theorem exercise_1_1a
  (x : ℝ) (y : ℚ) :
  ( irrational x ) -> irrational ( x + y ) :=

import .common

open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter

noncomputable theory

If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational.

\begin{proof}

    If $r$ and $r+x$ were both rational, then $x=r+x-r$ would also be rational.

\end{proof}


theorem exercise_1_2 : ¬ ∃ (x : ℚ), ( x ^ 2 = 12 ) :=

Prove that there is no rational number whose square is $12$.

\begin{proof}

    Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor. It follows that $m$ must be even, and therefore $n$ must be odd. Let $m=2 r$. Then we have $r^2=3 n^2$, so that $r$ is also odd. Let $r=2 s+1$ and $n=2 t+1$. Then

$$

4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3,

$$

so that

$$

4\left(s^2+s-3 t^2-3 t\right)=2 .

$$

But this is absurd, since 2 cannot be a multiple of 4 .

\end{proof}


theorem exercise_1_5 (A minus_A : set ℝ) (hA : A.nonempty)
  (hA_bdd_below : bdd_below A) (hminus_A : minus_A = {x | -x ∈ A}) :
  Inf A = Sup minus_A :=

Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$.

\begin{proof}

    We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$. For brevity, let $\alpha=-\sup (-A)$. We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.

Suppose $x \in A$. Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$. It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$. Thus $\alpha$ is a lower bound of $A$.

Now let $\beta$ be any lower bound of $A$. This means $\beta \leq x$ for all $x$ in $A$. Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$. This means $-\beta$ is an upper bound of $-A$. Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$, and so $-\sup (-A)$ is the greatest lower bound of $A$.

\end{proof}


theorem exercise_1_11a (z : ℂ) :
  ∃ (r : ℝ) (w : ℂ), abs w = 1 ∧ z = r * w :=

If $z$ is a complex number, prove that there exists an $r\geq 0$ and a complex number $w$ with $| w | = 1$ such that $z = rw$.

\begin{proof}

    If $z=0$, we take $r=0, w=1$. (In this case $w$ is not unique.) Otherwise we take $r=|z|$ and $w=z /|z|$, and these choices are unique, since if $z=r w$, we must have $r=r|w|=|r w|=|z|, z / r$

\end{proof}


theorem exercise_1_13 (x y : ℂ) :
  |(abs x) - (abs y)| ≤ abs (x - y) :=

If $x, y$ are complex, prove that $||x|-|y|| \leq |x-y|$.

\begin{proof}

    Since $x=x-y+y$, the triangle inequality gives

$$

|x| \leq|x-y|+|y|

$$

so that $|x|-|y| \leq|x-y|$. Similarly $|y|-|x| \leq|x-y|$. Since $|x|-|y|$ is a real number we have either ||$x|-| y||=|x|-|y|$ or ||$x|-| y||=|y|-|x|$. In either case, we have shown that ||$x|-| y|| \leq|x-y|$.

\end{proof}


theorem exercise_1_16a
  (n : ℕ)
  (d r : ℝ)
  (x y z : euclidean_space ℝ (fin n)) -- R^n
  (h₁ : n ≥ 3)
  (h₂ : ‖x - y‖ = d)
  (h₃ : d > 0)
  (h₄ : r > 0)
  (h₅ : 2 * r > d)
  : set.infinite {z : euclidean_space ℝ (fin n) | ‖z - x‖ = r ∧ ‖z - y‖ = r} :=

Suppose $k \geq 3, x, y \in \mathbb{R}^k, |x - y| = d > 0$, and $r > 0$. Prove that if $2r > d$, there are infinitely many $z \in \mathbb{R}^k$ such that $|z-x|=|z-y|=r$.

\begin{proof}

    (a) Let w be any vector satisfying the following two equations:

$$

\begin{aligned}

\mathbf{w} \cdot(\mathbf{x}-\mathbf{y}) &=0, \\

|\mathbf{w}|^2 &=r^2-\frac{d^2}{4} .

\end{aligned}

$$

From linear algebra it is known that all but one of the components of a solution $\mathbf{w}$ of the first equation can be arbitrary. The remaining component is then uniquely determined. Also, if $w$ is any non-zero solution of the first equation, there is a unique positive number $t$ such that $t$ w satisfies both equations. (For example, if $x_1 \neq y_1$, the first equation is satisfied whenever

$$

z_1=\frac{z_2\left(x_2-y_2\right)+\cdots+z_k\left(x_k-y_k\right)}{y_1-x_1} .

$$

If $\left(z_1, z_2, \ldots, z_k\right)$ satisfies this equation, so does $\left(t z_1, t z_2, \ldots, t z_k\right)$ for any real number $t$.) Since at least two of these components can vary independently, we can find a solution with these components having any prescribed ratio. This ratio does not change when we multiply by the positive number $t$ to obtain a solution of both equations. Since there are infinitely many ratios, there are infinitely many distinct solutions. For each such solution $\mathbf{w}$ the vector $\mathbf{z}=$ $\frac{1}{2} \mathrm{x}+\frac{1}{2} \mathrm{y}+\mathrm{w}$ is a solution of the required equation. For

$$

\begin{aligned}

|\mathrm{z}-\mathbf{x}|^2 &=\left|\frac{\mathbf{y}-\mathbf{x}}{2}+\mathbf{w}\right|^2 \\

&=\left|\frac{\mathbf{y}-\mathbf{x}}{2}\right|^2+2 \mathbf{w} \cdot \frac{\mathbf{x}-\mathbf{y}}{2}+|\mathbf{w}|^2 \\

&=\frac{d^2}{4}+0+r^2-\frac{d^2}{4} \\

&=r^2

\end{aligned}

$$

and a similar relation holds for $|z-y|^2$.

\end{proof}


theorem exercise_1_18a
  (n : ℕ)
  (h : n > 1)
  (x : euclidean_space ℝ (fin n)) -- R^n
  : ∃ (y : euclidean_space ℝ (fin n)), y ≠ 0 ∧ (inner x y) = (0 : ℝ) :=

If $k \geq 2$ and $\mathbf{x} \in R^{k}$, prove that there exists $\mathbf{y} \in R^{k}$ such that $\mathbf{y} \neq 0$ but $\mathbf{x} \cdot \mathbf{y}=0$

\begin{proof}

    If $\mathbf{x}$ has any components equal to 0 , then $\mathbf{y}$ can be taken to have the corresponding components equal to 1 and all others equal to 0 . If all the components of $\mathbf{x}$ are nonzero, $\mathbf{y}$ can be taken as $\left(-x_2, x_1, 0, \ldots, 0\right)$. This is, of course, not true when $k=1$, since the product of two nonzero real numbers is nonzero.

\end{proof}


theorem exercise_1_19
  (n : ℕ)
  (a b c x : euclidean_space ℝ (fin n))
  (r : ℝ)
  (h₁ : r > 0)
  (h₂ : 3 • c = 4 • b - a)
  (h₃ : 3 * r = 2 * ‖x - b‖)
  : ‖x - a‖ = 2 * ‖x - b‖ ↔ ‖x - c‖ = r :=

Suppose $a, b \in R^k$. Find $c \in R^k$ and $r > 0$ such that $|x-a|=2|x-b|$ if and only if $| x - c | = r$. Prove that $3c = 4b - a$ and $3r = 2 |b - a|$.

\begin{proof}

    Since the solution is given to us, all we have to do is verify it, i.e., we need to show that the equation

$$

|\mathrm{x}-\mathrm{a}|=2|\mathrm{x}-\mathrm{b}|

$$

is equivalent to $|\mathrm{x}-\mathbf{c}|=r$, which says

$$

\left|\mathbf{x}-\frac{4}{3} \mathbf{b}+\frac{1}{3} \mathbf{a}\right|=\frac{2}{3}|\mathbf{b}-\mathbf{a}| .

$$

If we square both sides of both equations, we an equivalent pair of equations, the first of which reduces to

$$

3|\mathbf{x}|^2+2 \mathbf{a} \cdot \mathbf{x}-8 \mathbf{b} \cdot \mathbf{x}-|\mathbf{a}|^2+4|\mathbf{b}|^2=0,

$$

and the second of which reduces to this equation divided by 3 . Hence these equations are indeed equivalent.

\end{proof}


theorem exercise_2_24 {X : Type*} [metric_space X]
  (hX : ∀ (A : set X), infinite A → ∃ (x : X), x ∈ closure A) :
  separable_space X :=

Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is separable.

\begin{proof}

    We observe that if the process of constructing $x_j$ did not terminate, the result would be an infinite set of points $x_j, j=1,2, \ldots$, such that $d\left(x_i, x_j\right) \geq \delta$ for $i \neq j$. It would then follow that for any $x \in X$, the open ball $B_{\frac{\delta}{2}}(x)$ contains at most one point of the infinite set, hence that no point could be a limit point of this set, contrary to hypothesis. Hence $X$ is totally bounded, i.e., for each $\delta>0$ there is a finite set $x_1, \ldots, x_{N\delta}$such that $X=\bigcup_{j / 1}^{N\delta} B_\delta\left(x_j\right)$

Let $x_{n_1}, \ldots, x_{n N_n}$ be such that $X=\bigcup_{j / 1}^{N_n} B_{\frac{1}{n}}\left(x_{n j}\right), n=1,2, \ldots$ We claim that $\left\{x_{n j}: 1 \leq j \leq N_n ; n=1,2, \ldots\right\}$ is a countable dense subset of $X$. Indeed

25

if $x \in X$ and $\delta>0$, then $x \in B_{\frac{1}{n}}\left(x_{n j}\right)$ for some $x_{n j}$ for some $n>\frac{1}{\delta}$, and hence $d\left(x, x_{n j}\right)<\delta$. By definition, this means that $\left\{x_{n j}\right\}$ is dense in $X$.

\end{proof}


theorem exercise_2_27a (k : ℕ) (E P : set (euclidean_space ℝ (fin k)))
  (hE : E.nonempty ∧ ¬ set.countable E)
  (hP : P = {x | ∀ U ∈ 𝓝 x, ¬ set.countable (P ∩ E)}) :
  is_closed P ∧ P = {x | cluster_pt x (𝓟 P)}  :=

Suppose $E\subset\mathbb{R}^k$ is uncountable, and let $P$ be the set of condensation points of $E$. Prove that $P$ is perfect.

\begin{proof}

    We see that $E \cap W$ is at most countable, being a countable union of at-most-countable sets. It remains to show that $P=W^c$, and that $P$ is perfect.

\end{proof}


theorem exercise_2_28 (X : Type*) [metric_space X] [separable_space X]
  (A : set X) (hA : is_closed A) :
  ∃ P₁ P₂ : set X, A = P₁ ∪ P₂ ∧
  is_closed P₁ ∧ P₁ = {x | cluster_pt x (𝓟 P₁)} ∧
  set.countable P₂ :=

Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable.

\begin{proof}

    If $E$ is closed, it contains all its limit points, and hence certainly all its condensation points. Thus $E=P \cup(E \backslash P)$, where $P$ is perfect (the set of all condensation points of $E$ ), and $E \backslash P$ is at most countable.

Since a perfect set in a separable metric space has the same cardinality as the real numbers, the set $P$ must be empty if $E$ is countable. The at-mostcountable set $E \backslash P$ cannot be perfect, hence must have isolated points if it is nonempty.

\end{proof}


theorem exercise_3_1a
  (f : ℕ → ℝ)
  (h : ∃ (a : ℝ), tendsto (λ (n : ℕ), f n) at_top (𝓝 a))
  : ∃ (a : ℝ), tendsto (λ (n : ℕ), |f n|) at_top (𝓝 a) :=

Prove that convergence of $\left\{s_{n}\right\}$ implies convergence of $\left\{\left|s_{n}\right|\right\}$.

\begin{proof}

    Let $\varepsilon>0$. Since the sequence $\left\{s_n\right\}$ is a Cauchy sequence, there exists $N$ such that $\left|s_m-s_n\right|<\varepsilon$ for all $m>N$ and $n>N$. We then have $\left| |s_m| - |s_n| \right| \leq\left|s_m-s_n\right|<\varepsilon$ for all $m>N$ and $n>N$. Hence the sequence $\left\{\left|s_n\right|\right\}$ is also a Cauchy sequence, and therefore must converge.

\end{proof}


theorem exercise_3_3
  : ∃ (x : ℝ), tendsto f at_top (𝓝 x) ∧ ∀ n, f n < 2 :=

If $s_{1}=\sqrt{2}$, and $s_{n+1}=\sqrt{2+\sqrt{s_{n}}} \quad(n=1,2,3, \ldots),$ prove that $\left\{s_{n}\right\}$ converges, and that $s_{n}<2$ for $n=1,2,3, \ldots$.

\begin{proof}

    Since $\sqrt{2}<2$, it is manifest that if $s_n<2$, then $s_{n+1}<\sqrt{2+2}=2$. Hence it follows by induction that $\sqrt{2}<s_n<2$ for all $n$. In view of this fact,it also follows that $\left(s_n-2\right)\left(s_n+1\right)<0$ for all $n>1$, i.e., $s_n>s_n^2-2=s_{n-1}$. Hence the sequence is an increasing sequence that is bounded above (by 2 ) and so converges. Since the limit $s$ satisfies $s^2-s-2=0$, it follows that the limit is 2.

\end{proof}


theorem exercise_3_6a
: tendsto (λ (n : ℕ), (∑ i in finset.range n, g i)) at_top at_top :=

Prove that $\lim_{n \rightarrow \infty} \sum_{i<n} a_i = \infty$, where $a_i = \sqrt{i + 1} -\sqrt{i}$.

\begin{proof}

    (a) Multiplying and dividing $a_n$ by $\sqrt{n+1}+\sqrt{n}$, we find that $a_n=\frac{1}{\sqrt{n+1}+\sqrt{n}}$, which is larger than $\frac{1}{2 \sqrt{n+1}}$. The series $\sum a_n$ therefore diverges by comparison with the $p$ series $\left(p=\frac{1}{2}\right)$.

\end{proof}


theorem exercise_3_8
  (a b : ℕ → ℝ)
  (h1 : ∃ y, (tendsto (λ n, (∑ i in (finset.range n), a i)) at_top (𝓝 y)))
  (h2 : monotone b)
  (h3 : metric.bounded (set.range b)) :
  ∃ y, tendsto (λ n, (∑ i in (finset.range n), (a i) * (b i))) at_top (𝓝 y) :=

If $\Sigma a_{n}$ converges, and if $\left\{b_{n}\right\}$ is monotonic and bounded, prove that $\Sigma a_{n} b_{n}$ converges.

\begin{proof}

    We shall show that the partial sums of this series form a Cauchy sequence, i.e., given $\varepsilon>0$ there exists $N$ such that $\left|\sum_{k=m+1}^n a_k b_k\right|\langle\varepsilon$ if $n\rangle$ $m \geq N$. To do this, let $S_n=\sum_{k=1}^n a_k\left(S_0=0\right)$, so that $a_k=S_k-S_{k-1}$ for $k=1,2, \ldots$ Let $M$ be an uppper bound for both $\left|b_n\right|$ and $\left|S_n\right|$, and let $S=\sum a_n$ and $b=\lim b_n$. Choose $N$ so large that the following three inequalities hold for all $m>N$ and $n>N$ :

$$

\left|b_n S_n-b S\right|<\frac{\varepsilon}{3} ; \quad\left|b_m S_m-b S\right|<\frac{\varepsilon}{3} ; \quad\left|b_m-b_n\right|<\frac{\varepsilon}{3 M} .

$$

Then if $n>m>N$, we have, from the formula for summation by parts,

$$

\sum_{k=m+1}^n a_n b_n=b_n S_n-b_m S_m+\sum_{k=m}^{n-1}\left(b_k-b_{k+1}\right) S_k

$$

Our assumptions yield immediately that $\left|b_n S_n-b_m S_m\right|<\frac{2 \varepsilon}{3}$, and

$$

\left|\sum_{k=m}^{n-1}\left(b_k-b_{k+1}\right) S_k\right| \leq M \sum_{k=m}^{n-1}\left|b_k-b_{k+1}\right| .

$$

Since the sequence $\left\{b_n\right\}$ is monotonic, we have

$$

\sum_{k=m}^{n-1}\left|b_k-b_{k+1}\right|=\left|\sum_{k=m}^{n-1}\left(b_k-b_{k+1}\right)\right|=\left|b_m-b_n\right|<\frac{\varepsilon}{3 M},

$$

from which the desired inequality follows.

\end{proof}


theorem exercise_3_20 {X : Type*} [metric_space X]
  (p : ℕ → X) (l : ℕ) (r : X)
  (hp : cauchy_seq p)
  (hpl : tendsto (λ n, p (l * n)) at_top (𝓝 r)) :
  tendsto p at_top (𝓝 r) :=

Suppose $\left\{p_{n}\right\}$ is a Cauchy sequence in a metric space $X$, and some sequence $\left\{p_{n l}\right\}$ converges to a point $p \in X$. Prove that the full sequence $\left\{p_{n}\right\}$ converges to $p$.

\begin{proof}

    Let $\varepsilon>0$. Choose $N_1$ so large that $d\left(p_m, p_n\right)<\frac{\varepsilon}{2}$ if $m>N_1$ and $n>N_1$. Then choose $N \geq N_1$ so large that $d\left(p_{n_k}, p\right)<\frac{\varepsilon}{2}$ if $k>N$. Then if $n>N$, we have

$$

d\left(p_n, p\right) \leq d\left(p_n, p_{n_{N+1}}\right)+d\left(p_{n_{N+1}}, p\right)<\varepsilon

$$

For the first term on the right is less than $\frac{\varepsilon}{2}$ since $n>N_1$ and $n_{N+1}>N+1>$ $N_1$. The second term is less than $\frac{\varepsilon}{2}$ by the choice of $N$.

\end{proof}


theorem exercise_3_22 (X : Type*) [metric_space X] [complete_space X]
  (G : ℕ → set X) (hG : ∀ n, is_open (G n) ∧ dense (G n)) :
  ∃ x, ∀ n, x ∈ G n :=

Suppose $X$ is a nonempty complete metric space, and $\left\{G_{n}\right\}$ is a sequence of dense open sets of $X$. Prove Baire's theorem, namely, that $\bigcap_{1}^{\infty} G_{n}$ is not empty.

\begin{proof}

    Let $F_n$ be the complement of $G_n$, so that $F_n$ is closed and contains no open sets. We shall prove that any nonempty open set $U$ contains a point not in any $F_n$, hence in all $G_n$. To this end, we note that $U$ is not contained in $F_1$, so that there is a point $x_1 \in U \backslash F_1$. Since $U \backslash F_1$ is open, there exists $r_1>0$ such that $B_1$, defined as the open ball of radius $r_1$ about $x_1$, is contained in $U \backslash F_1$. Let $E_1$ be the open ball of radius $\frac{r_1}{2}$ about $x_1$, so that the closure of $E_1$ is contained in $B_1$. Now $F_2$ does not contain $E_1$, and so we can find a point $x_2 \in E_1 \backslash F_2$. Since $E_1 \backslash F_2$ is an open set, there exists a positive number $r_2$ such that $B_2$, the open ball of radius $R_2$ about $x_2$, is contained in $E_1 \backslash F_2$, which in turn is contained in $U \backslash\left(F_1 \cup F_2\right)$. We let $E_2$ be the open ball of radius $\frac{r_2}{2}$ about $x_2$, so that $\bar{E}_2 \subseteq B_2$. Proceeding in this way, we construct a sequence of open balls $E_j$, such that $E_j \supseteq \bar{E}_{j+1}$, and the diameter of $E_j$ tends to zero. By the previous exercise, there is a point $x$ belonging to all the sets $\bar{E}_j$, hence to all the sets $U \backslash\left(F_1 \cup F_2 \cup \cdots \cup F_n\right)$. Thus the point $x$ belongs to $U \cap\left(\cap_1^{\infty} G_n\right)$.

\end{proof}


theorem exercise_4_2a
  {α : Type} [metric_space α]
  {β : Type} [metric_space β]
  (f : α → β)
  (h₁ : continuous f)
  : ∀ (x : set α), f '' (closure x) ⊆ closure (f '' x) :=

If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, prove that $f(\overline{E}) \subset \overline{f(E)}$ for every set $E \subset X$. ($\overline{E}$ denotes the closure of $E$).

\begin{proof}

    Let $x \in \bar{E}$. We need to show that $f(x) \in \overline{f(E)}$. To this end, let $O$ be any neighborhood of $f(x)$. Since $f$ is continuous, $f^{-1}(O)$ contains (is) a neighborhood of $x$. Since $x \in \bar{E}$, there is a point $u$ of $E$ in $f^{-1}(O)$. Hence $\frac{f(u)}{f(E)} \in O \cap f(E)$. Since $O$ was any neighborhood of $f(x)$, it follows that $f(x) \in \overline{f(E)}$

\end{proof}


theorem exercise_4_4a
  {α : Type} [metric_space α]
  {β : Type} [metric_space β]
  (f : α → β)
  (s : set α)
  (h₁ : continuous f)
  (h₂ : dense s)
  : f '' set.univ ⊆ closure (f '' s) :=

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 $f(E)$ is dense in $f(X)$.

\begin{proof}

    To prove that $f(E)$ is dense in $f(X)$, simply use that $f(X)=f(\bar{E}) \subseteq \overline{f(E)}$.

\end{proof}


theorem exercise_4_5a
  (f : ℝ → ℝ)
  (E : set ℝ)
  (h₁ : is_closed E)
  (h₂ : continuous_on f E)
  : ∃ (g : ℝ → ℝ), continuous g ∧ ∀ x ∈ E, f x = g x :=

If $f$ is a real continuous function defined on a closed set $E \subset \mathbb{R}$, prove that there exist continuous real functions $g$ on $\mathbb{R}$ such that $g(x)=f(x)$ for all $x \in E$.

\begin{proof}

Following the hint, let the complement of $E$ consist of a countable collection of finite open intervals $\left(a_k, b_k\right)$ together with possibly one or both of the the semi-infinite intervals $(b,+\infty)$ and $(-\infty, a)$. The function $f(x)$ is already defined at $a_k$ and $b_k$, as well as at $a$ and $b$ (if these last two points exist). Define $g(x)$ to be $f(b)$ for $x>b$ and $f(a)$ for $x<a$ if $a$ and $b$ exist. On the interval $\left(a_k, b_k\right)$ let

$$

g(x)=f\left(a_k\right)+\frac{x-a_k}{b_k-a_k}\left(f\left(b_k\right)-f\left(a_k\right)\right) .

$$

Of course we let $g(x)=f(x)$ for $x \in E$. It is now fairly clear that $g(x)$ is continuous. A rigorous proof proceeds as follows. Let $\varepsilon>0$. To choose $\delta>0$ such that $|x-u|<\delta$ implies $|g(x)-g(u)|<\varepsilon$, we consider three cases.

i. If $x>b$, let $\delta=x-b$. Then if $|x-u|<\delta$, it follows that $u>b$ also, so that $g(u)=f(b)=g(x)$, and $|g(u)-g(x)|=0<\varepsilon$. Similarly if $x<a$, let $\delta=a-x$

ii. If $a_k<x<b_k$ and $f\left(a_k\right)=f\left(b_k\right)$, let $\delta=\min \left(x-a_k, b_k-x\right)$. Since $|x-u|<\delta$ implies $a_k<u<b_k$, so that $g(u)=f\left(a_k\right)=f\left(b_k\right)=g(x)$, we again have $|g(x)-g(u)|=0<\varepsilon$. If $a_k<x<b_k$ and $f\left(a_k\right) \neq f\left(b_k\right)$, let $\delta=\min \left(x-a_k, b_k-x, \frac{\left(b_k-a_k\right) \varepsilon}{\left|f\left(b_k\right)-f\left(a_k\right)\right|}\right)$. Then if $|x-u|<\delta$, we again have $a_k<u<b_k$ and so

$$

|g(x)-g(u)|=\frac{|x-u|}{b_k-a_k}\left|f\left(b_k\right)-f\left(a_k\right)\right|<\varepsilon .

$$

iii. If $x \in E$, let $\delta_1$ be such that $|f(u)-f(x)|<\varepsilon$ if $u \in E$ and $|x-u|<\delta_1$. (Subcase a). If there are points $x_1 \in E \cap\left(x-\delta_1, x\right)$ and $x_2 \in E \cap\left(x, x+\delta_1\right)$, let $\delta=\min \left(x-x_1, x_2-x\right)$. If $|u-x|<\delta$ and $u \in E$, then $|f(u)-f(x)|<\varepsilon$ by definition of $\delta_1$. if $u \notin E$, then, since $x_1, x$, and $x_2$ are all in $E$, it follows that $u \in\left(a_k, b_k\right)$, where $a_k \in E, b_k \in E$, and $\left|a_k-x\right|<\delta$ and $\left|b_k-x\right|<\delta$, so that $\left|f\left(a_k\right)-f(x)\right|<\varepsilon$ and $\left|f\left(b_k\right)-f(x)\right|<\varepsilon$. If $f\left(a_k\right)=f\left(b_k\right)$, then $f(u)=f\left(a_k\right)$ also, and we have $|f(u)-f(x)|<\varepsilon$. If $f\left(a_k\right) \neq f\left(b_k\right)$, then

$$

\begin{aligned}

|f(u)-f(x)| & =\left|f\left(a_k\right)-f(x)+\frac{u-a_k}{b_k-a_k}\left(f\left(b_k\right)-f\left(a_k\right)\right)\right| \\

& =\left|\frac{b_k-u}{b_k-a_k}\left(f\left(a_k\right)-f(x)\right)+\frac{u-a_k}{b_k-a_k}\left(f\left(b_k\right)-f(x)\right)\right| \\

& <\frac{b_k-u}{b_k-a_k} \varepsilon+\frac{u-a_k}{b_k-a_k} \varepsilon \\

& =\varepsilon

\end{aligned}

$$

(Subcase b). Suppose $x_2$ does not exist, i.e., either $x=a_k$ or $x=a_k$ and $b_k>a_k+\delta_1$. Let us consider the second of these cases and show how to get $|f(u)-f(x)|<\varepsilon$ for $x<u<x+\delta$. If $f\left(a_k\right)=f\left(b_k\right)$, let $\delta=\delta_1$. If $u>x$ we have $a_k<u<b_k$ and $f(u)=f\left(a_k\right)=f(x)$. If $f\left(a_k\right) \neq f\left(b_k\right)$, let $\delta=$ $\min \left(\delta_1, \frac{\left(b_k-a_k\right) \varepsilon}{\left|f\left(b_k\right)-f\left(a_k\right)\right|}\right)$. Then, just as in Subcase a, we have $|f(u)-f(x)|<\varepsilon$.

The case when $x=b_k$ for some $k$ and $a_k<x-\delta_1$ is handled in exactly the same way.

\end{proof}


theorem exercise_4_6
  (f : ℝ → ℝ)
  (E : set ℝ)
  (G : set (ℝ × ℝ))
  (h₁ : is_compact E)
  (h₂ : G = {(x, f x) | x ∈ E})
  : continuous_on f E ↔ is_compact G :=

If $f$ is defined on $E$, the graph of $f$ is the set of points $(x, f(x))$, for $x \in E$. In particular, if $E$ is a set of real numbers, and $f$ is real-valued, the graph of $f$ is a subset of the plane. Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact.

\begin{proof}

    Let $Y$ be the co-domain of the function $f$. We invent a new metric space $E \times Y$ as the set of pairs of points $(x, y), x \in E, y \in Y$, with the metric $\rho\left(\left(x_1, y_1\right),\left(x_2, y_2\right)\right)=d_E\left(x_1, x_2\right)+d_Y\left(y_1, y_2\right)$. The function $\varphi(x)=(x, f(x))$ is then a mapping of $E$ into $E \times Y$.

We claim that the mapping $\varphi$ is continuous if $f$ is continuous. Indeed, let $x \in X$ and $\varepsilon>0$ be given. Choose $\eta>0$ so that $d_Y(f(x), f(u))<\frac{\varepsilon}{2}$ if $d_E(x, y)<\eta$. Then let $\delta=\min \left(\eta, \frac{\varepsilon}{2}\right)$. It is easy to see that $\rho(\varphi(x), \varphi(u))<\varepsilon$ if $d_E(x, u)<\delta$. Conversely if $\varphi$ is continuous, it is obvious from the inequality $\rho(\varphi(x), \varphi(u)) \geq d_Y(f(x), f(u))$ that $f$ is continuous.

From these facts we deduce immediately that the graph of a continuous function $f$ on a compact set $E$ is compact, being the image of $E$ under the continuous mapping $\varphi$. Conversely, if $f$ is not continuous at some point $x$, there is a sequence of points $x_n$ converging to $x$ such that $f\left(x_n\right)$ does not converge to $f(x)$. If no subsequence of $f\left(x_n\right)$ converges, then the sequence $\left\{\left(x_n, f\left(x_n\right)\right\}_{n=1}^{\infty}\right.$ has no convergent subsequence, and so the graph is not compact. If some subsequence of $f\left(x_n\right)$ converges, say $f\left(x_{n_k}\right) \rightarrow z$, but $z \neq f(x)$, then the graph of $f$ fails to contain the limit point $(x, z)$, and hence is not closed. A fortiori it is not compact.

\end{proof}


theorem exercise_4_8b
  (E : set ℝ) :
  ∃ f : ℝ → ℝ, uniform_continuous_on f E ∧ ¬ metric.bounded (set.image f E) :=

Let $E$ be a bounded set in $R^{1}$. Prove that there exists a real function $f$ such that $f$ is uniformly continuous and is not bounded on $E$.

\begin{proof}

    The function $f(x)=x$ is uniformly continuous on the entire line, but not bounded.

\end{proof}


theorem exercise_4_12
  {α β γ : Type*} [uniform_space α] [uniform_space β] [uniform_space γ]
  {f : α → β} {g : β → γ}
  (hf : uniform_continuous f) (hg : uniform_continuous g) :
  uniform_continuous (g ∘ f) :=

A uniformly continuous function of a uniformly continuous function is uniformly continuous.

\begin{proof}

    Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be uniformly continuous. Then $g \circ f: X \rightarrow Z$ is uniformly continuous, where $g \circ f(x)=g(f(x))$ for all $x \in X$.

To prove this fact, let $\varepsilon>0$ be given. Then, since $g$ is uniformly continuous, there exists $\eta>0$ such that $d_Z(g(u), g(v))<\varepsilon$ if $d_Y(u, v)<\eta$. Since $f$ is uniformly continuous, there exists $\delta>0$ such that $d_Y(f(x), f(y))<\eta$ if $d_X(x, y)<\delta$

It is then obvious that $d_Z(g(f(x)), g(f(y)))<\varepsilon$ if $d_X(x, y)<\delta$, so that $g \circ f$ is uniformly continuous.

\end{proof}


theorem exercise_4_19
  {f : ℝ → ℝ} (hf : ∀ a b c, a < b → f a < c → c < f b → ∃ x, a < x ∧ x < b ∧ f x = c)
  (hg : ∀ r : ℚ, is_closed {x | f x = r}) : continuous f :=

Suppose $f$ is a real function with domain $R^{1}$ which has the intermediate value property: if $f(a)<c<f(b)$, then $f(x)=c$ for some $x$ between $a$ and $b$. Suppose also, for every rational $r$, that the set of all $x$ with $f(x)=r$ is closed. Prove that $f$ is continuous.

\begin{proof}

    The contradiction is evidently that $x_0$ is a limit point of the set of $t$ such that $f(t)=r$, yet, $x_0$ does not belong to this set. This contradicts the hypothesis that the set is closed.

\end{proof}


theorem exercise_4_24 {f : ℝ → ℝ}
  (hf : continuous f) (a b : ℝ) (hab : a < b)
  (h : ∀ x y : ℝ, a < x → x < b → a < y → y < b → f ((x + y) / 2) ≤ (f x + f y) / 2) :
  convex_on ℝ (set.Ioo a b) f :=

Assume that $f$ is a continuous real function defined in $(a, b)$ such that $f\left(\frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}$ for all $x, y \in(a, b)$. Prove that $f$ is convex.

\begin{proof}

    We shall prove that

$$

f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)

$$

for all "dyadic rational" numbers, i.e., all numbers of the form $\lambda=\frac{k}{2^n}$, where $k$ is a nonnegative integer not larger than $2^n$. We do this by induction on $n$. The case $n=0$ is trivial (since $\lambda=0$ or $\lambda=1$ ). In the case $n=1$ we have $\lambda=0$ or $\lambda=1$ or $\lambda=\frac{1}{2}$. The first two cases are again trivial, and the third is precisely the hypothesis of the theorem. Suppose the result is proved for $n \leq r$, and consider $\lambda=\frac{k}{2^{r+1}}$. If $k$ is even, say $k=2 l$, then $\frac{k}{2^{r+1}}=\frac{l}{2^r}$, and we can appeal to the induction hypothesis. Now suppose $k$ is odd. Then $1 \leq k \leq 2^{r+1}-1$, and so the numbers $l=\frac{k-1}{2}$ and $m=\frac{k+1}{2}$ are integers with $0 \leq l<m \leq 2^r$. We can now write

$$

\lambda=\frac{s+t}{2},

$$

where $s=\frac{k-1}{2^{r+1}}=\frac{l}{2^r}$ and $t=\frac{k+1}{2^{r+1}}=\frac{m}{2^r}$. We then have

$$

\lambda x+(1-\lambda) y=\frac{[s x+(1-s) y]+[t x+(1-t) y]}{2}

$$

Hence by the hypothesis of the theorem and the induction hypothesis we have

$$

\begin{aligned}

f(\lambda x+(1-\lambda) y) & \leq \frac{f(s x+(1-s) y)+f(t x+(1-t) y)}{2} \\

& \leq \frac{s f(x)+(1-s) f(y)+t f(x)+(1-t) f(y)}{2} \\

&=\left(\frac{s+t}{2}\right) f(x)+\left(1-\frac{s+t}{2}\right) f(y) \\

&=\lambda f(x)+(1-\lambda) f(y)

\end{aligned}

$$

This completes the induction.

Now for each fixed $x$ and $y$ both sides of the inequality

$$

f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)

$$

are continuous functions of $\lambda$. Hence the set on which this inequality holds (the inverse image of the closed set $[0, \infty)$ under the mapping $\lambda \mapsto \lambda f(x)+(1-$ $\lambda) f(y)-f(\lambda x+(1-\lambda) y))$ is a closed set. Since it contains all the points $\frac{k}{2^n}$, $0 \leq k \leq n, n=1,2, \ldots$, it must contain the closure of this set of points, i.e., it must contain all of $[0,1]$. Thus $f$ is convex.

\end{proof}


theorem exercise_5_2 {a b : ℝ}
  {f g : ℝ → ℝ} (hf : ∀ x ∈ set.Ioo a b, deriv f x > 0)
  (hg : g = f⁻¹)
  (hg_diff : differentiable_on ℝ g (set.Ioo a b)) :
  differentiable_on ℝ g (set.Ioo a b) ∧
  ∀ x ∈ set.Ioo a b, deriv g x = 1 / deriv f x :=

Suppose $f^{\prime}(x)>0$ in $(a, b)$. Prove that $f$ is strictly increasing in $(a, b)$, and let $g$ be its inverse function. Prove that $g$ is differentiable, and that $g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \quad(a<x<b)$.

\begin{proof}

   For any $c, d$ with $a<c<d<b$ there exists a point $p \in(c, d)$ such that $f(d)-f(c)=f^{\prime}(p)(d-c)>0$. Hence $f(c)<f(d)$

We know from Theorem $4.17$ that the inverse function $g$ is continuous. (Its restriction to each closed subinterval $[c, d]$ is continuous, and that is sufficient.) Now observe that if $f(x)=y$ and $f(x+h)=y+k$, we have

$$

\frac{g(y+k)-g(y)}{k}-\frac{1}{f^{\prime}(x)}=\frac{1}{\frac{f(x+h)-f(x)}{h}}-\frac{1}{f^{\prime}(x)}

$$

Since we know $\lim \frac{1}{\varphi(t)}=\frac{1}{\lim \varphi(t)}$ provided $\lim \varphi(t) \neq 0$, it follows that for any $\varepsilon>0$ there exists $\eta>0$ such that

$$

\left|\frac{1}{\frac{f(x+h)-f(x)}{h}}-\frac{1}{f^{\prime}(x)}\right|<\varepsilon

$$

if $0<|h|<\eta$. Since $h=g(y+k)-g(y)$, there exists $\delta>0$ such that $0<|h|<\eta$ if $0<|k|<\delta$. The proof is now complete.

\end{proof}


theorem exercise_5_4 {n : ℕ}
  (C : ℕ → ℝ)
  (hC : ∑ i in (finset.range (n + 1)), (C i) / (i + 1) = 0) :
  ∃ x, x ∈ (set.Icc (0 : ℝ) 1) ∧ ∑ i in finset.range (n + 1), (C i) * (x^i) = 0 :=

If $C_{0}+\frac{C_{1}}{2}+\cdots+\frac{C_{n-1}}{n}+\frac{C_{n}}{n+1}=0,$ where $C_{0}, \ldots, C_{n}$ are real constants, prove that the equation $C_{0}+C_{1} x+\cdots+C_{n-1} x^{n-1}+C_{n} x^{n}=0$ has at least one real root between 0 and 1.

\begin{proof}

    Consider the polynomial

$$

p(x)=C_0 x+\frac{C_1}{2} x^2+\cdots+\frac{C_{n-1}}{n} x^n+\frac{C_n}{n+1} x^{n+1},

$$

whose derivative is

$$

p^{\prime}(x)=C_0+C_1 x+\cdots+C_{n-1} x^{n-1}+C_n x^n .

$$

It is obvious that $p(0)=0$, and the hypothesis of the problem is that $p(1)=0$. Hence Rolle's theorem implies that $p^{\prime}(x)=0$ for some $x$ between 0 and 1 .

\end{proof}


theorem exercise_5_6
  {f : ℝ → ℝ}
  (hf1 : continuous f)
  (hf2 : ∀ x, differentiable_at ℝ f x)
  (hf3 : f 0 = 0)
  (hf4 : monotone (deriv f)) :
  monotone_on (λ x, f x / x) (set.Ioi 0) :=

Suppose (a) $f$ is continuous for $x \geq 0$, (b) $f^{\prime}(x)$ exists for $x>0$, (c) $f(0)=0$, (d) $f^{\prime}$ is monotonically increasing. Put $g(x)=\frac{f(x)}{x} \quad(x>0)$ and prove that $g$ is monotonically increasing.

\begin{proof}

    Put

$$

g(x)=\frac{f(x)}{x} \quad(x>0)

$$

and prove that $g$ is monotonically increasing.

By the mean-value theorem

$$

f(x)=f(x)-f(0)=f^{\prime}(c) x

$$

for some $c \in(0, x)$. Since $f^{\prime}$ is monotonically increasing, this result implies that $f(x)<x f^{\prime}(x)$. It therefore follows that

$$

g^{\prime}(x)=\frac{x f^{\prime}(x)-f(x)}{x^2}>0,

$$

so that $g$ is also monotonically increasing.

\end{proof}


theorem exercise_5_15 {f : ℝ → ℝ} (a M0 M1 M2 : ℝ)
  (hf' : differentiable_on ℝ f (set.Ici a))
  (hf'' : differentiable_on ℝ (deriv f) (set.Ici a))
  (hM0 : M0 = Sup {(| f x | )| x ∈ (set.Ici a)})
  (hM1 : M1 = Sup {(| deriv f x | )| x ∈ (set.Ici a)})
  (hM2 : M2 = Sup {(| deriv (deriv f) x | )| x ∈ (set.Ici a)}) :
  (M1 ^ 2) ≤ 4 * M0 * M2 :=

Suppose $a \in R^{1}, f$ is a twice-differentiable real function on $(a, \infty)$, and $M_{0}, M_{1}, M_{2}$ are the least upper bounds of $|f(x)|,\left|f^{\prime}(x)\right|,\left|f^{\prime \prime}(x)\right|$, respectively, on $(a, \infty)$. Prove that $M_{1}^{2} \leq 4 M_{0} M_{2} .$

\begin{proof}

    The inequality is obvious if $M_0=+\infty$ or $M_2=+\infty$, so we shall assume that $M_0$ and $M_2$ are both finite. We need to show that

$$

\left|f^{\prime}(x)\right| \leq 2 \sqrt{M_0 M_2}

$$

for all $x>a$. We note that this is obvious if $M_2=0$, since in that case $f^{\prime}(x)$ is constant, $f(x)$ is a linear function, and the only bounded linear function is a constant, whose derivative is zero. Hence we shall assume from now on that $0<M_2<+\infty$ and $0<M_0<+\infty$.

Following the hint, we need only choose $h=\sqrt{\frac{M_0}{M_2}}$, and we obtain

$$

\left|f^{\prime}(x)\right| \leq 2 \sqrt{M_0 M_2},

$$

which is precisely the desired inequality.

The case of equality follows, since the example proposed satisfies

$$

f(x)=1-\frac{2}{x^2+1}

$$

for $x \geq 0$. We see easily that $|f(x)| \leq 1$ for all $x>-1$. Now $f^{\prime}(x)=\frac{4 x}{\left(x^2+1\right)^2}$ for $x>0$ and $f^{\prime}(x)=4 x$ for $x<0$. It thus follows from Exercise 9 above that $f^{\prime}(0)=0$, and that $f^{\prime}(x)$ is continuous. Likewise $f^{\prime \prime}(x)=4$ for $x<0$ and $f^{\prime \prime}(x)=\frac{4-4 x^2}{\left(x^2+1\right)^3}=-4 \frac{x^2-1}{\left(x^2+1\right)^3}$. This shows that $\left|f^{\prime \prime}(x)\right|<4$ for $x>0$ and also that $\lim _{x \rightarrow 0} f^{\prime \prime}(x)=4$. Hence Exercise 9 again implies that $f^{\prime \prime}(x)$ is continuous and $f^{\prime \prime}(0)=4$.

On $n$-dimensional space let $\mathbf{f}(x)=\left(f_1(x), \ldots, f_n(x)\right), M_0=\sup |\mathbf{f}(x)|$, $M_1=\sup \left|\mathbf{f}^{\prime}(x)\right|$, and $M_2=\sup \left|\mathbf{f}^{\prime \prime}(x)\right|$. Just as in the numerical case, there is nothing to prove if $M_2=0$ or $M_0=+\infty$ or $M_2=+\infty$, and so we assume $0<M_0<+\infty$ and $0<M_2<\infty$. Let $a$ be any positive number less than $M_1$, let $x_0$ be such that $\left|\mathbf{f}^{\prime}\left(x_0\right)\right|>a$, and let $\mathbf{u}=\frac{1}{\left|\mathbf{f}^{\prime}\left(x_0\right)\right|} \mathbf{f}^{\prime}\left(x_0\right)$. Consider the real-valued function $\varphi(x)=\mathrm{u} \cdot \mathrm{f}(x)$. Let $N_0, N_1$, and $N_2$ be the suprema of $|\varphi(x)|,\left|\varphi^{\prime}(x)\right|$, and $\left|\varphi^{\prime \prime}(x)\right|$ respectively. By the Schwarz inequality we have (since $|\mathbf{u}|=1) N_0 \leq M_0$ and $N_2 \leq M_2$, while $N_1 \geq \varphi\left(x_0\right)=\left|\mathbf{f}^{\prime}\left(x_0\right)\right|>a$. We therefore have $a^2<4 N_0 N_2 \leq 4 M_0 M_2$. Since $a$ was any positive number less than $M_1$, we have $M_1^2 \leq 4 M_0 M_2$, i.e., the result holds also for vector-valued functions.

Equality can hold on any $R^n$, as we see by taking $\mathbf{f}(x)=(f(x), 0, \ldots, 0)$ or $\mathbf{f}(x)=(f(x), f(x), \ldots, f(x))$, where $f(x)$ is a real-valued function for which equality holds.

\end{proof}


theorem exercise_13_1 (X : Type*) [topological_space X] (A : set X)
  (h1 : ∀ x ∈ A, ∃ U : set X, x ∈ U ∧ is_open U ∧ U ⊆ A) :
  is_open A :=

import .common

open set topological_space filter
open_locale classical topology filter
noncomputable theory

Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x \in A$ there is an open set $U$ containing $x$ such that $U \subset A$. Show that $A$ is open in $X$.

\begin{proof}

    Since, from the given hypothesis given any $x \in A$ there exists an open set containing $x$ say, $U_x$ such that $U_x \subset A$. Thus, we claim that

$$

A=\bigcup_{x \in A} U_x

$$

Observe that if we prove the above claim, then $A$ will be open, being a union of arbitrary open sets. Since, for each $x \in A, U_x \subset A \Longrightarrow \cup U_x \subset A$. For the converse, observe that given any $x \in A, x \in U_x$ and hence in the union. Thus we proved our claim, and hence $A$ is an open set.

\end{proof}


theorem exercise_13_4a1 (X I : Type*) (T : I → set (set X)) (h : ∀ i, is_topology X (T i)) :
  is_topology X (⋂ i : I, T i) :=

If $\mathcal{T}_\alpha$ is a family of topologies on $X$, show that $\bigcap \mathcal{T}_\alpha$ is a topology on $X$.

\begin{proof}

    Since $\emptyset$ and $X$ belong to $\mathcal{T}_\alpha$ for each $\alpha$, they belong to $\bigcap_\alpha \mathcal{T}_\alpha$. Let $\left\{V_\beta\right\}_\beta$ be a collection of open sets in $\bigcap_\alpha \mathcal{T}_\alpha$. For any fixed $\alpha$ we have $\cup_\beta V_\beta \in \mathcal{T}_\alpha$ since $\mathcal{T}_\alpha$ is a topology on $X$, so $\bigcup_\beta V_\beta \in \bigcap_\alpha \mathcal{T}_\alpha$. Similarly, if $U_1, \ldots, U_n$ are elements of $\bigcap_\alpha \mathcal{T}_\alpha$, then for each $\alpha$ we have $\bigcup_{i=1}^n U_i \in \mathcal{T}_\alpha$ and therefore $\bigcup_{i=1}^n U_i \in \bigcap_\alpha \mathcal{T}_\alpha$. It follows that $\bigcap_\alpha \mathcal{T}_\alpha$ is a topology on $X$.

\end{proof}


theorem exercise_13_4b1 (X I : Type*) (T : I → set (set X)) (h : ∀ i, is_topology X (T i)) :
  ∃! T', is_topology X T' ∧ (∀ i, T i ⊆ T') ∧
  ∀ T'', is_topology X T'' → (∀ i, T i ⊆ T'') → T'' ⊆ T' :=

Let $\mathcal{T}_\alpha$ be a family of topologies on $X$. Show that there is a unique smallest topology on $X$ containing all the collections $\mathcal{T}_\alpha$.

\begin{proof}

    (b) First we prove that there is a unique smallest topology on $X$ containing all the collections $\mathcal{T}_\alpha$. Uniqueness of such topology is clear. For each $\alpha$ let $\mathcal{B}_\alpha$ be a basis for $\mathcal{T}_\alpha$. Let $\mathcal{T}$ be the topology generated by the subbasis $\mathcal{S}=\bigcup_\alpha \mathcal{B}_\alpha$. Then the collection $\mathcal{B}$ of all finite intersections of elements of $\mathcal{S}$ is a basis for $\mathcal{T}$. Clearly $\mathcal{T}_\alpha \subset \mathcal{T}$ for all $\alpha$. We now prove that if $\mathcal{O}$ is a topology on $X$ such that $\mathcal{T}_\alpha \subset \mathcal{O}$ for all $\alpha$, then $\mathcal{T} \subset \mathcal{O}$. Given such $\mathcal{O}$, we have $\mathcal{B}_\alpha \subset \mathcal{O}$ for all $\alpha$, so $\mathcal{S} \subset \mathcal{O}$. Since $\mathcal{O}$ is a topology, it must contain all finite intersections of elements of $\mathcal{S}$, so $\mathcal{B} \subset \mathcal{O}$ and hence $\mathcal{T} \subset \mathcal{O}$. We conclude that the topology $\mathcal{T}$ generated by the subbasis $\mathcal{S}=\cup_\alpha \mathcal{B}_\alpha$ is the unique smallest topology on $X$ containing all the collections $\mathcal{T}_\alpha$.

\end{proof}


theorem exercise_13_5a {X : Type*}
  [topological_space X] (A : set (set X)) (hA : is_topological_basis A) :
  generate_from A = generate_from (sInter {T | is_topology X T ∧ A ⊆ T}) :=

Show that if $\mathcal{A}$ is a basis for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\mathcal{A}$.

\begin{proof}

    Let $\mathcal{T}$ be the topology generated by $\mathcal{A}$ and let $\mathcal{O}$ be the intersection of all topologies on $X$ that contains $\mathcal{A}$. Clearly $\mathcal{O} \subset \mathcal{T}$ since $\mathcal{T}$ is a topology on $X$ that contain $\mathcal{A}$. Conversely, let $U \in \mathcal{T}$, so that $U$ is a union of elements of $\mathcal{A}$. Since each of this elements is also an element of $\mathcal{O}$, their union $U$ belongs to $\mathcal{O}$. Thus $\mathcal{T} \subset \mathcal{O}$ and the equality holds.

\end{proof}


theorem exercise_13_6 :
  ¬ (∀ U, Rl.is_open U → K_topology.is_open U) ∧ ¬ (∀ U, K_topology.is_open U → Rl.is_open U) :=

Show that the lower limit topology $\mathbb{R}_l$ and $K$-topology $\mathbb{R}_K$ are not comparable.

\begin{proof}

    Let $\mathcal{T}_{\ell}$ and $\mathcal{T}_K$ denote the topologies of $\mathbb{R}_{\ell}$ and $\mathbb{R}_K$ respectively. Given the basis element $[0,1)$ for $\mathcal{T}_{\ell}$, there is no basis element for $\mathcal{T}_K$ containing 0 and contained in $[0,1)$, so $\mathcal{T}_{\ell} \not \subset \mathcal{T}_K$. Similarly, given the basis element $(-1,1) \backslash K$ for $\mathcal{T}_K$, there is no basis element for $\mathcal{T}_{\ell}$ containing 0 contained in $(-1,1) \backslash K$, so $\mathcal{T}_K \not \subset \mathcal{T}_{\ell}$.

\end{proof}


theorem exercise_13_8b :
  (topological_space.generate_from {S : set ℝ | ∃ a b : ℚ, a < b ∧ S = Ico a b}).is_open ≠
  (lower_limit_topology ℝ).is_open :=

Show that the collection $\{(a,b) \mid a < b, a \text{ and } b \text{ rational}\}$ is a basis that generates a topology different from the lower limit topology on $\mathbb{R}$.

\begin{proof}

    (b) $\mathcal{C}$ is a basis for a topology on $\mathbb{R}$ since the union of its elements is $\mathbb{R}$ and the intersection of two elements of $\mathcal{C}$ is either empty or another element of $\mathcal{C}$. Now consider $[r, s)$ where $r$ is any irrational number and $s$ is any real number greater than $r$. Then $[r, s)$ is a basis element for the topology of $\mathbb{R}_{\ell}$, but $[r, s)$ is not a union of elements of $\mathcal{C}$. Indeed, suppose that $[r, s)=\cup_\alpha\left[a_\alpha, b_\alpha\right)$ for rationals $a_\alpha, b_\alpha$. Then $r \in\left[a_\alpha, b_\alpha\right)$ for some $\alpha$. Since $r$ is irrational we must have $a_\alpha<r$, but then $a_\alpha \notin[r, s)$, a contradiction. It follows that the topology generated by $\mathcal{C}$ is strictly coarser than the lower limit topology on $\mathbb{R}$.

\end{proof}


theorem exercise_16_4 {X Y : Type*} [topological_space X] [topological_space Y]
  (π₁ : X × Y → X)
  (π₂ : X × Y → Y)
  (h₁ : π₁ = prod.fst)
  (h₂ : π₂ = prod.snd) :
  is_open_map π₁ ∧ is_open_map π₂ :=

A map $f: X \rightarrow Y$ is said to be an open map if for every open set $U$ of $X$, the set $f(U)$ is open in $Y$. Show that $\pi_{1}: X \times Y \rightarrow X$ and $\pi_{2}: X \times Y \rightarrow Y$ are open maps.

\begin{proof}

    Exercise 16.4. Let $U \times V$ be a (standard) basis element for $X \times Y$, so that $U$ is open in $X$ and $V$ is open in $Y$. Then $\pi_1(U \times V)=U$ is open in $X$ and $\pi_2(U \times V)=V$ is open in $Y$. Since arbitrary maps and unions satisfy $f\left(\bigcup_\alpha W_\alpha\right)=\bigcup_\alpha f\left(W_\alpha\right)$, it follows that $\pi_1$ and $\pi_2$ are open maps.

\end{proof}


theorem exercise_17_4 {X : Type*} [topological_space X]
  (U A : set X) (hU : is_open U) (hA : is_closed A) :
  is_open (U \ A) ∧ is_closed (A \ U) :=

Show that if $U$ is open in $X$ and $A$ is closed in $X$, then $U-A$ is open in $X$, and $A-U$ is closed in $X$.

\begin{proof}

Since

$$

X \backslash(U \backslash A)=(X \backslash U) \cup A \text { and } \quad X \backslash(A \backslash U)=(X \backslash A) \cup U,

$$

it follows that $X \backslash(U \backslash A)$ is closed in $X$ and $X \backslash(A \backslash U)$ is open in $X$.

\end{proof}


theorem exercise_18_8b {X Y : Type*} [topological_space X] [topological_space Y]
  [linear_order Y] [order_topology Y] {f g : X → Y}
  (hf : continuous f) (hg : continuous g) :
  continuous (λ x, min (f x) (g x)) :=

Let $Y$ be an ordered set in the order topology. Let $f, g: X \rightarrow Y$ be continuous. Let $h: X \rightarrow Y$ be the function $h(x)=\min \{f(x), g(x)\}.$ Show that $h$ is continuous.

\begin{proof}

    Let $A=\{x \mid f(x) \leq g(x)\}$ and $B=\{x \mid g(x) \leq f(x)\}$. Then $A$ and $B$ are closed in $X$ by (a), $A \cap B=\{x \mid f(x)=g(x)\}$, and $X=A \cup B$. Since $f$ and $g$ are continuous, their restrictions $f^{\prime}: A \rightarrow Y$ and $g^{\prime}: B \rightarrow Y$ are continuous. It follows from the pasting lemma that

$$

h: X \rightarrow Y, \quad h(x)=\min \{f(x), g(x)\}= \begin{cases}f^{\prime}(x) & \text { if } x \in A \\ g^{\prime}(x) & \text { if } x \in B\end{cases}

$$

is continuous

\end{proof}


theorem exercise_19_6a
  {n : ℕ}
  {f : fin n → Type*} {x : ℕ → Πa, f a}
  (y : Πi, f i)
  [Πa, topological_space (f a)] :
  tendsto x at_top (𝓝 y) ↔ ∀ i, tendsto (λ j, (x j) i) at_top (𝓝 (y i)) :=

Let $\mathbf{x}_1, \mathbf{x}_2, \ldots$ be a sequence of the points of the product space $\prod X_\alpha$.  Show that this sequence converges to the point $\mathbf{x}$ if and only if the sequence $\pi_\alpha(\mathbf{x}_i)$ converges to $\pi_\alpha(\mathbf{x})$ for each $\alpha$.

\begin{proof}

    For each $n \in \mathbb{Z}_{+}$, we write $\mathbf{x}_n=\left(x_n^\alpha\right)_\alpha$, so that $\pi_\alpha\left(\mathbf{x}_n\right)=x_n^\alpha$ for each $\alpha$.

First assume that the sequence $\mathbf{x}_1, \mathbf{x}_2, \ldots$ converges to $\mathbf{x}=\left(x_\alpha\right)_\alpha$ in the product space $\prod_\alpha X_\alpha$. Fix an index $\beta$ and let $U$ be a neighbourhood of $\pi_\beta(\mathbf{x})=x_\beta$. Let $V=\prod_\alpha U_\alpha$, where $U_\alpha=X_\alpha$ for each $\alpha \neq \beta$ and $U_\beta=U$. Then $V$ is a neighbourhood of $\mathbf{x}$, so there exists $N \in \mathbb{Z}_{+}$such that $\mathbf{x}_n \in V$ for all $n \geq N$. Therefore $\pi_\beta\left(\mathbf{x}_n\right)=x_n^\beta \in U$ for all $n \geq N$. Since $U$ was arbitrary, it follows that $\pi_\beta\left(\mathbf{x}_1\right), \pi_\beta\left(\mathbf{x}_2\right), \ldots$ converges to $\pi_\beta(\mathbf{x})$. Since $\beta$ was arbitrary, this holds for all indices $\alpha$.

\end{proof}


theorem exercise_21_6a
  (f : ℕ → I → ℝ )
  (h : ∀ x n, f n x = x ^ n) :
  ∀ x, ∃ y, tendsto (λ n, f n x) at_top (𝓝 y) :=

Define $f_{n}:[0,1] \rightarrow \mathbb{R}$ by the equation $f_{n}(x)=x^{n}$. Show that the sequence $\left(f_{n}(x)\right)$ converges for each $x \in[0,1]$.

\begin{proof}

If $0 \leq x<1$ is fixed, then $f_n(x) \rightarrow 0$ as $n \rightarrow \infty$. As $f_n(1)=1$ for all $n, f_n(1) \rightarrow 1$. Thus $\left(f_n\right)_n$ converges to $f:[0,1] \rightarrow \mathbb{R}$ given by $f(x)=0$ if $x=0$ and $f(1)=1$. The sequence

\end{proof}


theorem exercise_21_8
  {X : Type*} [topological_space X] {Y : Type*} [metric_space Y]
  {f : ℕ → X → Y} {x : ℕ → X}
  (hf : ∀ n, continuous (f n))
  (x₀ : X)
  (hx : tendsto x at_top (𝓝 x₀))
  (f₀ : X → Y)
  (hh : tendsto_uniformly f f₀ at_top) :
  tendsto (λ n, f n (x n)) at_top (𝓝 (f₀ x₀)) :=

Let $X$ be a topological space and let $Y$ be a metric space. Let $f_{n}: X \rightarrow Y$ be a sequence of continuous functions. Let $x_{n}$ be a sequence of points of $X$ converging to $x$. Show that if the sequence $\left(f_{n}\right)$ converges uniformly to $f$, then $\left(f_{n}\left(x_{n}\right)\right)$ converges to $f(x)$.

\begin{proof}

    Let $d$ be the metric on $Y$. Let $V$ be a neighbourhood of $f(x)$, and let $\varepsilon>0$ be such that $f(x) \in B_d(f(x), \varepsilon) \subset V$. Since $\left(f_n\right)_n$ converges uniformly to $f$, there exists $N_1 \in \mathbb{Z}_{+}$such that $d\left(f_n(x), f(x)\right)<\varepsilon / 2$ for all $x \in X$ and all $n \geq N_1$, so that $d\left(f_n\left(x_n\right), f\left(x_n\right)\right)<\varepsilon / 2$ for all $n \geq N_1$. Moreover, $f$ is continuous, so there exists $N_2 \in \mathbb{Z}_{+}$such that $d\left(f\left(x_n\right), f(x)\right)<\varepsilon / 2$ for all $n \geq N_2$. Thus, if $N>\max \left\{N_1, N_2\right\}$, then

$$

d\left(f_n\left(x_n\right), f(x)\right) \leq d\left(f_n\left(x_n\right), f\left(x_n\right)\right)+d\left(f\left(x_n\right), f(x)\right)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon

$$

for all $n \geq N$, so $f_n\left(x_n\right) \in V$ for all $n \geq N$. It follows that $\left(f_n\left(x_n\right)\right)_n$ converges to $f(x)$.

\end{proof}


theorem exercise_22_2b {X : Type*} [topological_space X]
  {A : set X} (r : X → A) (hr : continuous r) (h : ∀ x : A, r x = x) :
  quotient_map r :=

If $A \subset X$, a retraction of $X$ onto $A$ is a continuous map $r: X \rightarrow A$ such that $r(a)=a$ for each $a \in A$. Show that a retraction is a quotient map.

\begin{proof}

The inclusion map $i: A \rightarrow X$ is continuous and $r \circ i=1_A$ is the identity. Thus $r$ is a quotient map by (a).

\end{proof}


theorem exercise_23_2 {X : Type*}
  [topological_space X] {A : ℕ → set X} (hA : ∀ n, is_connected (A n))
  (hAn : ∀ n, A n ∩ A (n + 1) ≠ ∅) :
  is_connected (⋃ n, A n) :=

Let $\left\{A_{n}\right\}$ be a sequence of connected subspaces of $X$, such that $A_{n} \cap A_{n+1} \neq \varnothing$ for all $n$. Show that $\bigcup A_{n}$ is connected.

\begin{proof}

    Suppose that $\bigcup_n A_n=B \cup C$, where $B$ and $C$ are disjoint open subsets of $\bigcup_n A_n$. Since $A_1$ is connected and a subset of $B \cup C$, by Lemma $23.2$ it lies entirely within either $B$ or $C$. Without any loss of generality, we may assume $A_1 \subset B$. Note that given $n$, if $A_n \subset B$ then $A_{n+1} \subset B$, for if $A_{n+1} \subset C$ then $A_n \cap A_{n+1} \subset B \cap C=\emptyset$, in contradiction with the assumption. By induction, $A_n \subset B$ for all $n \in \mathbb{Z}_{+}$, so that $\bigcup_n A_n \subset B$. It follows that $\bigcup_n A_n$ is connected.

\end{proof}


theorem exercise_23_4 {X : Type*} [topological_space X] [cofinite_topology X]
  (s : set X) : set.infinite s → is_connected s :=

Show that if $X$ is an infinite set, it is connected in the finite complement topology.

\begin{proof}

    Suppose that $A$ is a non-empty subset of $X$ that is both open and closed, i.e., $A$ and $X \backslash A$ are finite or all of $X$. Since $A$ is non-empty, $X \backslash A$ is finite. Thus $A$ cannot be finite as $X \backslash A$ is infinite, so $A$ is all of $X$. Therefore $X$ is connected.

\end{proof}


theorem exercise_23_9 {X Y : Type*}
  [topological_space X] [topological_space Y]
  (A₁ A₂ : set X)
  (B₁ B₂ : set Y)
  (hA : A₁ ⊂ A₂)
  (hB : B₁ ⊂ B₂)
  (hA : is_connected A₂)
  (hB : is_connected B₂) :
  is_connected ({x | ∃ a b, x = (a, b) ∧ a ∈ A₂ ∧ b ∈ B₂} \
      {x | ∃ a b, x = (a, b) ∧ a ∈ A₁ ∧ b ∈ B₁}) :=

Let $A$ be a proper subset of $X$, and let $B$ be a proper subset of $Y$. If $X$ and $Y$ are connected, show that $(X \times Y)-(A \times B)$ is connected.

\begin{proof}

This is similar to the proof of Theorem 23.6. Take $c \times d \in(X \backslash A) \times(Y \backslash B)$. For each $x \in X \backslash A$, the set

$$

U_x=(X \times\{d\}) \cup(\{x\} \times Y)

$$

is connected since $X \times\{d\}$ and $\{x\} \times Y$ are connected and have the common point $x \times d$. Then $U=\bigcup_{x \in X \backslash A} U_x$ is connected because it is the union of the connected spaces $U_x$ which have the point $c \times d$ in common. Similarly, for each $y \in Y \backslash B$ the set

$$

V_y=(X \times\{y\}) \cup(\{c\} \times Y)

$$

is connected, so $V=\bigcup_{y \in Y \backslash B} V_y$ is connected. Thus $(X \times Y) \backslash(A \times B)=U \cup V$ is connected since $c \times d$ is a common point of $U$ and $V$.

\end{proof}


theorem exercise_24_2 {f : (metric.sphere 0 1 : set ℝ) → ℝ}
  (hf : continuous f) : ∃ x, f x = f (-x) :=

Let $f: S^{1} \rightarrow \mathbb{R}$ be a continuous map. Show there exists a point $x$ of $S^{1}$ such that $f(x)=f(-x)$.

\begin{proof}

    Let $f: S^1 \rightarrow \mathbb{R}$ be continuous. Let $x \in S^1$. If $f(x)=f(-x)$ we are done, so assume $f(x) \neq f(-x)$. Define $g: S^1 \rightarrow \mathbb{R}$ by setting $g(x)=f(x)-f(-x)$. Then $g$ is continuous. Suppose $f(x)>f(-x)$, so that $g(x)>0$. Then $-x \in S^1$ and $g(-x)<0$. By the intermediate value theorem, since $S^1$ is connected and $g(-x)<0<g(x)$, there exists $y \in S^1$ such that $g(y)=0$. i.e, $f(y)=f(-y)$. Similarly, if $f(x)<f(-x)$, then $g(x)<0<g(-x)$ and again the intermediate value theorem gives the result.

\end{proof}


theorem exercise_25_4 {X : Type*} [topological_space X]
  [loc_path_connected_space X] (U : set X) (hU : is_open U)
  (hcU : is_connected U) : is_path_connected U :=

Let $X$ be locally path connected. Show that every connected open set in $X$ is path connected.

\begin{proof}

    Let $U$ be a open connected set in $X$. By Theorem 25.4, each path component of $U$ is open in $X$, hence open in $U$. Thus, each path component in $U$ is both open and closed in $U$, so must be empty or all of $U$. It follows that $U$ is path-connected.

\end{proof}


theorem exercise_26_11
  {X : Type*} [topological_space X] [compact_space X] [t2_space X]
  (A : set (set X)) (hA : ∀ (a b : set X), a ∈ A → b ∈ A → a ⊆ b ∨ b ⊆ a)
  (hA' : ∀ a ∈ A, is_closed a) (hA'' : ∀ a ∈ A, is_connected a) :
  is_connected (⋂₀ A) :=

Let $X$ be a compact Hausdorff space. Let $\mathcal{A}$ be a collection of closed connected subsets of $X$ that is simply ordered by proper inclusion. Then $Y=\bigcap_{A \in \mathcal{A}} A$ is connected.

\begin{proof}

 Since each $A \in \mathcal{A}$ is closed, $Y$ is closed. Suppose that $C$ and $D$ form a separation of $Y$. Then $C$ and $D$ are closed in $Y$, hence closed in $X$. Since $X$ is compact, $C$ and $D$ are compact by Theorem 26.2. Since $X$ is Hausdorff, by Exercise 26.5, there exist $U$ and $V$ open in $X$ and disjoint containing $C$ and $D$, respectively. We show that

$$

\bigcap_{A \in \mathcal{A}}(A \backslash(U \cup V))

$$

is not empty. Let $\left\{A_1, \ldots, A_n\right\}$ be a finite subcollection of elements of $\mathcal{A}$. We may assume that $A_i \subsetneq A_{i+1}$ for all $i=1, \ldots, n-1$. Then

$$

\bigcap_{i=1}^n\left(A_i \backslash(U \cup V)\right)=A_1 \backslash(U \cup V) \text {. }

$$

Suppose that $A_1 \backslash(U \cup V)=\emptyset$. Then $A_1 \subset U \cup V$. Since $A_1$ is connected and $U \cap V=\emptyset, A_1$ lies within either $U$ or $V$, say $A_1 \subset U$. Then $Y \subset A_1 \subset U$, so that $C=Y \cap C \subset Y \cap V=\emptyset$, contradicting the fact that $C$ and $D$ form a separation of $Y$. Hence, $\bigcap_{i=1}^n\left(A_i \backslash(U \cup V)\right)$ is non-empty. Therefore, the collection $\{A \backslash(U \cup V) \mid A \in \mathcal{A}\}$ has the finite intersection property, so

$$

\bigcap_{A \in \mathcal{A}}(A \backslash(U \cup V))=\left(\bigcap_{A \in \mathcal{A}} A\right) \backslash(U \cup V)=Y \backslash(U \cup V)

$$

is non-empty. So there exists $y \in Y$ such that $y \notin U \cup V \supset C \cup D$, contradicting the fact that $C$ and $D$ form a separation of $Y$. We conclude that there is no such separation, so that $Y$ is connected.

\end{proof}


theorem exercise_27_4
  {X : Type*} [metric_space X] [connected_space X] (hX : ∃ x y : X, x ≠ y) :
  ¬ countable (univ : set X) :=

Show that a connected metric space having more than one point is uncountable.

\begin{proof}

    The distance function $d: X \times X \rightarrow \mathbb{R}$ is continuous by Exercise 20.3(a), so given $x \in X$, the function $d_x: X \rightarrow \mathbb{R}$ given by $d_x(y)=d(x, y)$ is continuous by Exercise 19.11. Since $X$ is connected, the image $d_x(X)$ is a connected subspace of $\mathbb{R}$, and contains 0 since $d_x(x)=0$. Thus, if $y \in X$ and $y \neq x$, then $d_x(X)$ contains the set $[0, \delta]$, where $\delta=d_x(y)>0$. Therefore $X$ must be uncountable.

\end{proof}


theorem exercise_28_5
  (X : Type*) [topological_space X] :
  countably_compact X ↔ ∀ (C : ℕ → set X), (∀ n, is_closed (C n)) ∧
  (∀ n, C n ≠ ∅) ∧ (∀ n, C n ⊆ C (n + 1)) → ∃ x, ∀ n, x ∈ C n :=

Show that X is countably compact if and only if every nested sequence $C_1 \supset C_2 \supset \cdots$ of closed nonempty sets of X has a nonempty intersection.

\begin{proof}

We could imitate the proof of Theorem 26.9, but we prove directly each direction. First let $X$ be countable compact and let $C_1 \supset C_2 \supset \cdots$ be a nested sequence of closed nonempty sets of $X$. For each $n \in \mathbb{Z}_{+}, U_n=X \backslash C_n$ is open in $X$. Then $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$is a countable collection of open sets with no finite subcollection covering $X$, for if $U_{i_1} \cup \cdots \cup U_{1_n}$ covers $X$, then $C_{i_1} \cap \cdots \cap C_{i_n}$ is empty, contrary to the assumption. Hence $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$does not cover $X$, so there exist $x \in X \backslash \bigcup_{n \in \mathbb{Z}_{+}} U_n=\bigcap_{n \in Z_{+}}\left(X \backslash U_n\right)=\bigcap_{n \in Z_{+}} C_n$.

Conversely, assume that every nested sequence $C_1 \supset C_2 \supset \cdots$ of closed non-empty sets of $X$ has a non-empty intersection and let $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$be a countable open covering of $X$. For each $n$, let $V_n=U_1 \cup \cdots \cup U_n$ and $C_n=X \backslash V_n$. Suppose that no finite subcollection of $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$covers $X$. Then each $C_n$ is non-empty, so $C_1 \supset C_2 \supset \cdots$ is a nested sequence of non-empty closed sets and $\bigcap_{n \in \mathbb{Z}_{+}} C_n$ is non-empty by assumption. Then there exists $x \in \bigcap_{n \in \mathbb{Z}_{+}} C_n$, so that $x \notin V_n$ for all $n$, contradicting the fact that $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$covers $X$. It follows that there exists $N \in \mathbb{Z}_{+}$such that $C_N=\emptyset$, so that $X=V_N$ and hence some finite subcollection of $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$covers $X$. We deduce that $X$ is countable compact.

\end{proof}


theorem exercise_29_1 : ¬ locally_compact_space ℚ :=

Show that the rationals $\mathbb{Q}$ are not locally compact.

\begin{proof}

    First, we prove that each set $\mathbb{Q} \cap[a, b]$, where $a, b$ are irrational numbers, is not compact. Indeed, since $\mathbb{Q} \cap[a, b]$ is countable, we can write $\mathbb{Q} \cap[a, b]=\left\{q_1, q_2, \ldots\right\}$. Then $\left\{U_i\right\}_{i \in \mathbb{Z}_{+}}$, where $U_i=\mathbb{Q} \cap\left[a, q_i\right)$ for each $i$, is an open covering of $\mathbb{Q} \cap[a, b]$ with no finite subcovering. Now let $x \in \mathbb{Q}$ and suppose that $\mathbb{Q}$ is locally compact at $x$. Then there exists a compact set $C$ containing a neighbourhood $U$ of $x$. Then $U$ contains a set $\mathbb{Q} \cap[a, b]$ where $a, b$ are irrational numbers. Since this set is closed and contained in the compact $C$, it follows $\mathbb{Q} \cap[a, b]$ is compact, a contradiction. Therefore, $\mathbb{Q}$ is not locally compact.

\end{proof}


theorem exercise_29_10 {X : Type*}
  [topological_space X] [t2_space X] (x : X)
  (hx : ∃ U : set X, x ∈ U ∧ is_open U ∧ (∃ K : set X, U ⊂ K ∧ is_compact K))
  (U : set X) (hU : is_open U) (hxU : x ∈ U) :
  ∃ (V : set X), is_open V ∧ x ∈ V ∧ is_compact (closure V) ∧ closure V ⊆ U :=

Show that if $X$ is a Hausdorff space that is locally compact at the point $x$, then for each neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ such that $\bar{V}$ is compact and $\bar{V} \subset U$.

\begin{proof}

    Let $U$ be a neighbourhood of $x$. Since $X$ is locally compact at $x$, there exists a compact subspace $C$ of $X$ containing a neighbourhood $W$ of $x$. Then $U \cap W$ is open in $X$, hence in $C$. Thus, $C \backslash(U \cap W)$ is closed in $C$, hence compact. Since $X$ is Hausdorff, there exist disjoint open sets $V_1$ and $V_2$ of $X$ containing $x$ and $C \backslash(U \cap W)$ respectively. Let $V=V_1 \cap U \cap W$. Since $\bar{V}$ is closed in $C$, it is compact. Furthermore, $\bar{V}$ is disjoint from $C \backslash(U \cap W) \supset C \backslash U$, so $\bar{V} \subset U$.

\end{proof}


theorem exercise_30_13 {X : Type*} [topological_space X]
  (h : ∃ (s : set X), countable s ∧ dense s) (U : set (set X))
  (hU : ∀ (x y : set X), x ∈ U → y ∈ U → x ≠ y → x ∩ y = ∅) :
  countable U :=

Show that if $X$ has a countable dense subset, every collection of disjoint open sets in $X$ is countable.

\begin{proof}

    Let $\mathcal{U}$ be a collection of disjoint open sets in $X$ and let $A$ be a countable dense subset of $X$.

Since $A$ is dense in $X$, every $U \in \mathcal{U}$ intesects $S$. Therefore, there exists a point $x_U \in U \cap S$.

Let $U_1, U_2 \in \mathcal{U}, U_1 \neq U_2$. Then $x_{U_1} \neq x_{U_2}$ since $U_1 \cap U_2=\emptyset$.

Thus, the function $\mathcal{U} \rightarrow S$ given by $U \mapsto x_U$ is injective and therefore, since $S$ is countable, it follows that $\mathcal{U}$ is countable.

\end{proof}


theorem exercise_31_2 {X : Type*}
  [topological_space X] [normal_space X] {A B : set X}
  (hA : is_closed A) (hB : is_closed B) (hAB : disjoint A B) :
  ∃ (U V : set X), is_open U ∧ is_open V ∧ A ⊆ U ∧ B ⊆ V ∧ closure U ∩ closure V = ∅ :=

Show that if $X$ is normal, every pair of disjoint closed sets have neighborhoods whose closures are disjoint.

\begin{proof}

    Let $A$ and $B$ be disjoint closed sets. Then there exist disjoint open sets $U$ and $V$ containing $A$ and $B$ respectively.

Since $X \backslash V$ is closed and contains $U$, the closure of $U$ is contained in $X \backslash V$ hence $B$ and closure of $U$ are disjoint.

Repeat steps 1 and 2 for $B$ and $\bar{U}$ instead of $A$ and $B$ respectively and you will have open set $V^{\prime}$ which contains $B$ and its closure doesn't intersect with $\bar{U}$.

\end{proof}


theorem exercise_32_1 {X : Type*} [topological_space X]
  (hX : normal_space X) (A : set X) (hA : is_closed A) :
  normal_space {x // x ∈ A} :=

Show that a closed subspace of a normal space is normal.

\begin{proof}

    Let $X$ be a normal space and $Y$ a closed subspace of $X$.

First we shows that $Y$ is a $T_1$-space.

Let $y \in Y$ be any point. Since $X$ is normal, $X$ is also a $T_1$ space and therefore $\{y\}$ is closed in $X$.

Then it follows that $\{y\}=\{y\} \cap Y$ is closed in $Y$ (in relative topology).

Now let's prove that $X$ is a $T_4$-space.

Let $F, G \subseteq Y$ be disjoint closed sets. Since $F$ and $G$ are closed in $Y$ and $Y$ is closed in $X$, it follows that $F$ and $G$ are closed in $X$.

Since $X$ is normal, $X$ is also a $T_4$-space and therefore there exist disjoint open sets $U, V \subseteq$ $X$ such that $F \subseteq U$ and $G \subseteq V$.

However, then $U \cap Y$ and $V \cap Y$ are open disjoint sets in $Y$ (in relative topology) which separate $F$ and $G$.

\end{proof}


theorem exercise_32_2b
  {ι : Type*} {X : ι → Type*} [∀ i, topological_space (X i)]
  (h : ∀ i, nonempty (X i)) (h2 : regular_space (Π i, X i)) :
  ∀ i, regular_space (X i) :=

Show that if $\prod X_\alpha$ is regular, then so is $X_\alpha$. Assume that each $X_\alpha$ is nonempty.

\begin{proof}

    Suppose that $X=\prod_\beta X_\beta$ is regular and let $\alpha$ be any index.

We have to prove that $X_\alpha$ satisfies the $T_1$ and the $T_3$ axiom.

Since $X$ is regular, it follows that $X$ is Hausdorff, which then implies that $X_\alpha$ is Hausdorff. However, this implies that $X_\alpha$ satisfies the $T_1$ axiom.

Let now $F \subseteq X_\alpha$ be a closed set and $x \in X_\alpha \backslash F$ a point.

Then $\prod_\beta F_\beta$, where $F_\alpha=F$ and $F_\beta=X_\beta$ for $\beta \neq \alpha$, is a closed set in $X$ since $\left(\prod_\beta F_\beta\right)^c=\prod_\beta U_\beta$, where $U_\alpha=F^c$ and $U_\beta=X_\beta$ for $\beta \neq \alpha$, which is an open set because it is a base element for the product topology.

Since all $X_\beta$ are nonempty, there exists a point $\mathbf{x} \in X$ such that $x_\alpha=x$. Then $\mathbf{x} \notin \prod_\beta F_\beta$.

Now since $X$ is regular (and therefore satisfies the $T_3$ axiom), there exist disjoint open sets $U, V \subseteq X$ such that $\mathbf{x} \in U$ and $\prod_\beta F_\beta \subseteq V$.

Now for every $\beta \neq \alpha$ we have that $x_\beta \in X_\beta=\pi_\beta(V)$. However, since $x_\beta \in \pi_\beta(U)$, it follows that $\pi_\beta(U) \cap \pi_\beta(V) \neq \emptyset$.

Then $U \cap V=\emptyset$ implies that $\pi_\alpha(U) \cap \pi_\alpha(V)=\emptyset$.. Also, $x \in \pi_\alpha(U)$ and $F \subseteq \pi_\alpha(V)$ and $\pi_\alpha(U), \pi_\alpha(V)$ are open sets since $\pi_\alpha$ is an open map.

Therefore, $X_\alpha$ satisfies the $T_3$ axiom.

\end{proof}


theorem exercise_32_3 {X : Type*} [topological_space X]
  (hX : locally_compact_space X) (hX' : t2_space X) :
  regular_space X :=

Show that every locally compact Hausdorff space is regular.

\begin{proof}

    Let $X$ be a LCH space.

Then it follows that for every $x \in X$ and for every open neighborhood $U \subseteq X$ of $x$ there exists an open neighborhood $V \subseteq X$ of $x$ such that $\bar{V} \subseteq U$ (and $\bar{V}$ is compact, but this is not important here).

Since $X$ is a Hausdorff space, it satisfies the $T_1$ axiom.

Then it follows that $X$ is regular.

\end{proof}


theorem exercise_33_8
  (X : Type*) [topological_space X] [regular_space X]
  (h : ∀ x A, is_closed A ∧ ¬ x ∈ A →
  ∃ (f : X → I), continuous f ∧ f x = (1 : I) ∧ f '' A = {0})
  (A B : set X) (hA : is_closed A) (hB : is_closed B)
  (hAB : disjoint A B)
  (hAc : is_compact A) :
  ∃ (f : X → I), continuous f ∧ f '' A = {0} ∧ f '' B = {1} :=

Let $X$ be completely regular, let $A$ and $B$ be disjoint closed subsets of $X$. Show that if $A$ is compact, there is a continuous function $f \colon X \rightarrow [0, 1]$ such that $f(A) = \{0\}$ and $f(B) = \{1\}$.

\begin{proof}

    Since $X$ is completely regular $\forall a \in A, \exists f_a: X \rightarrow[0,1]: f_a(a)=0$ and $f_a(B)=\{1\}$. For some $\epsilon_a \in(0,1)$ we have that $U_a:=f_a^{-1}([0, \epsilon))$ is an open neighborhood of $a$ that does not intersect $B$. We therefore have an open covering $\left\{U_a \mid a \in A\right\}$ of $A$, so since $A$ is compact we have a finite subcover $\left\{U_{a_i} \mid 1 \leq i \leq m\right\}$. For each $1 \leq i \leq m$ define

$$

\begin{aligned}

\tilde{f}_{a_i}: X & \rightarrow[0,1] \\

x & \mapsto \frac{\max \left(f_{a_i}(x), \epsilon_{a_i}\right)-\epsilon_{a_i}}{1-\epsilon_{a_i}}

\end{aligned}

$$

so that $\forall x \in U_{a_i}: \tilde{f}_{a_i}(x)=0$ and $\forall x \in B, \forall 1 \leq i \leq m: \tilde{f}_{a_i}(x)=1$, and define $f:=$ $\prod_{i=1}^m \tilde{f}_{a_i}$. Then since $A \subset \cup_{i=1}^m U_{a_i}$ we have that $f(A)=\{0\}$ and also we have $f(B)=\{1\}$.

\end{proof}


theorem exercise_38_6 {X : Type*}
  (X : Type*) [topological_space X] [regular_space X]
  (h : ∀ x A, is_closed A ∧ ¬ x ∈ A →
  ∃ (f : X → I), continuous f ∧ f x = (1 : I) ∧ f '' A = {0}) :
  is_connected (univ : set X) ↔ is_connected (univ : set (stone_cech X)) :=

Let $X$ be completely regular. Show that $X$ is connected if and only if the Stone-Čech compactification of $X$ is connected.

\begin{proof}

    The closure of a connected set is connected, so if $X$ is connected so is $\beta(X)$

Suppose $X$ is the union of disjoint open subsets $U, V \subset X$. Define the continuous map

$$

\begin{aligned}

& f: X \rightarrow\{0,1\} \\

& x \mapsto \begin{cases}0, & x \in U \\

1, & x \in V\end{cases}

\end{aligned}

$$

By the fact that $\{0,1\}$ is compact and Hausdorff we can extend $f$ to a surjective map $\bar{f}: \beta(X) \rightarrow\{0,1\}$ such that $\bar{f}^{-1}(\{0\})$ and $\bar{f}^{-1}(\{1\})$ are disjoint open sets that cover $\beta(X)$, which makes this space not-connected.

\end{proof}


theorem exercise_1_3 {F V : Type*} [add_comm_group V] [field F]
  [module F V] {v : V} : -(-v) = v :=

import .common

open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space

open_locale big_operators

Prove that $-(-v) = v$ for every $v \in V$.

\begin{proof}

    By definition, we have

$$

(-v)+(-(-v))=0 \quad \text { and } \quad v+(-v)=0 .

$$

This implies both $v$ and $-(-v)$ are additive inverses of $-v$, by the uniqueness of additive inverse, it follows that $-(-v)=v$.

\end{proof}


theorem exercise_1_6 : ∃ U : set (ℝ × ℝ),
  (U ≠ ∅) ∧
  (∀ (u v : ℝ × ℝ), u ∈ U ∧ v ∈ U → u + v ∈ U) ∧
  (∀ (u : ℝ × ℝ), u ∈ U → -u ∈ U) ∧
  (∀ U' : submodule ℝ (ℝ × ℝ), U ≠ ↑U') :=

Give an example of a nonempty subset $U$ of $\mathbf{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \in U$ whenever $u \in U$), but $U$ is not a subspace of $\mathbf{R}^2$.

\begin{proof}

    \[U=\mathbb{Z}^2=\left\{(x, y) \in \mathbf{R}^2: x, y \text { are integers }\right\}\]

$U=\mathbb{Z}^2$ satisfies the desired properties. To come up with this, note by assumption, $U$ must be closed under addition and subtraction, so in particular, it must contain 0 . We need to find a set which fails scalar multiplication. A discrete set like $\mathbb{Z}^2$ does this.

\end{proof}


theorem exercise_1_8 {F V : Type*} [add_comm_group V] [field F]
  [module F V] {ι : Type*} (u : ι → submodule F V) :
  ∃ U : submodule F V, (⋂ (i : ι), (u i).carrier) = ↑U :=

Prove that the intersection of any collection of subspaces of $V$ is a subspace of $V$.

\begin{proof}

Let $V_1, V_2, \ldots, V_n$ be subspaces of the vector space $V$ over the field $F$. We must show that their intersection $V_1 \cap V_2 \cap \ldots \cap V_n$ is also a subspace of $V$.

To begin, we observe that the additive identity $0$ of $V$ is in $V_1 \cap V_2 \cap \ldots \cap V_n$. This is because $0$ is in each subspace $V_i$, as they are subspaces and hence contain the additive identity.

Next, we show that the intersection of subspaces is closed under addition. Let $u$ and $v$ be vectors in $V_1 \cap V_2 \cap \ldots \cap V_n$. By definition, $u$ and $v$ belong to each of the subspaces $V_i$. Since each $V_i$ is a subspace and therefore closed under addition, it follows that $u+v$ belongs to each $V_i$. Thus, $u+v$ belongs to the intersection $V_1 \cap V_2 \cap \ldots \cap V_n$.

Finally, we show that the intersection of subspaces is closed under scalar multiplication. Let $a$ be a scalar in $F$ and let $v$ be a vector in $V_1 \cap V_2 \cap \ldots \cap V_n$. Since $v$ belongs to each $V_i$, we have $av$ belongs to each $V_i$ as well, as $V_i$ are subspaces and hence closed under scalar multiplication. Therefore, $av$ belongs to the intersection $V_1 \cap V_2 \cap \ldots \cap V_n$.

Thus, we have shown that $V_1 \cap V_2 \cap \ldots \cap V_n$ is a subspace of $V$.

\end{proof}


theorem exercise_3_1 {F V : Type*}
  [add_comm_group V] [field F] [module F V] [finite_dimensional F V]
  (T : V →ₗ[F] V) (hT : finrank F V = 1) :
  ∃ c : F, ∀ v : V, T v = c • v:=

Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\operatorname{dim} V=1$ and $T \in \mathcal{L}(V, V)$, then there exists $a \in \mathbf{F}$ such that $T v=a v$ for all $v \in V$.

\begin{proof}

    If $\operatorname{dim} V=1$, then in fact, $V=\mathbf{F}$ and it is spanned by $1 \in \mathbf{F}$.

Let $T$ be a linear map from $V$ to itself. Let $T(1)=\lambda \in V(=\mathbf{F})$.

Step 2

2 of 3

Every $v \in V$ is a scalar. Therefore,

$$

\begin{aligned}

T(v) & =T(v \cdot 1) \\

& =v T(1) \ldots .(\text { By the linearity of } T) \\

& =v \lambda

\end{aligned}

$$

Hence, $T v=\lambda v$ for every $v \in V$.

\end{proof}


theorem exercise_4_4 (p : polynomial ℂ) :
  p.degree = @card (root_set p ℂ) (polynomial.root_set_fintype p ℂ) ↔
  disjoint
  (@card (root_set p.derivative ℂ) (polynomial.root_set_fintype p.derivative ℂ))
  (@card (root_set p ℂ) (polynomial.root_set_fintype p ℂ)) :=

Suppose $p \in \mathcal{P}(\mathbf{C})$ has degree $m$. Prove that $p$ has $m$ distinct roots if and only if $p$ and its derivative $p^{\prime}$ have no roots in common.

\begin{proof}

    First, let $p$ have $m$ distinct roots. Since $p$ has the degree of $m$, then this could imply that $p$ can be actually written in the form of $p(z)=c\left(z-\lambda_1\right) \ldots\left(z-\lambda_m\right)$, which you have $\lambda_1, \ldots, \lambda_m$ being distinct.

To prove that both $p$ and $p^{\prime}$ have no roots in commons, we must now show that $p^{\prime}\left(\lambda_j\right) \neq 0$ for every $j$. So, to do so, just fix $j$. The previous expression for $p$ shows that we can now write $p$ in the form of $p(z)=\left(z-\lambda_j\right) q(z)$, which $q$ is a polynomial such that $q\left(\lambda_j\right) \neq 0$.

When you differentiate both sides of the previous equation, then you would then have $p^{\prime}(z)=(z-$ $\left.\lambda_j\right) q^{\prime}(z)+q(z)$

Therefore: $\left.=p^{\prime}\left(\lambda_j\right)=q \lambda_j\right)$

Equals: $p^{\prime}\left(\lambda_j\right) \neq 0$

Now, to prove the other direction, we would now prove the contrapositive, which means that we will be proving that if $p$ has actually less than $m$ distinct roots, then both $p$ and $p^{\prime}$ have at least one root in common.

Now, for some root of $\lambda$ of $p$, we can write $p$ is in the form of $\left.p(z)=(z-\lambda)^n q(z)\right)$, which is where both $n \geq 2$ and $q$ is a polynomial. When differentiating both sides of the previous equations, we would then have $p^{\prime}(z)=(z-\lambda)^n q^{\prime}(z)+n(z-\lambda)^{n-1} q(z)$.

Therefore, $p^{\prime}(\lambda)=0$, which would make $\lambda$ is a common root of both $p$ and $p^{\prime}$.

\end{proof}


theorem exercise_5_4 {F V : Type*} [add_comm_group V] [field F]
  [module F V] (S T : V →ₗ[F] V) (hST : S ∘ T = T ∘ S) (c : F):
  map S (T - c • id).ker = (T - c • id).ker :=

Suppose that $S, T \in \mathcal{L}(V)$ are such that $S T=T S$. Prove that $\operatorname{null} (T-\lambda I)$ is invariant under $S$ for every $\lambda \in \mathbf{F}$.

\begin{proof}

    First off, fix $\lambda \in F$. Secondly, let $v \in \operatorname{null}(T-\lambda I)$. If so, then $(T-\lambda I)(S v)=T S v-\lambda S v=$ $S T v-\lambda S v=S(T v-\lambda v)=0$. Therefore, $S v \in \operatorname{null}(T-\lambda I)$ since $n u l l(T-\lambda I)$ is actually invariant under $S$.

\end{proof}


theorem exercise_5_12 {F V : Type*} [add_comm_group V] [field F]
  [module F V] {S : End F V}
  (hS : ∀ v : V, ∃ c : F, v ∈ eigenspace S c) :
  ∃ c : F, S = c • id :=

Suppose $T \in \mathcal{L}(V)$ is such that every vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator.

\begin{proof}

    For every single $v \in V$, there does exist $a_v \in F$ such that $T v=a_v v$. Since $T 0=0$, then we have to make $a_0$ be the any number in F. However, for every single $v \in V\{0\}$, then the value of $a_V$ is uniquely determined by the previous equation of $T v=a_v v$.

Now, to show that $T$ is a scalar multiple of the identity, then me must show that $a_v$ is independent of $v$ for $v \in V\{0\}$. We would now want to show that $a_v=a_w$.

First, just make the case of where $(v, w)$ is linearly dependent. Then, there does exist $b \in F$ such that $w=b v$. Now, you would have the following: $a_W w=T w=T(b v)=b T v=b\left(a_v v\right)=a_v w$. This is showing that $a_v=a_w$.

Finally, make the consideration to make $(v, w)$ be linearly independent. Now, we would have the following: $\left.a_{(} v+w\right)(v+w)=T(v+w)=T v+T w=a_v v+a_w w$.

That previous equation implies the following: $\left.\left.\left(a_{(} v+w\right)-a_v\right) v+\left(a_{(} v+w\right)-a_w\right) w=0$. Since $(v, w)$ is linearly independent, this would imply that both $\left.a_{(} v+w\right)=a_v$ and $\left.a_{(} v+w\right)=a_w$. Therefore, $a_v=a_w$.

\end{proof}


theorem exercise_5_20 {F V : Type*} [add_comm_group V] [field F]
  [module F V] [finite_dimensional F V] {S T : End F V}
  (h1 : @card T.eigenvalues (eigenvalues.fintype T) = finrank F V)
  (h2 : ∀ v : V, ∃ c : F, v ∈ eigenspace S c ↔ ∃ c : F, v ∈ eigenspace T c) :
  S * T = T * S :=

Suppose that $T \in \mathcal{L}(V)$ has $\operatorname{dim} V$ distinct eigenvalues and that $S \in \mathcal{L}(V)$ has the same eigenvectors as $T$ (not necessarily with the same eigenvalues). Prove that $S T=T S$.

\begin{proof}

    First off, let $n=\operatorname{dim} V$. so, there is a basis of $\left(v_1, \ldots, v_j\right)$ of $V$ that consist of eigenvectors of $T$. Now, let $\lambda_1, \ldots, \lambda_n$ be the corresponding eigenvalues, then we would have $T v_j=\lambda_1 v_j$ for every single $j$.

Now, for every $v_j$ is also an eigenvector of S, so $S v_j=a_j v_j$ for some $a_j \in F$. For each $j$, we would then have $(S T) v_j=S\left(T v_j\right)=\lambda_j S v_j=a_j \lambda_j v_j$ and $(T S) v_j=T\left(S v_j\right)=a_j T v_j=a_j \lambda_j v_j$. Since both operators, which are $S T$ and $T S$, agree on a basis, then both are equal.

\end{proof}


theorem exercise_6_2 {V : Type*} [add_comm_group V] [module ℂ V]
  [inner_product_space ℂ V] (u v : V) :
  ⟪u, v⟫_ℂ = 0 ↔ ∀ (a : ℂ), ‖u‖  ≤ ‖u + a • v‖ :=

Suppose $u, v \in V$. Prove that $\langle u, v\rangle=0$ if and only if $\|u\| \leq\|u+a v\|$ for all $a \in \mathbf{F}$.

\begin{proof}

    First off, let us suppose that $(u, v)=0$.

Now, let $a \in \mathbb{F}$. Next, $u, a v$ are orthogonal.

The Pythagorean theorem thus implies that

$$

\begin{aligned}

\|u+a v\|^2 & =\|u\|^2+\|a v\|^2 \\

& \geq\|u\|^2

\end{aligned}

$$

So, by taking the square roots, this will now give us $\|u\| \leq\|u+a v\|$.

Now, to prove the implication in the other direction, we must now let $\|u\| \leq$ $\|u+a v\|$ for all $a \in \mathbb{F}$. Squaring this inequality, we get both:

$$

\begin{gathered}

\|u\|^2 a n d \leq\|u+a v\|^2 \\

=(u+a v, u+a v) \\

=(u, u)+(u, a v)+(a v, u)+(a v, a v) \\

=\|u\|^2+\bar{a}(u, v)+a \overline{(u, v)}+|a|^2\|v\|^2 \\

\|u\|^2+2 \Re \bar{a}(u, v)+|a|^2\|v\|^2

\end{gathered}

$$

for all $a \in \mathbb{F}$.

Therefore,

$$

-2 \Re \bar{a}(u, v) \leq|a|^2\|v\|^2

$$

for all $a \in \mathbb{F}$. In particular, we can let $a$ equal $-t(u, v)$ for $t>0$. Substituting this value for $a$ into the inequality above gives

$$

2 t|(u, v)|^2 \leq t^2|(u, v)|^2\|v\|^2

$$

for all $t>0$.

Step 4

4 of 4

Divide both sides of the inequality above by $t$, getting

$$

2|(u, v)|^2 \leq t \mid(u, v)^2\|v\|^2

$$

for all $t>0$. If $v=0$, then $(u, v)=0$, as desired. If $v \neq 0$, set $t$ equal to $1 /\|v\|^2$ in the inequality above, getting

$$

2|(u, v)|^2 \leq|(u, v)|^2,

$$

which implies that $(u, v)=0$.

\end{proof}


theorem exercise_6_7 {V : Type*} [inner_product_space ℂ V] (u v : V) :
  ⟪u, v⟫_ℂ = (‖u + v‖^2 - ‖u - v‖^2 + I*‖u + I•v‖^2 - I*‖u-I•v‖^2) / 4 :=

Prove that if $V$ is a complex inner-product space, then $\langle u, v\rangle=\frac{\|u+v\|^{2}-\|u-v\|^{2}+\|u+i v\|^{2} i-\|u-i v\|^{2} i}{4}$ for all $u, v \in V$.

\begin{proof}

Let $V$ be an inner-product space and $u, v\in V$. Then

$$

\begin{aligned}

\|u+v\|^2 & =\langle u+v, v+v\rangle \\

& =\|u\|^2+\langle u, v\rangle+\langle v, u\rangle+\|v\|^2 \\

-\|u-v\|^2 & =-\langle u-v, u-v\rangle \\

& =-\|u\|^2+\langle u, v\rangle+\langle v, u\rangle-\|v\|^2 \\

i\|u+i v\|^2 & =i\langle u+i v, u+i v\rangle \\

& =i\|u\|^2+\langle u, v\rangle-\langle v, u\rangle+i\|v\|^2 \\

-i\|u-i v\|^2 & =-i\langle u-i v, u-i v\rangle \\

& =-i\|u\|^2+\langle u, v\rangle-\langle v, u\rangle-i\|v\|^2 .

\end{aligned}

$$

Thus $\left(\|u+v\|^2\right)-\|u-v\|^2+\left(i\|u+i v\|^2\right)-i\|u-i v\|^2=4\langle u, v\rangle.$

\end{proof}


theorem exercise_6_16 {K V : Type*} [is_R_or_C K] [inner_product_space K V]
  {U : submodule K V} :
  U.orthogonal = ⊥  ↔ U = ⊤ :=

Suppose $U$ is a subspace of $V$. Prove that $U^{\perp}=\{0\}$ if and only if $U=V$

\begin{proof}

    $V=U \bigoplus U^{\perp}$, therefore $U^\perp = \{0\}$ iff $U=V$.

\end{proof}


theorem exercise_7_6 {V : Type*} [inner_product_space ℂ V]
  [finite_dimensional ℂ V] (T : End ℂ V)
  (hT : T * T.adjoint = T.adjoint * T) :
  T.range = T.adjoint.range :=

Prove that if $T \in \mathcal{L}(V)$ is normal, then $\operatorname{range} T=\operatorname{range} T^{*}.$

\begin{proof}

    Let $T \in \mathcal{L}(V)$ to be a normal operator.

Suppose $u \in \operatorname{null} T$. Then, by $7.20$,

$$

0=\|T u\|=\left\|T^* u\right\|,

$$

which implies that $u \in \operatorname{null} T^*$.

Hence

$$

\operatorname{null} T=\operatorname{null} T^*

$$

because $\left(T^*\right)^*=T$ and the same argument can be repeated.

Now we have

$$

\begin{aligned}

\text { range } T & =\left(\text { null } T^*\right)^{\perp} \\

& =(\text { null } T)^{\perp} \\

& =\operatorname{range} T^*,

\end{aligned}

$$

where the first and last equality follow from items (d) and (b) of 7.7.

Hence, range $T=$ range $T^*$.

\end{proof}


theorem exercise_7_10 {V : Type*} [inner_product_space ℂ V]
  [finite_dimensional ℂ V] (T : End ℂ V)
  (hT : T * T.adjoint = T.adjoint * T) (hT1 : T^9 = T^8) :
  is_self_adjoint T ∧ T^2 = T :=

Suppose $V$ is a complex inner-product space and $T \in \mathcal{L}(V)$ is a normal operator such that $T^{9}=T^{8}$. Prove that $T$ is self-adjoint and $T^{2}=T$.

\begin{proof}

    Based on the complex spectral theorem, there is an orthonormal basis of $\left(e_1, \ldots, e_n\right)$ of $V$ consisting of eigenvectors of $T$. Now, let $\lambda_1, \ldots, \lambda_n$ be the corresponding eigenvalues. Therefore,

$$

T e_1=\lambda_j e_j

$$

for $j=1 \ldots n$.

Next, by applying $T$ repeatedly to both sides of the equation above, we get $T^9 e_j=\left(\lambda_j\right)^9 e_j$ and rei =8ej. Thus $T^8 e_j=\left(\lambda_j\right)^8 e_j$, which implies that $\lambda_j$ equals 0 or 1 . In particular, all the eigenvalues of $T$ are real. This would then imply that $T$ is self-adjoint.

Now, by applying $T$ to both sides of the equation above, we get

$$

\begin{aligned}

T^2 e_j & =\left(\lambda_j\right)^2 e_j \\

& =\lambda_j e_j \\

& =T e_j

\end{aligned}

$$

which is where the second equality holds because $\lambda_j$ equals 0 or 1 . Because $T^2$ and $T$ agree on a basis, they must be equal.

\end{proof}


theorem exercise_7_14 {𝕜 V : Type*} [is_R_or_C 𝕜]
  [inner_product_space 𝕜 V] [finite_dimensional 𝕜 V]
  {T : End 𝕜 V} (hT : is_self_adjoint T)
  {l : 𝕜} {ε : ℝ} (he : ε > 0) : ∃ v : V, ‖v‖= 1 ∧ (‖T v - l • v‖ < ε →
  (∃ l' : T.eigenvalues, ‖l - l'‖ < ε)) :=

Suppose $T \in \mathcal{L}(V)$ is self-adjoint, $\lambda \in \mathbf{F}$, and $\epsilon>0$. Prove that if there exists $v \in V$ such that $\|v\|=1$ and $\|T v-\lambda v\|<\epsilon,$ then $T$ has an eigenvalue $\lambda^{\prime}$ such that $\left|\lambda-\lambda^{\prime}\right|<\epsilon$.

\begin{proof}

    Let $T \in \mathcal{L}(V)$ be a self-adjoint, and let $\lambda \in \mathbf{F}$ and $\epsilon>0$.

By the Spectral Theorem, there is $e_1, \ldots, e_n$ an orthonormal basis of $V$ consisting of eigenvectors of $T$ and let $\lambda_1, \ldots, \lambda_n$ denote their corresponding eigenvalues.

Choose an eigenvalue $\lambda^{\prime}$ of $T$ such that $\left|\lambda^{\prime}-\lambda\right|^2$ is minimized.

There are $a_1, \ldots, a_n \in \mathbb{F}$ such that

$$

v=a_1 e_1+\cdots+a_n e_n .

$$

Thus, we have

$$

\begin{aligned}

\epsilon^2 & >|| T v-\left.\lambda v\right|^2 \\

& =\left|\left\langle T v-\lambda v, e_1\right\rangle\right|^2+\cdots+\left|\left\langle T v-\lambda v, e_n\right\rangle\right|^2 \\

& =\left|\lambda_1 a_1-\lambda a_1\right|^2+\cdots+\left|\lambda_n a_n-\lambda a_n\right|^2 \\

& =\left|a_1\right|^2\left|\lambda_1-\lambda\right|^2+\cdots+\left|a_n\right|^2\left|\lambda_n-\lambda\right|^2 \\

& \geq\left|a_1\right|^2\left|\lambda^{\prime}-\lambda\right|^2+\cdots+\left|a_n\right|^2\left|\lambda^{\prime}-\lambda\right|^2 \\

& =\left|\lambda^{\prime}-\lambda\right|^2

\end{aligned}

$$

where the second and fifth lines follow from $6.30$ (the fifth because $\|v\|=1$ ). Now, we taking the square root.

Hence, $T$ has an eigenvalue $\lambda^{\prime}$ such that $\left|\lambda^{\prime}-\lambda\right|<\epsilon$

\end{proof}


theorem exercise_1_27 {n : ℕ} (hn : odd n) : 8 ∣ (n^2 - 1) :=

import .common

open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function

open_locale big_operators
noncomputable theory

For all odd $n$ show that $8 \mid n^{2}-1$.

\begin{proof}

    We have $n^2-1=(n+1)(n-1)$. Since $n$ is odd, both $n+1, n-1$ are even, and moreso, one of these must be divisible by 4 , as one of the two consecutive odd numbers is divisible by 4 . Thus, their product is divisible by 8 . Similarly, if 3 does not divide $n$, it must divide one of $n-1, n+1$, otherwise it wouldn't divide three consecutive integers, which is impossible. As $n$ is odd, $n+1$ is even, so $(n+1)(n-1)$ is divisible by both 2 and 3 , so it is divisible by 6 .

\end{proof}


theorem exercise_1_31  : (⟨1, 1⟩ : gaussian_int) ^ 2 ∣ 2 :=

Show that 2 is divisible by $(1+i)^{2}$ in $\mathbb{Z}[i]$.


theorem exercise_2_21 {l : ℕ → ℝ}
  (hl : ∀ p n : ℕ, p.prime → l (p^n) = log p )
  (hl1 : ∀ m : ℕ, ¬ is_prime_pow m → l m = 0) :
  l = λ n, ∑ d : divisors n, moebius (n/d) * log d  :=

Define $\wedge(n)=\log p$ if $n$ is a power of $p$ and zero otherwise. Prove that $\sum_{A \mid n} \mu(n / d) \log d$ $=\wedge(n)$.

\begin{proof}

$$

\left\{

\begin{array}{cccl}

    \land(n)& =  & \log p & \mathrm{if}\  n =p^\alpha,\ \alpha \in \mathbb{N}^*  \\

  &  = &   0 & \mathrm{otherwise }.

\end{array}

\right.

$$

Let $n = p_1^{\alpha_1}\cdots p_t^{\alpha_t}$ the decomposition of $n$ in prime factors. As $\land(d) = 0$ for all divisors of $n$, except for $d = p_j^i, i>0, j=1,\ldots t$,

\begin{align*}

\sum_{d \mid n} \land(d)&= \sum_{i=1}^{\alpha_1} \land(p_1^{i}) + \cdots+ \sum_{i=1}^{\alpha_t} \land(p_t^{i})\\

&= \alpha_1 \log p_1+\cdots + \alpha_t \log p_t\\

&= \log n

\end{align*}

By Mobius Inversion Theorem,

$$\land(n) = \sum_{d \mid n} \mu\left (\frac{n}{d}\right ) \log d.$$

\end{proof}


theorem exercise_3_1 : infinite {p : primes // p ≡ -1 [ZMOD 6]} :=

Show that there are infinitely many primes congruent to $-1$ modulo 6 .

\begin{proof}

Let $n$ any integer such that $n\geq 3$, and $N = n! -1 =   2 \times 3 \times\cdots\times n - 1 >1$.

Then $N \equiv -1 \pmod 6$. As $6k +2, 6k +3, 6k +4$ are composite for all integers $k$, every prime factor of $N$ is congruent to $1$ or $-1$ modulo $6$.  If every prime factor of $N$ was congruent to 1, then $N \equiv 1 \pmod 6$ : this is a contradiction because $-1 \not \equiv 1 \pmod 6$.  So there exists a prime factor $p$ of $N$ such that $p\equiv -1 \pmod 6$.

If $p\leq n$, then $p \mid n!$, and $p \mid N = n!-1$, so $p \mid 1$. As $p$ is prime, this is a contradiction, so $p>n$.

Conclusion :

 for any integer $n$, there exists a prime $p >n$ such that $p \equiv -1 \pmod 6$ : there are infinitely many primes congruent to $-1$ modulo $6$.

\end{proof}


theorem exercise_3_5 : ¬ ∃ x y : ℤ, 7*x^3 + 2 = y^3 :=

Show that the equation $7 x^{3}+2=y^{3}$ has no solution in integers.

\begin{proof}

    If $7x^2 + 2 = y^3,\ x,y \in \mathbb{Z}$, then $y^3 \equiv 2 \pmod 7$ (so $y \not \equiv 0 \pmod 7$)

From Fermat's Little Theorem, $y^6 \equiv 1 \pmod 7$, so $2^2 \equiv y^6 \equiv 1 \pmod 7$, which implies $7 \mid 2^2-1 = 3$ : this is a contradiction. Thus the equation $7x^2 + 2 = y^3$ has no solution in integers.

\end{proof}


theorem exercise_3_14 {p q n : ℕ} (hp0 : p.prime ∧ p > 2)
  (hq0 : q.prime ∧ q > 2) (hpq0 : p ≠ q) (hpq1 : p - 1 ∣ q - 1)
  (hn : n.gcd (p*q) = 1) :
  n^(q-1) ≡ 1 [MOD p*q] :=

Let $p$ and $q$ be distinct odd primes such that $p-1$ divides $q-1$. If $(n, p q)=1$, show that $n^{q-1} \equiv 1(p q)$.

\begin{proof}

As $n \wedge pq = 1, n\wedge p=1, n \wedge q = 1$, so from Fermat's Little Theorem

$$n^{q-1} \equiv 1 \pmod q,\qquad n^{p-1} \equiv 1 \pmod p.$$

$p-1 \mid q-1$, so there exists $k \in \mathbb{Z}$ such that $q-1 = k(p-1)$.

Thus

$$n^{q-1} = (n^{p-1})^k \equiv 1 \pmod p.$$

$p \mid n^{q-1} - 1, q \mid n^{q-1} - 1$, and $p\wedge q = 1$, so $pq \mid n^{q-1} - 1$ :

$$n^{q-1} \equiv 1 \pmod{pq}.$$

\end{proof}


theorem exercise_4_5 {p t : ℕ} (hp0 : p.prime) (hp1 : p = 4*t + 3)
  (a : zmod p) :
  is_primitive_root a p ↔ ((-a) ^ ((p-1)/2) = 1 ∧ ∀ (k : ℕ), k < (p-1)/2 → (-a)^k ≠ 1) :=

Consider a prime $p$ of the form $4 t+3$. Show that $a$ is a primitive root modulo $p$ iff $-a$ has order $(p-1) / 2$.

\begin{proof}

    Let $a$ a primitive root modulo $p$.

As $a^{p-1} \equiv 1(\bmod p), p \mid\left(a^{(p-1) / 2}-1\right)\left(a^{(p-1) / 2}+1\right)$, so $p \mid a^{(p-1) / 2}-1$ or $p \mid$ $a^{(p-1) / 2}+1$. As $a$ is a primitive root modulo $p, a^{(p-1) / 2} \not \equiv 1(\bmod p)$, so

$$

a^{(p-1) / 2} \equiv-1 \quad(\bmod p) .

$$

Hence $(-a)^{(p-1) / 2}=(-1)^{2 t+1} a^{(p-1) / 2} \equiv(-1) \times(-1)=1(\bmod p)$.

Suppose that $(-a)^n \equiv 1(\bmod p)$, with $n \in \mathbb{N}$.

Then $a^{2 n}=(-a)^{2 n} \equiv 1(\bmod p)$, so $p-1\left|2 n, \frac{p-1}{2}\right| n$.

So $-a$ has order $(p-1) / 2$ modulo $p$.

Conversely, suppose that $-a$ has order $(p-1) / 2=2 t+1$ modulo $p$. Let $2, p_1, \ldots p_k$ the prime factors of $p-1$, where $p_i$ are odd.

$a^{(p-1) / 2}=a^{2 t+1}=-(-a)^{2 t+1}=-(-a)^{(p-1) / 2} \equiv-1$, so $a^{(p-1) / 2} \not \equiv 1(\bmod 2)$.

As $p-1$ is even, $(p-1) / p_i$ is even, so $a^{(p-1) / p_i}=(-a)^{(p-1) / p_i} \not \equiv 1(\bmod p)($ since $-a$ has order $p-1)$.

So the order of $a$ is $p-1$ (see Ex. 4.8) : $a$ is a primitive root modulo $p$.

\end{proof}


theorem exercise_4_8 {p a : ℕ} (hp : odd p) :
  is_primitive_root a p ↔ (∀ q ∣ (p-1), q.prime → ¬ a^(p-1) ≡ 1 [MOD p]) :=

Let $p$ be an odd prime. Show that $a$ is a primitive root modulo $p$ iff $a^{(p-1) / q} \not \equiv 1(p)$ for all prime divisors $q$ of $p-1$.

\begin{proof}

$\bullet$ If $a$ is a primitive root, then $a^k \not \equiv 1$ for all $k, 1\leq k < p-1$, so $a^{(p-1)/q} \not \equiv 1 \pmod p$ for all prime divisors $q$ of $p - 1$.

$\bullet$ In the other direction, suppose $a^{(p-1)/q} \not \equiv 1 \pmod p$ for all prime divisors $q$ of $p - 1$.

Let $\delta$ the order of $a$, and $p-1 = q_1^{a_1}q_2^{a_2}\cdots q_k^{a_k}$ the decomposition of $p-1$ in prime factors. As $\delta \mid p-1, \delta = q_1^{b_1}p_2^{b_2}\cdots q_k^{b_k}$, with $b_i \leq a_i, i=1,2,\ldots,k$. If $b_i < a_i$ for some index $i$, then $\delta \mid (p-1)/q_i$, so $a^{(p-1)/q_i} \equiv 1 \pmod p$, which is in contradiction with the hypothesis. Thus $b_i = a_i$ for all $i$, and $\delta = q-1$ : $a$ is a primitive root modulo $p$.

\end{proof}


theorem exercise_5_13 {p x: ℤ} (hp : prime p)
  (hpx : p ∣ (x^4 - x^2 + 1)) : p ≡ 1 [ZMOD 12] :=

Show that any prime divisor of $x^{4}-x^{2}+1$ is congruent to 1 modulo 12 .


theorem exercise_5_37 {p q : ℕ} [fact(p.prime)] [fact(q.prime)] {a : ℤ}
  (ha : a < 0) (h0 : p ≡ q [ZMOD 4*a]) (h1 : ¬ ((p : ℤ) ∣ a)) :
  legendre_sym p a = legendre_sym q a :=

Show that if $a$ is negative then $p \equiv q(4 a) together with p\not | a$ imply $(a / p)=(a / q)$.


theorem exercise_18_4 {n : ℕ} (hn : ∃ x y z w : ℤ,
  x^3 + y^3 = n ∧ z^3 + w^3 = n ∧ x ≠ z ∧ x ≠ w ∧ y ≠ z ∧ y ≠ w) :
  n ≥ 1729 :=

Show that 1729 is the smallest positive integer expressible as the sum of two different integral cubes in two ways.

\begin{proof}

    Let $n=a^3+b^3$, and suppose that $\operatorname{gcd}(a, b)=1$. If a prime $p \mid a^3+b^3$, then

$$

\left(a b^{-1}\right)^3 \equiv_p-1

$$

Thus $3 \mid \frac{p-1}{2}$, that is, $p \equiv_6 1$.

If we have $n=a^3+b^3=c^3+d^3$, then we can factor $n$ as

$$

\begin{aligned}

& n=(a+b)\left(a^2-a b+b^2\right) \\

& n=(c+d)\left(c^2-c d+d^2\right)

\end{aligned}

$$

Thus we need $n$ to have atleast 3 disctinct prime factors, and so the smallest taxicab number is on the form

$$

n=(6 k+1)(12 k+1)(18 k+1)

$$

\end{proof}


theorem exercise_1_13a {f : ℂ → ℂ} (Ω : set ℂ) (a b : Ω) (h : is_open Ω)
  (hf : differentiable_on ℂ f Ω) (hc : ∃ (c : ℝ), ∀ z ∈ Ω, (f z).re = c) :
  f a = f b :=

import .common

open complex filter function interval_integral metric

open_locale big_operators
open_locale filter
open_locale topology

Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $\text{Re}(f)$ is constant, then $f$ is constant.

\begin{proof}

Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.

Since $\operatorname{Re}(f)=$ constant,

$$

\frac{\partial u}{\partial x}=0, \frac{\partial u}{\partial y}=0 .

$$

By the Cauchy-Riemann equations,

$$

\frac{\partial v}{\partial x}=-\frac{\partial u}{\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 .

$$

3

Thus $f(z)$ is constant.

\end{proof}


theorem exercise_1_13c {f : ℂ → ℂ} (Ω : set ℂ) (a b : Ω) (h : is_open Ω)
  (hf : differentiable_on ℂ f Ω) (hc : ∃ (c : ℝ), ∀ z ∈ Ω, abs (f z) = c) :
  f a = f b :=

Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $|f|$ is constant, then $f$ is constant.

\begin{proof}

Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.

We first give a mostly correct argument; the reader should pay attention to find the difficulty. Since $|f|=\sqrt{u^2+v^2}$ is constant,

$$

\left\{\begin{array}{l}

0=\frac{\partial\left(u^2+v^2\right)}{\partial x}=2 u \frac{\partial u}{\partial x}+2 v \frac{\partial v}{\partial x} . \\

0=\frac{\partial\left(u^2+v^2\right)}{\partial y}=2 u \frac{\partial u}{\partial y}+2 v \frac{\partial v}{\partial y} .

\end{array}\right.

$$

Plug in the Cauchy-Riemann equations and we get

$$

\begin{gathered}

u \frac{\partial v}{\partial y}+v \frac{\partial v}{\partial x}=0 \\

-u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}=0 \\

(1.14) \Rightarrow \frac{\partial v}{\partial x}=\frac{v}{u} \frac{\partial v}{\partial y}

\end{gathered}

$$

Plug (1.15) into (1.13) and we get

$$

\frac{u^2+v^2}{u} \frac{\partial v}{\partial y}=0 .

$$

So $u^2+v^2=0$ or $\frac{\partial v}{\partial y}=0$.

If $u^2+v^2=0$, then, since $u, v$ are real, $u=v=0$, and thus $f=0$ which is constant.

Thus we may assume $u^2+v^2$ equals a non-zero constant, and we may divide by it. We multiply both sides by $u$ and find $\frac{\partial v}{\partial y}=0$, then by (1.15), $\frac{\partial v}{\partial x}=0$, and by Cauchy-Riemann, $\frac{\partial u}{\partial x}=0$.

$$

f^{\prime}=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=0 .

$$

Thus $f$ is constant.

Why is the above only mostly a proof? The problem is we have a division by $u$, and need to make sure everything is well-defined. Specifically, we need to know that $u$ is never zero. We do have $f^{\prime}=0$ except at points where $u=0$, but we would need to investigate that a bit more.

Let's return to

$$

\left\{\begin{array}{l}

0=\frac{\partial\left(u^2+v^2\right)}{\partial x}=2 u \frac{\partial u}{\partial x}+2 v \frac{\partial v}{\partial x} . \\

0=\frac{\partial\left(u^2+v^2\right)}{\partial y}=2 u \frac{\partial u}{\partial y}+2 v \frac{\partial v}{\partial y} .

\end{array}\right.

$$

Plug in the Cauchy-Riemann equations and we get

$$

\begin{array}{r}

u \frac{\partial v}{\partial y}+v \frac{\partial v}{\partial x}=0 \\

-u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}=0 .

\end{array}

$$

We multiply the first equation $u$ and the second by $v$, and obtain

$$

\begin{aligned}

u^2 \frac{\partial v}{\partial y}+u v \frac{\partial v}{\partial x} & =0 \\

-u v \frac{\partial v}{\partial x}+v^2 \frac{\partial v}{\partial y} & =0 .

\end{aligned}

$$

Adding the two yields

$$

u^2 \frac{\partial v}{\partial y}+v^2 \frac{\partial v}{\partial y}=0,

$$

or equivalently

$$

\left(u^2+v^2\right) \frac{\partial v}{\partial y}=0 .

$$

We now argue in a similar manner as before, except now we don't have the annoying $u$ in the denominator. If $u^2+v^2=0$ then $u=v=0$, else we can divide by $u^2+v^2$ and find $\partial v / \partial y=0$. Arguing along these lines finishes the proof.

\end{proof}


theorem exercise_1_19b (z : ℂ) (hz : abs z = 1) (s : ℕ → ℂ)
    (h : s = (λ n, ∑ i in (finset.range n), i * z / i ^ 2)) :
    ∃ y, tendsto s at_top (𝓝 y) :=

Prove that the power series $\sum zn/n^2$ converges at every point of the unit circle.

\begin{proof}

    Since $\left|z^n / n^2\right|=1 / n^2$ for all $|z|=1$, then $\sum z^n / n^2$ converges at every point in the unit circle as $\sum 1 / n^2$ does ( $p$-series $p=2$.)

\end{proof}


theorem exercise_1_26
  (f F₁ F₂ : ℂ → ℂ) (Ω : set ℂ) (h1 : is_open Ω) (h2 : is_connected Ω)
  (hF₁ : differentiable_on ℂ F₁ Ω) (hF₂ : differentiable_on ℂ F₂ Ω)
  (hdF₁ : ∀ x ∈ Ω, deriv F₁ x = f x) (hdF₂ : ∀ x ∈ Ω, deriv F₂ x = f x)
  : ∃ c : ℂ, ∀ x, F₁ x = F₂ x + c :=

Suppose $f$ is continuous in a region $\Omega$. Prove that any two primitives of $f$ (if they exist) differ by a constant.

\begin{proof}

    Suppose $F_1$ adn $F_2$ are primitives of $F$. Then $(F_1-F_2)^\prime = f - f = 0$, therefore $F_1$ and $F_2$ differ by a constant.

\end{proof}


theorem exercise_2_9
  {f : ℂ → ℂ} (Ω : set ℂ) (b : metric.bounded Ω) (h : is_open Ω)
  (hf : differentiable_on ℂ f Ω) (z ∈ Ω) (hz : f z = z) (h'z : deriv f z = 1) :
  ∃ (f_lin : ℂ →L[ℂ] ℂ), ∀ x ∈ Ω, f x = f_lin x :=

Let $\Omega$ be a bounded open subset of $\mathbb{C}$, and $\varphi: \Omega \rightarrow \Omega$ a holomorphic function. Prove that if there exists a point $z_{0} \in \Omega$ such that $\varphi\left(z_{0}\right)=z_{0} \quad \text { and } \quad \varphi^{\prime}\left(z_{0}\right)=1$ then $\varphi$ is linear.

\begin{proof}

    When $\Omega$ is connected, if $\varphi$ is not linear, then there exists $n \geq 2$ and $a_n \neq 0$, such that

$$

\varphi(z)=z+a_n\left(z-z_0\right)^n+O\left(\left(z-z_0\right)^{n+1}\right) .

$$

As you have noticed, by induction, it follows that for every $k \geq 1$,

$$

\varphi^k(z)=z+k a_n\left(z-z_0\right)^n+O\left(\left(z-z_0\right)^{n+1}\right) .

$$

Let $r>0$ be such that when $\left|z-z_0\right| \leq r$, then $z \in \Omega$. Then by (1),

$$

k a_n=\frac{1}{2 \pi i} \int_{\left|z-z_0\right|=r} \frac{\varphi^k(z)}{\left(z-z_0\right)^{n+1}} d z .

$$

Since $\varphi^k(\Omega) \subset \Omega$ and since $\Omega$ is bounded, there exists $M>0$, independent of $k$, such that $\left|\varphi^k\right| \leq M$ on $\Omega$. Then by (2),

$$

k\left|a_n\right| \leq M r^{-n} .

$$

Since $k$ is arbitrary, $a_n=0$, a contradiction.

\end{proof}


theorem exercise_3_3 (a : ℝ) (ha : 0 < a) :
    tendsto (λ y, ∫ x in -y..y, real.cos x / (x ^ 2 + a ^ 2))
    at_top (𝓝 (real.pi * (real.exp (-a) / a))) :=

Show that $ \int_{-\infty}^{\infty} \frac{\cos x}{x^2 + a^2} dx = \pi \frac{e^{-a}}{a}$ for $a > 0$.

\begin{proof}

    $\cos x=\frac{e^{i x}+e^{-i x}}{2}$. changing $x \rightarrow-x$ we see that we can just integrate $e^{i x} /\left(x^2+a^2\right)$ and we'll get the same answer. Again, we use the same semicircle and part of the real line. The only pole is $x=i a$, it has order 1 and the residue at it is $\lim _{x \rightarrow i a} \frac{e^{i x}}{x^2+a^2}(x-i a)=\frac{e^{-a}}{2 i a}$, which multiplied by $2 \pi i$ gives the answer.

\end{proof}


theorem exercise_3_9 : ∫ x in 0..1, real.log (real.sin (real.pi * x)) = - real.log 2 :=

Show that $\int_0^1 \log(\sin \pi x) dx = - \log 2$.

\begin{proof}

Consider

$$

\begin{gathered}

f(z)=\log \left(1-e^{2 \pi z i}\right)=\log \left(e^{\pi z i}\left(e^{-\pi z i}-e^{\pi z i}\right)\right)=\log (-2 i)+\pi z i+\log \\

(\sin (\pi z))

\end{gathered}

$$

Then we have

$$

\begin{aligned}

\int_0^1 f(z) d z & =\log (-2 i)+\frac{i \pi}{2}+\int_0^1 \log (\sin (\pi z)) d z \\

& =\int_0^1 \log (\sin (\pi z)) d z+\log (-2 i)+\log (i) \\

& =\log (2)+\int_0^1 \log (\sin (\pi z)) d z

\end{aligned}

$$

Now it suffices to show that $\int_0^1 f(z) d z=0$. Consider the contour $C(\epsilon, R)$ (which is the contour given in your question) given by the following.

1. $C_1(\epsilon, R)$ : The vertical line along the imaginary axis from $i R$ to $i \epsilon$.

2. $C_2(\epsilon)$ : The quarter turn of radius $\epsilon$ about 0 .

3. $C_3(\epsilon)$ : Along the real axis from $(\epsilon, 1-\epsilon)$.

4. $C_4(\epsilon)$ : The quarter turn of radius $\epsilon$ about 1 .

5. $C_5(\epsilon, R)$ : The vertical line from $1+i \epsilon$ to $1+i R$.

6. $C_6(R)$ : The horizontal line from $1+i R$ to $i R$.

$f(z)$ is analytic inside the contour $C$ and hence $\oint_C f(z)=0$. This gives us

$$

\begin{aligned}

\int_{C_1(\epsilon, R)} f d z+\int_{C_2(\epsilon)} f d z+\int_{C_3(\epsilon)} f d z+\int_{C_4(\epsilon)} f d z+\int_{C_5(\epsilon, R)} f d z+\int_{C_6(R)} f d z \\

=0

\end{aligned}

$$

Now the integral along 1 cancels with the integral along 5 due to symmetry. Integrals along 2 and 4 scale as $\epsilon \log (\epsilon)$. Integral along 6 goes to 0 as $R \rightarrow \infty$. This gives us

$$

\lim _{\epsilon \rightarrow 0} \int_{C_3(\epsilon)} f d z=0

$$

which is what we need.

\end{proof}