question stringlengths 705 1.18k | answer stringlengths 0 3.93k |
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<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $rx$ is irrational.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
If $r x$ were rational, then $x=\frac{r x}{r}$ would also be rational.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that no order can be defined in the complex field that turns it into an ordered field.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \begin{proof}
Not true when $k=1$, since the product of two nonzero real numbers is nonzero.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: If $A$ and $B$ are disjoint closed sets in some metric space $X$, prove that they are separated.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that every compact metric space $K$ has a countable base.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that $\lim_{n \rightarrow \infty}\sqrt{n^2 + n} -n = 1/2$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that the Cauchy product of two absolutely convergent series converges absolutely.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that every continuous open mapping of $R^{1}$ into $R^{1}$ is monotonic.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that $\mathbb{R} \times \mathbb{R}$ in the dictionary order topology is metrizable.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that $[0, 1]^\omega$ is not locally compact in the uniform topology.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that if $X$ is regular, every pair of points of $X$ have neighborhoods whose closures are disjoint.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that every order topology is regular.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that every locally compact Hausdorff space is completely regular.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that if $a \in \mathbf{F}$, $v \in V$, and $av = 0$, then $a = 0$ or $v = 0$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Suppose $S, T \in \mathcal{L}(V)$. Prove that $S T$ and $T S$ have the same eigenvalues.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: 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
<think>
</think>
| \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} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that $\left(\sum_{j=1}^{n} a_{j} b_{j}\right)^{2} \leq\left(\sum_{j=1}^{n} j a_{j}{ }^{2}\right)\left(\sum_{j=1}^{n} \frac{b_{j}{ }^{2}}{j}\right)$ for all real numbers $a_{1}, \ldots, a_{n}$ and $b_{1}, \ldots, b_{n}$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in R$.
We have that
$$
\left(\sum_{j=1}^n a_j b_j\right)^2
$$
is equal to the
$$
\left(\sum_{j=1}^n a_j b_j \frac{\sqrt{j}}{\sqrt{j}}\right)^2=\left(\sum_{j=1}^n\left(\sqrt{j} a_j\right)\left(b_j \frac{1}{\sqrt{j}}\right)\right)^2
$$
This can be observed as an inner product, and using the Cauchy-Schwarz Inequality, we get
$$
\begin{aligned}
&\left(\sum_{j=1}^n a_j b_j\right)^2=\left(\sum_{j=1}^n\left(\sqrt{j} a_j\right)\left(b_j \frac{1}{\sqrt{j}}\right)\right)^2 \\
&=\left\langle\left(a, \sqrt{2} a_2, \ldots, \sqrt{n} a_n\right),\left(b_1, \frac{b_2}{\sqrt{2}}, \ldots, \frac{b_n}{\sqrt{n}}\right)\right\rangle \\
& \leq\left\|\left(a, \sqrt{2} a_2, \ldots, \sqrt{n} a_n\right)\right\|^2\left\|\left(b_1, \frac{b_2}{\sqrt{2}}, \ldots, \frac{b_n}{\sqrt{n}}\right)\right\|^2 \\
&=\left(\sum_{j=1}^n j a_j^2\right)\left(\sum_{j=1}^n \frac{b_j^2}{j}\right) \\
& \text { Hence, }\left(\sum_{j=1}^n a_j b_j\right)^2=\left(\sum_{j=1}^n j a_j^2\right)\left(\sum_{j=1}^n \frac{b_j^2}{j}\right) .
\end{aligned}
$$
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Suppose $\left(e_{1}, \ldots, e_{m}\right)$ is an or thonormal list of vectors in $V$. Let $v \in V$. Prove that $\|v\|^{2}=\left|\left\langle v, e_{1}\right\rangle\right|^{2}+\cdots+\left|\left\langle v, e_{m}\right\rangle\right|^{2}$ if and only if $v \in \operatorname{span}\left(e_{1}, \ldots, e_{m}\right)$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
If $v \in \operatorname{span}\left(e_1, \ldots, e_m\right)$, it means that
$$
v=\alpha_1 e_1+\ldots+\alpha_m e_m .
$$
for some scalars $\alpha_i$. We know that $\alpha_k=\left\langle v, e_k\right\rangle, \forall k \in\{1, \ldots, m\}$. Therefore,
$$
\begin{aligned}
\|v\|^2 & =\langle v, v\rangle \\
& =\left\langle\alpha_1 e_1+\ldots+\alpha_m e_m, \alpha_1 e_1+\ldots+\alpha_m e_m\right\rangle \\
& =\left|\alpha_1\right|^2\left\langle e_1, e_1\right\rangle+\ldots+\left|\alpha_m\right|^2\left\langle e_m, e_m\right\rangle \\
& =\left|\alpha_1\right|^2+\ldots+\left|\alpha_m\right|^2 \\
& =\left|\left\langle v, e_1\right\rangle\right|^2+\ldots+\left|\left\langle v, e_m\right\rangle\right|^2 .
\end{aligned}
$$
$\Rightarrow$ Assume that $v \notin \operatorname{span}\left(e_1, \ldots, e_m\right)$. Then, we must have
$$
v=v_{m+1}+\frac{\left\langle v, v_0\right\rangle}{\left\|v_0\right\|^2} v_0,
$$
where $v_0=\alpha_1 e_1+\ldots+\alpha_m e_m, \alpha_k=\left\langle v, e_k\right\rangle, \forall k \in\{1, \ldots, m\}$, and $v_{m+1}=v-$ $\frac{\left\langle v, v_0\right\rangle}{\left\|v_0\right\|^2} v_0 \neq 0$.
We have $\left\langle v_0, v_{m+1}\right\rangle=0$ (from which we get $\left\langle v, v_0\right\rangle=\left\langle v_0, v_0\right\rangle$ and $\left\langle v, v_{m+1}\right\rangle=$ $\left.\left\langle v_{m+1}, v_{m+1}\right\rangle\right)$. Now,
$$
\begin{aligned}
\|v\|^2 & =\langle v, v\rangle \\
& =\left\langle v, v_{m+1}+\frac{\left\langle v, v_0\right\rangle}{\left\|v_0\right\|^2} v_0\right\rangle \\
& =\left\langle v, v_{m+1}\right\rangle+\left\langle v, \frac{\left\langle v, v_0\right\rangle}{\left\|v_0\right\|^2} v_0\right\rangle \\
& =\left\langle v_{m+1}, v_{m+1}\right\rangle+\frac{\left\langle v_0, v_0\right\rangle}{\left\|v_0\right\|^2}\left\langle v_0, v_0\right\rangle \\
& =\left\|v_{m+1}\right\|^2+\left\|v_0\right\|^2 \\
& >\left\|v_0\right\|^2 \\
& =\left|\alpha_1\right|^2+\ldots+\left|\alpha_m\right|^2 \\
& =\left|\left\langle v, e_1\right\rangle\right|^2+\ldots+\left|\left\langle v, e_m\right\rangle\right|^2 .
\end{aligned}
$$
By contrapositive, if $\left\|v_1\right\|^2=\left|\left\langle v, e_1\right\rangle\right|^2+\ldots+\left|\left\langle v, e_m\right\rangle\right|^2$, then $v \in \operatorname{span}\left(e_1, \ldots, e_m\right)$.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that if $\operatorname{dim} V \geq 2$, then the set of normal operators on $V$ is not a subspace of $\mathcal{L}(V)$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
First off, suppose that $\operatorname{dim} V \geq 2$. Next let $\left(e_1, \ldots, e_n\right)$ be an orthonormal basis of $V$. Now, define $S, T \in L(V)$ by both $S\left(a_1 e_1+\ldots+a_n e_n\right)=a_2 e_1-a_1 e_2$ and $T\left(a_1 e_1+\ldots+\right.$ $\left.a_n e_n\right)=a_2 e_1+a_1 e_2$. So, just by now doing a simple calculation verifies that $S^*\left(a_1 e_1+\right.$ $\left.\ldots+a_n e_n\right)=-a_2 e_1+a_1 e_2$
Now, based on this formula, another calculation would show that $S S^*=S^* S$. Another simple calculation would that that $T$ is self-adjoint. Therefore, both $S$ and $T$ are normal. However, $S+T$ is given by the formula of $(S+T)\left(a_1 e_1+\ldots+a_n e_n\right)=2 a_2 e_1$. In this case, a simple calculator verifies that $(S+T)^*\left(a_1 e_1+\ldots+a_n e_n\right)=2 a_1 e_2$.
Therefore, there is a final simple calculation that shows that $(S+T)(S+T)^* \neq(S+$ $T)^*(S+T)$. So, in other words, $S+T$ isn't normal. Thereofre, the set of normal operators on $V$ isn't closed under addition and hence isn't a subspace of $L(V)$.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that a normal operator on a complex inner-product space is self-adjoint if and only if all its eigenvalues are real.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
First off, suppose $V$ is a complex inner product space and $T \in L(V)$ is normal. If $T$ is self-adjoint, then all its eigenvalues are real. So, conversely, let all of the eigenvalues of $T$ be real. By the complex spectral theorem, there's an orthonormal basis $\left(e_1, \ldots, e_n\right)$ of $V$ consisting of eigenvectors of $T$. Thus, there exists real numbers $\lambda_1, \ldots, \lambda_n$ such that $T e_j=\lambda_j e_j$ for $j=$ $1, \ldots, n$.
The matrix of $T$ with respect to the basis of $\left(e_1, \ldots, e_n\right)$ is the diagonal matrix with $\lambda_1, \ldots, \lambda_n$ on the diagonal. So, the matrix equals its conjugate transpose. Therefore, $T=T^*$. In other words, $T$ s self-adjoint.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Suppose $V$ is a complex inner-product space. Prove that every normal operator on $V$ has a square root. (An operator $S \in \mathcal{L}(V)$ is called a square root of $T \in \mathcal{L}(V)$ if $S^{2}=T$.)<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Let $V$ be a complex inner product space.
It is known that an operator $S \in \mathcal{L}(V)$ is called a square root of $T \in \mathcal{L}(V)$ if
$$
S^2=T
$$
Now, suppose that $T$ is a normal operator on $V$.
By the Complex Spectral Theorem, there is $e_1, \ldots, e_n$ an orthonormal basis of $V$ consisting of eigenvalues of $T$ and let $\lambda_1, \ldots, \lambda_n$ denote their corresponding eigenvalues.
Define $S$ by
$$
S e_j=\sqrt{\lambda_j} e_j,
$$
for each $j=1, \ldots, n$.
Obviously, $S^2 e_j=\lambda_j e_j=T e_j$.
Hence, $S^2=T$ so there exist a square root of $T$.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that $\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ is not an integer.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Let $2^s$ be the largest power of 2 occuring as a denominator in $H_n$, say $2^s=k \leqslant n$. Write $H_n=$ $\frac{1}{2^s}+\left(1+1 / 2+\ldots+1 /(k-1)+1 /(k+1)+\ldots+1 / n\right.$. The sum in parentheses can be written as $1 / 2^{s-1}$ times sum of fractions with odd denominators, so the denominator of the sum in parentheses will not be divisible by $2^s$, but it must equal $2^s$ by Ex $1.29$.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: If $a$ is a nonzero integer, then for $n>m$ show that $\left(a^{2^{n}}+1, a^{2^{m}}+1\right)=1$ or 2 depending on whether $a$ is odd or even.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
\begin{align*}
\operatorname{ord}_p\, n! &= \sum_{k\geq 1} \left \lfloor \frac{n}{p^{k}}\right \rfloor \leq \sum_{k\geq 1} \frac{n}{p^{k}} = \frac{n}{p} \frac{1}{1 - \frac{1}{p}} = \frac{n}{p-1}
\end{align*}
The decomposition of $n!$ in prime factors is
$n! = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$
where $\alpha_i = \operatorname{ord}_{p_i}\, n! \leq \frac{n}{p_i-1}$, and $p_i \leq n, \ i=1,2,\cdots,k$.
Then
\begin{align*}
n! &\leq p_1^{\frac{n}{p_1-1}}p_2^{\frac{n}{p_2-1}}\cdots p_k^{\frac{n}{p_n-1}}\\
\sqrt[n]{n!} &\leq p_1^{\frac{1}{p_1-1}}p_2^{\frac{1}{p_2-1}}\cdots p_k^{\frac{1}{p_n-1}}\\
&\leq \prod_{p\leq n} p^{\frac{1}{p-1}}
\end{align*}
(the values of $p$ in this product describe all prime numbers $p\leq n$.)
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that $\sum^{\prime} 1 / n$, the sum being over square free integers, diverges.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Let $S \subset \mathbb{N}^*$ the set of square free integers.
Let $N \in \mathbb{N}^*$. Every integer $n, \, 1\leq n \leq N$ can be written as $n = a b^2$, where $a,b$ are integers and $a$ is square free. Then $1\leq a \leq N$, and $1\leq b \leq \sqrt{N}$, so
$$\sum_{n\leq N} \frac{1}{n} \leq \sum_{a \in S, a\leq N}\ \sum_{1\leq b \leq \sqrt{N}} \frac{1}{ab^2} \leq \sum_{a \in S, a\leq N}\ \frac{1}{a} \, \sum_{b=1}^\infty \frac{1}{b^2} = \frac{\pi^2}{6} \sum_{a \in S, a\leq N}\ \frac{1}{a}.$$
So $$\sum_{a \in S, a\leq N} \frac{1}{a} \geq \frac{6}{\pi^2} \sum_{n\leq N} \frac{1}{n}.$$
As $\sum_{n=1}^\infty \frac{1}{n}$ diverges, $\lim\limits_{N \to \infty} \sum\limits_{a \in S, a\leq N} \frac{1}{a} = +\infty$, so the family $\left(\frac{1}{a}\right)_{a\in S}$ of the inverse of square free integers is not summable.
Let $S_N = \prod_{p<N}(1+1/p)$ , and $p_1,p_2,\ldots, p_l\ (l = l(N))$ all prime integers less than $N$. Then
\begin{align*}
S_N &= \left(1+\frac{1}{p_1}\right) \cdots \left(1+\frac{1}{p_l}\right)\\
&=\sum_{(\varepsilon_1,\cdots,\varepsilon_l) \in \{0,1\}^l } \frac{1}{p_1^{\varepsilon_1} \cdots p_l^{\varepsilon_l}}
\end{align*}
We prove this last formula by induction. This is true for $l=1$ : $\sum_{\varepsilon \in \{0,1\}} 1/p_1^\varepsilon = 1 + 1/p_1$.
If it is true for the integer $l$, then
\begin{align*}
\left(1+\frac{1}{p_1}\right) \cdots \left(1+\frac{1}{p_l}\right)\left(1+\frac{1}{p_{l+1}}\right) &= \sum_{(\varepsilon_1,\ldots,\varepsilon_l) \in \{0,1\}^l } \frac{1}{p_1^{\varepsilon_1} \cdots p_l^{\varepsilon_l}} \left(1+\frac{1}{p_{l+1}}\right)\\
&=\sum_{(\varepsilon_1,\ldots,\varepsilon_l) \in \{0,1\}^l } \frac{1}{p_1^{\varepsilon_1} \cdots p_l^{\varepsilon_l}} + \sum_{(\varepsilon_1,\ldots,\varepsilon_l) \in \{0,1\}^l } \frac{1}{p_1^{\varepsilon_1} \cdots p_l^{\varepsilon_l}p_{l+1}}\\
&=\sum_{(\varepsilon_1,\ldots,\varepsilon_l,\varepsilon_{l+1}) \in \{0,1\}^{l+1} } \frac{1}{p_1^{\varepsilon_1} \cdots p_l^{\varepsilon_l}p_{l+1}^{\varepsilon_{l+1}}}
\end{align*}
So it is true for all $l$.
Thus $S_N = \sum_{n\in \Delta} \frac{1}{n}$, where $\Delta$ is the set of square free integers whose prime factors are less than $N$.
As $\sum 1/n$, the sum being over square free integers, diverges, $\lim\limits_{N\to \infty} S_N = + \infty$ :
$$\lim_{N \to \infty} \prod_{p<N} \left(1+\frac{1}{p}\right) = +\infty.$$
$e^x \geq 1+x, x \geq \log (1+x)$ for $x>0$, so
$$\log S_N = \sum_{k=1}^{l(N)} \log\left(1+\frac{1}{p_k}\right) \leq \sum_{k=1}^{l(N)} \frac{1}{p_k}.$$
$\lim\limits_{N\to \infty} \log S_N = +\infty$ and $\lim\limits_{N\to \infty} l(N) = +\infty$, so
$$\lim_{N\to \infty} \sum_{p<N} \frac{1}{p} = +\infty.$$
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that the equation $3 x^{2}+2=y^{2}$ has no solution in integers.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
If $3x^2+2 = y^2$, then $\overline{y}^2 = \overline{2}$ in $\mathbb{Z}/3\mathbb{Z}$.
As $\{-1,0,1\}$ is a complete set of residues modulo $3$, the squares in $\mathbb{Z}/3\mathbb{Z}$ are $\overline{0} = \overline{0}^2$ and $\overline{1} = \overline{1}^2 = (\overline{-1})^2$, so $\overline{2}$ is not a square in $\mathbb{Z}/3\mathbb{Z}$ : $\overline{y}^2 = \overline{2}$ is impossible in $\mathbb{Z}/3\mathbb{Z}$.
Thus $3x^2+2 = y^2$ has no solution in integers.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: If $n$ is not a prime, show that $(n-1) ! \equiv 0(n)$, except when $n=4$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Suppose that $n >1$ is not a prime. Then $n = uv$, where $2 \leq u \leq v \leq n-1$.
$\bullet$ If $u \neq v$, then $n = uv \mid (n-1)! = 1\times 2 \times\cdots \times u \times\cdots \times v \times \cdots \times (n-1)$ (even if $u\wedge v \neq 1$ !).
$\bullet$ If $u=v$, $n = u^2$ is a square.
If $u$ is not prime, $u =st,\ 2\leq s \leq t \leq u-1 \leq n-1$, and $n = u' v'$, where $u' =s,v' =st^2$ verify $2 \leq u' < v' \leq n-1$. As in the first case, $n = u'v' \mid (n-1)!$.
If $u = p$ is a prime, then $n =p^2$.
In the case $p = 2$, $n = 4$ and $n=4 \nmid (n-1)! = 6$. In the other case $p >2$, and $(n-1)! = (p^2 - 1)!$ contains the factors $p < 2p < p^2$, so $p^2 \mid (p^2-1)!, n \mid (n-1)!$.
Conclusion : if $n$ is not a prime, $(n - 1)! \equiv 0 \pmod n$, except when $n=4$.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Consider a prime $p$ of the form $4 t+1$. Show that $a$ is a primitive root modulo $p$ iff $-a$ is a primitive root modulo $p$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Suppose that $a$ is a primitive root modulo $p$. As $p-1$ is even, $(-a)^{p-1}=a^{p-1} \equiv 1$ $(\bmod p)$
If $(-a)^n \equiv 1(\bmod p)$, with $n \in \mathbb{N}$, then $a^n \equiv(-1)^n(\bmod p)$.
Therefore $a^{2 n} \equiv 1(\bmod p)$. As $a$ is a primitive root modulo $p, p-1|2 n, 2 t| n$, so $n$ is even.
Hence $a^n \equiv 1(\bmod p)$, and $p-1 \mid n$. So the least $n \in \mathbb{N}^*$ such that $(-a)^n \equiv 1$ $(\bmod p)$ is $p-1:$ the order of $-a$ modulo $p$ is $p-1,-a$ is a primitive root modulo $p$. Conversely, if $-a$ is a primitive root modulo $p$, we apply the previous result at $-a$ to to obtain that $-(-a)=a$ is a primitive root.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: If $p=2^{n}+1$ is a Fermat prime, show that 3 is a primitive root modulo $p$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
\newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
Write $p = 2^k + 1$, with $k = 2^n$.
We suppose that $n>0$, so $k\geq 2, p \geq 5$. As $p$ is prime, $3^{p-1} \equiv 1 \pmod p$.
In other words, $3^{2^k} \equiv 1 \pmod p$ : the order of $3$ is a divisor of $2^k$, a power of $2$.
$3$ has order $2^k$ modulo $p$ iff $3^{2^{k-1}} \not \equiv 1 \pmod p$. As $\left (3^{2^{k-1}} \right)^2 \equiv 1 \pmod p$, where $p$ is prime, this is equivalent to $3^{2^{k-1}} \equiv -1 \pmod p$, which remains to prove.
$3^{2^{k-1}} = 3^{(p-1)/2} \equiv \legendre{3}{p} \pmod p$.
As the result is true for $p=5$, we can suppose $n\geq 2$.
From the law of quadratic reciprocity :
$$\legendre{3}{p} \legendre{p}{3} = (-1)^{(p-1)/2} = (-1)^{2^{k-1}} = 1.$$
So $\legendre{3}{p} = \legendre{p}{3}$
\begin{align*}
p = 2^{2^n}+1 &\equiv (-1)^{2^n} + 1 \pmod 3\\
&\equiv 2 \equiv -1 \pmod 3,
\end{align*}
so $\legendre{3}{p} = \legendre {p}{3} = -1$, that is to say
$$3^{2^{k-1}} \equiv -1 \pmod p.$$
The order of $3$ modulo $p = 2^{2^n} + 1$ is $p-1 = 2^{2^n}$ : $3$ is a primitive root modulo $p$.
(On the other hand, if $3$ is of order $p-1$ modulo $p$, then $p$ is prime, so
$$ F_n = 2^{2^n} + 1 \ \mathrm{is}\ \mathrm{prime}\ \iff 3^{(F_n-1)/2} = 3^{2^{2^n - 1}} \equiv -1 \pmod {F_n}.)$$
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that $1^{k}+2^{k}+\cdots+(p-1)^{k} \equiv 0(p)$ if $p-1 \nmid k$ and $-1(p)$ if $p-1 \mid k$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Let $S_k = 1^k+2^k+\cdots+(p-1)^k$.
Let $g$ a primitive root modulo $p$ : $\overline{g}$ a generator of $\mathbb{F}_p^*$.
As $(\overline{1},\overline{g},\overline{g}^{2}, \ldots, \overline{g}^{p-2}) $ is a permutation of $ (\overline{1},\overline{2}, \ldots,\overline{p-1})$,
\begin{align*}
\overline{S_k} &= \overline{1}^k + \overline{2}^k+\cdots+ \overline{p-1}^k\\
&= \sum_{i=0}^{p-2} \overline{g}^{ki} =
\left\{
\begin{array}{ccc}
\overline{ p-1} = -\overline{1} & \mathrm{if} & p-1 \mid k \\
\frac{ \overline{g}^{(p-1)k} -1}{ \overline{g}^k -1} = \overline{0}& \mathrm{if} & p-1 \nmid k
\end{array}
\right.
\end{align*}
since $p-1 \mid k \iff \overline{g}^k = \overline{1}$.
Conclusion :
\begin{align*}
1^k+2^k+\cdots+(p-1)^k&\equiv 0 \pmod p\ \mathrm{if} \ p-1 \nmid k\\
1^k+2^k+\cdots+(p-1)^k&\equiv -1 \pmod p\ \mathrm{if} \ p-1 \mid k\\
\end{align*}
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that $x^{4} \equiv 2(p)$ has a solution for $p \equiv 1(4)$ iff $p$ is of the form $A^{2}+64 B^{2}$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
If $p\equiv 1\ [4]$ and if there exists $x \in \mathbb{Z}$ such that $x^4 \equiv 2\ [p]$, then
$$2^{\frac{p-1}{4} }\equiv x^{p-1} \equiv 1 \ [p].$$
From Ex. 5.27, where $p = a^2 +b^2, a$ odd, we know that $$f^{\frac{ab}{2}} \equiv 2^{\frac{p-1}{4} } \equiv 1 \ [p].$$
Since $f^2 \equiv -1\ [p]$, the order of $f$ modulo $p$ is 4, thus $4 \mid \frac{ab}{2}$, so $8\mid ab$.
As $a$ is odd, $8 | b$, then $p = A^2 + 64 B^2$ (with $A = a, B = b/8$).
\bigskip
Conversely, if $p=A^2+64 B^2$, then $p\equiv A^2 \equiv 1 \ [4]$.
Let $a=A,b=8B$. Then $$2^{\frac{p-1}{4} } \equiv f^{\frac{ab}{2}} \equiv f^{4AB} \equiv (-1)^{2AB} \equiv 1 \ [p].$$
As $2^{\frac{p-1}{4} } \equiv 1 \ [p]$, $x^4 \equiv 2 \ [p]$ has a solution in $\mathbb{Z}$ (Prop. 4.2.1) : $2$ is a biquadratic residue modulo $p$.
Conclusion :
$$\exists A \in \mathbb{Z}, \exists B \in \mathbb{Z}\,, p = A^2+64 B^2 \iff( p\equiv 1 \ [4] \ \mathrm{and} \ \exists x \in \mathbb{Z}, \, x^4 \equiv 2 \ [p]).$$
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that $\sin (\pi / 12)$ is an algebraic number.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
$$
\begin{aligned}
\sin \pi/12=\sin \left(\pi/4-\pi/6\right) & =\sin \pi/4 \cos \pi/6-\cos \pi/4 \sin \pi/6 \\
& =\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}} \\
& =\frac{\sqrt{3}-1}{2 \sqrt{2}}
\end{aligned}
$$
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $\text{Im}(f)$ is constant, then $f$ is constant.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.
Since $\operatorname{Im}(f)=$ constant,
$$
\frac{\partial v}{\partial x}=0, \frac{\partial v}{\partial y}=0 .
$$
By the Cauchy-Riemann equations,
$$
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}=0 .
$$
Thus in $\Omega$,
$$
f^{\prime}(z)=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=0+0=0 .
$$
Thus $f$ is constant.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that the power series $\sum nz^n$ does not converge on any point of the unit circle.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
For $z \in S:=\{z \in \mathbb{C}:|z|=1\}$ it also holds $z^n \in S$ for all $n \in \mathbb{N}$ (since in this case $\left.\left|z^n\right|=|z|^n=1^n=1\right)$
Thus, the sequence $\left(a_n\right)_{n \in \mathbb{N}}$ with $a_n=n z^n$ does not converge to zero which is necessary for the corresponding sum $\sum_{n \in \mathbb{N}} a_n$ to be convergent. Hence this sum does not converge.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that the power series $\sum zn/n$ converges at every point of the unit circle except $z = 1$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
If $z=1$ then $\sum z^n / n=\sum 1 / n$ is divergent (harmonic series). If $|z|=1$ and $z \neq 1$, write $z=e^{2 \pi i t}$ with $t \in(0,1)$ and apply Dirichlet's test: if $\left\{a_n\right\}$ is a sequence of real numbers and $\left\{b_n\right\}$ a sequence of complex numbers satisfying
- $a_{n+1} \leq a_n$
- $\lim _{n \rightarrow \infty} a_n=0$
- $\left|\sum_{n=1}^N b_n\right| \leq M$ for every positive integer $N$ and some $M>0$,
then $\sum a_n b_n$ converges. Let $a_n=1 / n$, so $a_n$ satisfies $a_{n+1} \leq a_n$ and $\lim _{n \rightarrow \infty} a_n=0$. Let $b_n=e^{2 \pi i n t}$, then
$$
\left|\sum_{n=1}^N b_n\right|=\left|\sum_{n=1}^N e^{2 \pi i n t}\right|=\left|\frac{e^{2 \pi i t}-e^{2 \pi i(N+1) t}}{1-e^{2 \pi i t}}\right| \leq \frac{2}{\left|1-e^{2 \pi i t}\right|}=M \text { for all } N
$$
Thus $\sum a_n b_n=\sum z^n / n$ converges for every point in the unit circle except $z=1$.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that $\int_{0}^{\infty} \frac{\sin x}{x} d x=\frac{\pi}{2}$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
We have $\int_0^{\infty} \frac{\sin x}{x} d x=\frac{1}{2 * i} \int_0^{\infty} \frac{e^{i * x}-e^{-i * x}}{x} d x=\frac{1}{2 * i}\left(\int_0^{\infty} \frac{e^{i * x}-1}{x} d x-\int_0^{\infty} \frac{e^{-i * x}-1}{x} d x=\right.$ $\frac{1}{2 * i} \int_{-\infty}^{\infty} \frac{e^{i * x}-1}{x} d x$. Now integrate along the big and small semicircles $C_0$ and $C_1$ shown below. For $C_0$ : we have that $\int_{C_0} \frac{1}{x} d x=\pi * i$ and $\left|\int_{C_0} \frac{e^{i * x}}{x} d x\right| \leq$ $2 *\left|\int_{C_{00}} \frac{e^{i * x}}{x} d x\right|+\left|\int_{C_{01}} \frac{e^{i * x}}{x} d x\right|$ where $C_{00}$ and $C_{01}$ are shown below $\left(C_{01}\right.$ contains the part of $C_0$ that has points with imaginary parts more than $a$ and $C_{00}$ is one of the other 2 components). We have $\left|\int_{C_{00}} \frac{e^{i * x}}{x} d x\right| \leq$ $\sup _{x \in C_{00}}\left(e^{i * x}\right) / R * \int_{C_{00}}|d x| \leq e^{-a} * \pi$ and $\left|\int_{C_{01}} \frac{e^{i * x}}{x} d x\right| \leq\left|\int_{C_{01}} \frac{1}{x} d x\right| \leq$ $\frac{1}{R} * C * a$ for some constant $C$ (the constant $C$ exists because the length of the curve approaches $a$ as $a / R \rightarrow 0)$. Thus, the integral of $e^{i * x} / x$ over $C_0$ is bounded by $A * e^{-a}+B * a / R$ for some constants $A$ and $B$. Pick $R$ large and $a=\sqrt{R}$ and note that the above tends to 0 . About the integral over $C_1$ : We have $e^{i * x}-1=1+O(x)$ for $x \rightarrow 0$ (this is again from $\sin (x) / x \rightarrow 1$ ),
4
so $\left|\int_{C_1} \frac{e^{i * x}-1}{x} d x\right| \leq O(1) *\left|\int_{C_1} d x\right| \rightarrow 0$ as $x \rightarrow 0$. Thus, we only care about the integral over $C_{00}$ which is $-\pi * i$. Using Cauchy's theorem we get that our integral equals $\frac{1}{2 * i}(-(\pi * i))=\pi / 2$.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Suppose $f$ is an analytic function defined everywhere in $\mathbb{C}$ and such that for each $z_0 \in \mathbb{C}$ at least one coefficient in the expansion $f(z) = \sum_{n=0}^\infty c_n(z - z_0)^n$ is equal to 0. Prove that $f$ is a polynomial.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Say that at least one of the coefficients of the Taylor series vanishes is the same as saying that for every $a \in \mathbb{C}$ there is $m \in \mathbb{N}$ such that $f^{(m)}(a)=0$.
Consider $A_n:=\left\{z \in \mathbb{C}: f^{(n)}(z)=0\right\}$ for each $n \in \mathbb{N}$. Note that:
$f$ is polynomial iff $A_n$ is not countable for some $n \in \mathbb{N}$.
Indeed, if $f$ is polynomial of degree $n$, then $f^{(n+1)}(z)=0$ for all $z \in \mathbb{C}$, then $A_{n+1}=\mathbb{C}$, so, $A_{n+1}$ is not countable. Conversely, if there is $n \in \mathbb{C}$ such that $A_n$ is not countable, then $A_n$ has a limit point, then by Identity principle we have $f^{(n)}(z)=0$ for all $z \in \mathbb{C}$, so, $f$ is a polynomial of degree at most $n-1$.
Therefore, tt suffices to show that there is $n \in \mathbb{N}$ such that $A_n$ is not countable. Indeed, consider $\bigcup_{n \in \mathbb{N}} A_n$, by hypothesis for each $a \in \mathbb{C}$ there is $m \in \mathbb{N}$ such that $f^{(m)}(a)=0$, then $\mathbb{C} \subseteq \bigcup_{n \in \mathbb{N}} A_n$. Therefore, $\bigcup_{n \in \mathbb{N}} A_n$ is not countable, then there is $n \in \mathbb{N}$ such that $A_n$ is not countable.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that $ \int_{-\infty}^{\infty} \frac{x \sin x}{x^2 + a^2} dx = \pi e^{-a}$ for $a > 0$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
$$
x /\left(x^2+a^2\right)=x / 2 i a(1 /(x-i a)-1 /(x+i a))=1 / 2 i a(i a /(x-i a)+i a /(x+
$$
$i a))=(1 /(x-i a)+1 /(x+i a)) / 2$. So we care about $\sin (x)(1 /(x-i a)+$ $1 /(x+i a)) / 2$. Its residue at $x=i a$ is $\sin (i a) / 2=\left(e^{-a}-e^a\right) / 4 i$.?
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that all entire functions that are also injective take the form $f(z) = az + b$, $a, b \in \mathbb{C}$ and $a \neq 0$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Look at $f(1 / z)$. If it has an essential singularity at 0 , then pick any $z_0 \neq 0$. Now we know that the range of $f$ is dense as $z \rightarrow 0$. We also know that the image of $f$ in some small ball around $z_0$ contains a ball around $f\left(z_0\right)$. But this means that the image of $f$ around this ball intersects the image of $f$ in any arbitrarily small ball around 0 (because of the denseness). Thus, $f$ cannot be injective. So the singularity at 0 is not essential, so $f(1 / z)$ is some polynomial of $1 / z$, so $f$ is some polynomial of $z$. If its degree is more than 1 it is not injective (fundamental theorem of algebra), so the degree of $f$ is 1 .
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that if $f$ is holomorphic in the unit disc, bounded and not identically zero, and $z_{1}, z_{2}, \ldots, z_{n}, \ldots$ are its zeros $\left(\left|z_{k}\right|<1\right)$, then $\sum_{n}\left(1-\left|z_{n}\right|\right)<\infty$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Fix $\mathrm{N}$ and let $D(0, R)$ contains the first $\mathrm{N}$ zeroes of f. Let $S_N=\sum_{k=1}^N\left(1-\left|z_k\right|\right)=$ $\sum_{k=1}^N \int_{\left|z_k\right|}^1 1 d r$. Let $\eta_k$ be the characteristic function of the interval $\left.\| z_k \mid, 1\right]$. We have $S_N=\sum_{k=1}^N \int_0^1 \eta(r) d r=\int_0^1\left(\sum_{k=1}^N \eta_k(r)\right) d r \leq \int_0^1 n(r) d r$, where $n(r)$ is the number of zeroes of $f$ at the disk $D(0, r)$. For $r \leq 1$ we have $n(r) \leq \frac{n(r)}{r}$. This means that $S_N \leq \int_0^1 n(r) \frac{d r}{r}$. If $f(0)=0$ then we have $f(z)=z^m g(z)$ for some integer $\mathrm{m}$ and some holomorphic $g$ with $g(0) \neq 0$. The other zeroes of $\mathrm{f}$ are precisely the zeroes of $g$. Thus we have reduced the problem to $f(0) \neq 0$. By the Corollary of the Jensen's equality we get $S_N \leq \int_0^1 n(r) \frac{d r}{r}=\frac{1}{2 \pi} \int_0^{2 \pi} \log \left|f\left(R e^{i \pi}\right)\right| d \phi-\log |f(0)|<M$ since $f$ is bounded. The partial sums of the series are boundend and therefore the series converges.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be an infinitely differentiable function satisfying $f(0)=0, f(1)=1$, and $f(x) \geq 0$ for all $x \in$ $\mathbb{R}$. Show that there exist a positive integer $n$ and a real number $x$ such that $f^{(n)}(x)<0$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Call a function $f\colon \mathbb{R} \to \mathbb{R}$ \textit{ultraconvex} if $f$ is infinitely differentiable and $f^{(n)}(x) \geq 0$ for all $n \geq 0$ and all $x \in \mathbb{R}$, where $f^{(0)}(x) = f(x)$;
note that if $f$ is ultraconvex, then so is $f'$.
Define the set
\[
S = \{ f :\thinspace \mathbb{R} \to \mathbb{R} \,|\,f \text{ ultraconvex and } f(0)=0\}.
\]
For $f \in S$, we must have $f(x) = 0$ for all $x < 0$: if $f(x_0) > 0$ for some $x_0 < 0$, then
by the mean value theorem there exists $x \in (0,x_0)$ for which $f'(x) = \frac{f(x_0)}{x_0} < 0$.
In particular, $f'(0) = 0$, so $f' \in S$ also.
We show by induction that for all $n \geq 0$,
\[
f(x) \leq \frac{f^{(n)}(1)}{n!} x^n \qquad (f \in S, x \in [0,1]).
\]
We induct with base case $n=0$, which holds because any $f \in S$ is nondecreasing. Given the claim for $n=m$,
we apply the induction hypothesis to $f' \in S$ to see that
\[
f'(t) \leq \frac{f^{(n+1)}(1)}{n!} t^n \qquad (t \in [0,1]),
\]
then integrate both sides from $0$ to $x$ to conclude.
Now for $f \in S$, we have $0 \leq f(1) \leq \frac{f^{(n)}(1)}{n!}$ for all $n \geq 0$.
On the other hand, by Taylor's theorem with remainder,
\[
f(x) \geq \sum_{k=0}^n \frac{f^{(k)}(1)}{k!}(x-1)^k \qquad (x \geq 1).
\]
Applying this with $x=2$, we obtain $f(2) \geq \sum_{k=0}^n \frac{f^{(k)}(1)}{k!}$ for all $n$;
this implies that $\lim_{n\to\infty} \frac{f^{(n)}(1)}{n!} = 0$.
Since $f(1) \leq \frac{f^{(n)}(1)}{n!}$, we must have $f(1) = 0$.
For $f \in S$, we proved earlier that $f(x) = 0$ for all $x\leq 0$, as well as for $x=1$. Since
the function $g(x) = f(cx)$ is also ultraconvex for $c>0$, we also have $f(x) = 0$ for all $x>0$;
hence $f$ is identically zero.
To sum up, if $f\colon \mathbb{R} \to \mathbb{R}$ is infinitely differentiable, $f(0)=0$, and $f(1) = 1$,
then $f$ cannot be ultraconvex. This implies the desired result.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Given a real number $a$, we define a sequence by $x_{0}=1$, $x_{1}=x_{2}=a$, and $x_{n+1}=2 x_{n} x_{n-1}-x_{n-2}$ for $n \geq 2$. Prove that if $x_{n}=0$ for some $n$, then the sequence is periodic.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
We first rule out the case $|a|>1$. In this case, we prove that $|x_{n+1}| \geq |x_n|$ for all $n$, meaning that we cannot have $x_n = 0$. We proceed by induction; the claim is true for $n=0,1$ by hypothesis. To prove the claim for $n \geq 2$, write
\begin{align*}
|x_{n+1}| &= |2x_nx_{n-1}-x_{n-2}| \\
&\geq 2|x_n||x_{n-1}|-|x_{n-2}| \\
&\geq |x_n|(2|x_{n-1}|-1) \geq |x_n|,
\end{align*}
where the last step follows from $|x_{n-1}| \geq |x_{n-2}| \geq \cdots \geq |x_0| = 1$.
We may thus assume hereafter that $|a|\leq 1$. We can then write $a = \cos b$ for some $b \in [0,\pi]$.
Let $\{F_n\}$ be the Fibonacci sequence, defined as usual by $F_1=F_2=1$ and $F_{n+1}=F_n+F_{n-1}$. We show by induction that
\[
x_n = \cos(F_n b) \qquad (n \geq 0).
\]
Indeed, this is true for $n=0,1,2$; given that it is true for $n \leq m$, then
\begin{align*}
2x_mx_{m-1}&=2\cos(F_mb)\cos(F_{m-1}b) \\
&= \cos((F_m-F_{m-1})b)+\cos((F_m+F_{m-1})b) \\
&= \cos(F_{m-2}b)+\cos(F_{m+1}b)
\end{align*}
and so
$x_{m+1} = 2x_mx_{m-1}-x_{m-2} = \cos(F_{m+1}b)$. This completes the induction.
Since $x_n = \cos(F_n b)$, if $x_n=0$ for some $n$ then $F_n b = \frac{k}{2} \pi$ for some odd integer $k$. In particular, we can write $b = \frac{c}{d}(2\pi)$ where $c = k$ and $d = 4F_n$ are integers.
Let $x_n$ denote the pair $(F_n,F_{n+1})$, where each entry in this pair is viewed as an element of $\mathbb{Z}/d\mathbb{Z}$. Since there are only finitely many possibilities for $x_n$, there must be some $n_2>n_1$ such that $x_{n_1}=x_{n_2}$. Now $x_n$ uniquely determines both $x_{n+1}$ and $x_{n-1}$, and it follows that the sequence $\{x_n\}$ is periodic: for $\ell = n_2-n_1$, $x_{n+\ell} = x_n$ for all $n \geq 0$. In particular, $F_{n+\ell} \equiv F_n \pmod{d}$ for all $n$. But then $\frac{F_{n+\ell}c}{d}-\frac{F_n c}{d}$ is an integer, and so
\begin{align*}
x_{n+\ell} &= \cos\left(\frac{F_{n+\ell}c}{d}(2\pi)\right)\\
& = \cos\left(\frac{F_n c}{d}(2\pi)\right) = x_n
\end{align*}
for all $n$. Thus the sequence $\{x_n\}$ is periodic, as desired.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Let<|im_end|>
<|im_start|>assistant
<think>
</think>
| |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that there are unique positive integers $a, n$ such that $a^{n+1}-(a+1)^n=2001$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Suppose $a^{n+1} - (a+1)^n = 2001$.
Notice that $a^{n+1} + [(a+1)^n - 1]$ is a multiple of $a$; thus
$a$ divides $2002 = 2 \times 7 \times 11 \times 13$.
Since $2001$ is divisible by 3, we must have $a \equiv 1 \pmod{3}$,
otherwise one of $a^{n+1}$ and $(a+1)^n$ is a multiple of 3 and the
other is not, so their difference cannot be divisible by 3. Now
$a^{n+1} \equiv 1 \pmod{3}$, so we must have $(a+1)^n \equiv 1
\pmod{3}$, which forces $n$ to be even, and in particular at least 2.
If $a$ is even, then $a^{n+1} - (a+1)^n \equiv -(a+1)^n \pmod{4}$.
Since $n$ is even, $-(a+1)^n \equiv -1 \pmod{4}$. Since $2001 \equiv 1
\pmod{4}$, this is impossible. Thus $a$ is odd, and so must divide
$1001 = 7 \times 11 \times 13$. Moreover, $a^{n+1} - (a+1)^n \equiv a
\pmod{4}$, so $a \equiv 1 \pmod{4}$.
Of the divisors of $7 \times 11 \times 13$, those congruent to 1 mod 3
are precisely those not divisible by 11 (since 7 and 13 are both
congruent to 1 mod 3). Thus $a$ divides $7 \times 13$. Now
$a \equiv 1 \pmod{4}$ is only possible if $a$ divides $13$.
We cannot have $a=1$, since $1 - 2^n \neq 2001$ for any $n$. Thus
the only possibility is $a = 13$. One easily checks that $a=13, n=2$ is a
solution; all that remains is to check that no other $n$ works. In fact,
if $n > 2$, then $13^{n+1} \equiv 2001 \equiv 1 \pmod{8}$.
But $13^{n+1} \equiv 13 \pmod{8}$ since $n$ is even, contradiction.
Thus $a=13, n=2$ is the unique solution.
Note: once one has that $n$ is even, one can use that $2002
=a^{n+1} + 1 - (a+1)^n$ is divisible by $a+1$ to rule out cases.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Let $f$ be a real function with a continuous third derivative such that $f(x), f^{\prime}(x), f^{\prime \prime}(x), f^{\prime \prime \prime}(x)$ are positive for all $x$. Suppose that $f^{\prime \prime \prime}(x) \leq f(x)$ for all $x$. Show that $f^{\prime}(x)<2 f(x)$ for all $x$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
\setcounter{equation}{0}
We make repeated use of the following fact: if $f$ is a differentiable function on all of
$\mathbb{R}$, $\lim_{x \to -\infty} f(x) \geq 0$, and $f'(x) > 0$ for all $x \in \mathbb{R}$, then
$f(x) > 0$ for all $x \in \mathbb{R}$. (Proof: if $f(y) < 0$ for some $x$, then $f(x)< f(y)$ for all
$x<y$ since $f'>0$, but then $\lim_{x \to -\infty} f(x) \leq f(y) < 0$.)
From the inequality $f'''(x) \leq f(x)$ we obtain
\[
f'' f'''(x) \leq f''(x) f(x) < f''(x) f(x) + f'(x)^2
\]
since $f'(x)$ is positive. Applying the fact to the difference between the right and left sides,
we get
\begin{equation}
\frac{1}{2} (f''(x))^2 < f(x) f'(x).
\end{equation}
On the other hand, since $f(x)$ and $f'''(x)$ are both positive for all $x$,
we have
\[
2f'(x) f''(x) < 2f'(x)f''(x) + 2f(x) f'''(x).
\]
Applying the fact to the difference between the sides yields
\begin{equation}
f'(x)^2 \leq 2f(x) f''(x).
\end{equation}
Combining (1) and (2), we obtain
\begin{align*}
\frac{1}{2} \left( \frac{f'(x)^2}{2f(x)} \right)^2
&< \frac{1}{2} (f''(x))^2 \\
&< f(x) f'(x),
\end{align*}
or $(f'(x))^3 < 8 f(x)^3$. We conclude $f'(x) < 2f(x)$, as desired.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that, for any integers $a, b, c$, there exists a positive integer $n$ such that $\sqrt{n^3+a n^2+b n+c}$ is not an integer.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
We prove more generally that for any polynomial $P(z)$ with integer
coefficients which is not a perfect square, there exists a positive
integer $n$ such that $P(n)$ is not a perfect square. Of course it
suffices to assume $P(z)$ has no repeated factors, which is to say $P(z)$
and its derivative $P'(z)$ are relatively prime.
In particular, if we carry out the Euclidean algorithm on $P(z)$ and $P'(z)$
without dividing, we get an integer $D$ (the discriminant of $P$) such that
the greatest common divisor of $P(n)$ and $P'(n)$ divides $D$ for any $n$.
Now there exist infinitely many primes $p$ such that $p$ divides $P(n)$ for
some $n$: if there were only finitely many, say, $p_1, \dots, p_k$, then
for any $n$ divisible by $m = P(0) p_1 p_2 \cdots p_k$, we have $P(n)
\equiv P(0) \pmod{m}$, that is, $P(n)/P(0)$ is not divisible by $p_1,
\dots, p_k$, so must be $\pm 1$, but then $P$ takes some value infinitely
many times, contradiction. In particular, we can choose some such $p$ not
dividing $D$, and choose $n$ such that $p$ divides $P(n)$. Then $P(n+kp)
\equiv P(n) + kp P'(n) (\mathrm{mod}\,p)$
(write out the Taylor series of the left side);
in particular, since $p$ does not divide $P'(n)$, we can find some $k$
such that $P(n+kp)$ is divisible by $p$ but not by $p^2$, and so
is not a perfect square.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Prove that a set $U \subset M$ is open if and only if none of its points are limits of its complement.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Assume that none of the points of $U$ are limits of its complement, and let us prove that $U$ is open. Assume by contradiction that $U$ is not open, so there exists $p \in M$ so that $\forall r>0$ there exists $q \in M$ with $d(p, q)<r$ but $q \notin U$. Applying this to $r=1 / n$ we obtain $q_n \in U^c$ such that $d\left(q_n, p\right)<1 / n$. But then $q_n \rightarrow p$ and $p$ is a limit of a sequence of points in $U^c$, a contradiction.
Assume now that $U$ is open. Assume by contradiction there exists $p \in U$ and $p_n \in U^c$ such that $p_n \rightarrow p$. Since $U$ is open, there exists $r>0$ such that $d(p, x)<r$ for $x \in M$ implies $x \in U$. But since $p_n \rightarrow p$, there exists $n_0 \in \mathbb{N}$ such that $n \geq n_0$ implies $d\left(p_n, p\right)<r$, therefore $p_n \in U$ for $n \geq n_0$, a contradiction since $p_n \in U^c$.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Show that every subset of $\mathbb{N}$ is clopen.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
32. The one-point set $\{n\} \subset \mathbb{N}$ is open, since it contains all $m \in \mathbb{N}$ that satisfy $d(m, n)<\frac{1}{2}$. Every subset of $\mathbb{N}$ is a union of one-point sets, hence is open. Then every set it closed, since its complement is necessarily open.
\end{proof} |
<|im_start|>user
You will be presented with a <Question>. Before providing the [Answer], you should first think step-by-step carefully.
Your response format:
<start of response>
Reasoning Steps:
- Step 1: Your reasoning step 1
- Step 2: Your reasoning step 2
- Step 3: Your reasoning step 3
...
- Step N: Your reasoning step N
<Answer>: Your final answer
<end of response>
Follow the above output format STRICTLY! Do not add any other additional texts outside the template.
Keep each reasoning step concise (single steps should not be too long).
Each reasoning step must be on a single line (no line breaks within a step).
Now answer:
<Question>: Assume that $A, B$ are compact, disjoint, nonempty subsets of $M$. Prove that there are $a_0 \in A$ and $b_0 \in B$ such that for all $a \in A$ and $b \in B$ we have $d(a_0, b_0) \leq d(a, b)$.<|im_end|>
<|im_start|>assistant
<think>
</think>
| \begin{proof}
Let $A$ and $B$ be compact, disjoint and non-empty subsets of $M$. We want to show that there exist $a_0 \in A, b_0 \in B$ such that for all $a \in A, b \in B$,
$$
d\left(a_0, b_0\right) \leq d(a, b) .
$$
We saw in class that the distance function $d: M \times M \rightarrow \mathbb{R}$ is continuous. We also saw in class that any continuous, real-valued function assumes maximum and minimum values on a compact set. Since $A$ and $B$ are compact, $A \times B$ is (non-empty) compact in $M \times M$. Therefore there exists $\left(a_0, b_0\right) \in A \times B$ such that $d\left(a_0, b_0\right) \leq d(a, b), \forall(a, b) \in A \times B$.
\end{proof} |
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