$ with dimension $\operatorname{dim} V-1$ is invariant under $T$. Prove that $T$ is a scalar multiple of the identity operator.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
First off, let $T$ isn't a scalar multiple of the identity operator. So, there does exists that $v \in V$ such that $u$ isn't an eigenvector of $T$. Therefore, $(u, T u)$ is linearly independent.
Next, you should extend $(u, T u)$ to a basis of $\left(u, T u, v_1, \ldots, v_n\right)$ of $V$. So, let $U=\operatorname{span}\left(u, v_1, \ldots, v_n\right)$. Then, $U$ is a subspace of $V$ and $\operatorname{dim} U=\operatorname{dim} V-1$. However, $U$ isn't invariant under $T$ since both $u \in U$ and $T u \in U$. This given contradiction to our hypothesis about $T$ actually shows us that our guess that $T$ is not a scalar multiple of the identity must have been false.
\end{proof}
|
<|im_start|>system
Please reason step by step.
<|im_end|>
<|im_start|>user
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.
<|im_end|>
<|im_start|>assistant
$r$ and $r+x$ were both rational.
Then $x=r+x-r$ would also be rational.
<|im_end|>
<|im_start|>user
Prove that there is no rational number whose square is $12$.
<|im_end|>
<|im_start|>assistant
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor.
It follows that $m$ must be even, and therefore $n$ must be odd.
Let $m=2 r$.
Then we have $r^2=3 n^2$, so that $r$ is also odd.
Let $r=2 s+1$ and $n=2 t+1$.
Then $$ 4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3 $$.
So $$ 4\left(s^2+s-3 t^2-3 t\right)=2 $$.
But this is absurd, since 2 cannot be a multiple of 4.
<|im_end|>
<|im_start|>user
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.
<|im_end|>
<|im_start|>assistant
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$.
For brevity, let $\alpha=-\sup (-A)$.
We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$.
Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$.
It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$.
Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$.
This means $\beta \leq x$ for all $x$ in $A$.
Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$.
This means $-\beta$ is an upper bound of $-A$.
Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$.
And so $-\sup (-A)$ is the greatest lower bound of $A$.
<|im_end|>
<|im_start|>user
Suppose $V$ is a real vector space and $T \in \mathcal{L}(V)$ has no eigenvalues. Prove that every subspace of $V$ invariant under $T$ has even dimension.
<|im_end|>
<|im_start|>assistant
|
\begin{proof}
First off, let us assume that $U$ is a subspace of $V$ that is invariant under $T$. Therefore, $\left.T\right|_U \in \mathcal{L}(U)$. If $\operatorname{dim}$ $U$ were odd, then $\left.T\right|_U$ would have an eigenvalue $\lambda \in \mathbb{R}$, so there would exist a nonzero vector $u \in U$ such that
$$
\left.T\right|_U u=\lambda u .
$$
So, this would imply that $T_u=\lambda u$, which would imply that $\lambda$ is an eigenvalue of $T$. But $T$ has no eigenvalues, so $\operatorname{dim} U$ must be even.
\end{proof}
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