26th March 2019

# Homework set 4

## Question 1

Let $$A, B\subset \mathbb{R}$$ be compact. We show that $$A\times B$$ the product space is sequentially compact in $$\mathbb{R}^2$$. Consider any sequence $$((a_i, b_i))_{i \in \mathbb{N}}$$. The projected sequence $$(a_i)_{i\in\mathbb{N}}$$ has a convergent subsequence because $$A$$ is a compact set, so call it $$(a_{i_k})_{k\in\mathbb{N}}$$. Now we have a subsequence $$((a_{i_k}, b_{i_k}))_{k\in\mathbb{N}}$$, the next step is to project its second coordinate, so we consider the sequence $$(b_{i_k})_{k\in\mathbb{N}}$$ in $$B$$. Now because $$B$$ is compact, $$(b_{i_k})_{k\in\mathbb{N}}$$ has a further convergent subsequence $$(b_{i_{k_l}})_{l\in\mathbb{N}}$$. Finally we can see that $$((a_{i_{k_l}}, b_{i_{k_l}}))_{l\in\mathbb{N}}$$ is a convergent subsequence of $$A\times B$$. $$\square$$

## Question 2

Fix $$x\in X$$, we have $$d(x,A) := \inf_{y\in A} d(x,y)$$. Either we already have $$y^*\in A$$ such that $$d(x,A) = d(x,y^*)$$ in which case we are done, or we have a sequence of points $$(y_n)_{n\in\mathbb{N}}$$ such that $$\lim_{n\to\infty} d(x,y_n) = d(x, A)$$. By compactness of $$A$$, we find a subsequence $$n_1 < n_2 < \dots$$ such that $$y_{n_k} \to y^*$$ for some $$y^*\in A$$. Claim is that $$d(x,y^*) = d(x, A)$$. Note that because $$y^*\in A$$, it is already established that $$d(x,y^*) \geq d(x,A)$$. For each $$k\in\mathbb{N}$$, by triangle inequality $d(x,y^*) \leq d(x, y_{n_k}) + d(y_{n_k}, y^*)$ taking limit as $$k\to\infty$$ we obtain $d(x,y^*) \leq d(x, A) + 0$ which shows that $$d(x,y^*) = d(x, A)$$.$$\square$$

## Question 3

Let $$A,B \in \mathcal{H}(X)$$ and we can define the Hausdorff distance as $d_H(A,B) := \max{\left\{ \sup_{x\in A} d(x,B), \sup_{y\in B} d(y,A) \right\}}.$

By reading off the definition we can see that it’s symmetric, so $$d_H(A,B) = d_H(B,A)$$.

The property that $$d_H(A,B)\geq 0$$ is inherited from that of $$d(-,-)$$.

Now we show $$A = B$$ iff $$d_H(A,B) = 0$$. Suppose $$A = B$$, then it is easy to see that for any $$x\in A$$ $$d(x,A) = \inf_{y\in A} d(x,y) = d(x,x) = 0$$. This means $$d_H(A,B) = \max{\left\{ 0,0 \right\}} = 0$$. Conversely assume $$d_H(A,B) = 0$$. By positiveness of $$d(-,-)$$, we have $\sup_{x\in A} d(x,B) = \sup_{y\in B} d(y, A) = 0$ again using positiveness we can deduce that $\forall x\in A.~ d(x,B) = 0 \text{ and } \forall y\in B.~ d(y,A) = 0.$ Now we have enough results to prove the equality. Let $$x\in A$$, then $$d(x,B) = \inf_{y\in B} d(x,y) = 0$$. Use Q2 to get existence of $$y^* \in B$$ such that $$d(x,y^*) = 0$$. By definiteness of $$d(-,-)$$ we have $$x = y^*$$ which shows $$x\in B$$. Therefore $$A\subset B$$. The proof of the reverse containment is the same, so $$A = B$$.

Note that this property breaks down if $$A$$ or $$B$$ is not compact. Let $$A$$ be the open unit ball in $$R^k$$ and $$B$$ be the closure of $$A$$. We have for any $$x\in A$$, $$d(x, B) = 0$$ because $$A\subset B$$. Also for any $$x\in B$$, $$d(x, A) = \inf_{y\in B} d(x,y) = 0$$ in both cases where $$x\in A$$ and $$x$$ is a limit point of $$A$$. Then in this example $$d_H(A,B) = 0$$ but by construction $$A \ne B$$.

Now we show the triangle inequality holds. Let $$A,B,C\subset X$$ be nonempty and bounded so that their distance is real. Note that compactness was not assumed. We want to show that $d_H(A,B) \leq d_H(A,C) + d_H(C,B)$ although $$d_H(A,B) := \max{\left\{ \sup_{x\in A} d(x,B), \sup_{y\in B} d(y,A) \right\}}$$, we assume without loss of generality that we are in the case $$d_H(A,B) = \sup_{x\in A} d(x,B)$$, as interchanging role of $$A,B$$ will cover the other case. Let $$a\in A$$ be arbitrary, we have $d(a,B) = \inf_{b\in B} d(a,b)$ for all $$c\in C$$, by triangle inequality each $$d(a,b) \leq d(a,c) + d(c,b)$$, so for all $$c\in C$$, \begin{align*} d(a,B) &\leq \inf_{b\in B} ( d(a,c) + d(c,b) ) \\ &= d(a,c) + \inf_{b\in B} d(c,b) \\ &= d(a,c) + d(c,B) \\ &\leq d(a,c) + d(C,B) \end{align*} since this bound holds for all $$c\in C$$, \begin{align*} d(a,B) &\leq (\inf_{c\in C} d(a,c) + d(C,B)) \\ &= \inf_{c\in C} d(a,c) + d(C,B) \\ &= d(a,C) + d(C,B) \\ &\leq d(A,C) + d(C,B) \end{align*} Since $$a\in A$$ was arbitrary, $\sup_{a\in A} d(a,B) = d(A,B) \leq d(A,C) + d(C,B). \tag*{\square}$ As compactness was not invoked in the proof triangle inequality holds as long as $$d_H$$ between the sets are defined.

# Homework set 5

## Rudin Chapter 3 Q1

We can assume the sequence is complex. Suppose $$(s_n)$$ converges, to see that $$(\left\lvert s_n \right\rvert)$$ converges, let $$\varepsilon>0$$, by convergence of $$(s_n)$$ there exists $$s\in\mathbb{C}, N\in\mathbb{N}$$ such that $n\geq N\implies \left\lvert s_n - s \right\rvert < \varepsilon.$ We now claim that $$(\left\lvert s_n \right\rvert) \to \left\lvert s \right\rvert$$, to see why, note that whenever $$n\in\mathbb{N}, n > N$$, we have by Chapter 1 Q13 (from Homework 2) $\left\lvert \left\lvert s_n \right\rvert - \left\lvert s \right\rvert \right\rvert \leq \left\lvert s_n - s \right\rvert < \varepsilon. \tag*{\square}$

The converse can be seen to be false if we consider the sequence $$(s_n)$$ where for each $$n\in\mathbb{N}$$, $$s_n = (-1)^n$$. In this case $$\left\lvert s_n \right\rvert \to 1$$ but $$s_n$$ oscillates.

## Rudin Chapter 3 Q20

Suppose $$(p_n)_{n\in\mathbb{N}}$$ is a Cauchy sequence in a metric space $$X$$ and we have a subsequence $$n_1 < n_2 < \dots$$ such that $$(p_{n_i})_{i\in\mathbb{N}} \to p$$ in $$X$$. We need to show that the original sequence converges to $$p$$, so fix $$\varepsilon>0$$. By convergence of the subsequence, we can find $$M_0\in\mathbb{N}$$ such that $m\geq M_0 \implies d(p_{n_m}, p) < \varepsilon/5.$ Also because the sequence $$(p_n)$$ itself is Cauchy, we can find $$M_1\in \mathbb{N}$$ such that $n,m\geq M_1 \implies d(p_n, p_m) < \varepsilon/5.$ Now let $$N := \max(n_{M_0}, M_1)$$, whenever $$n\in\mathbb{N}, n\geq N$$, we can define $$k$$ to be the smallest number such that $$n_k \geq n$$. Note that by our choice of $$N$$, $$k \geq M_0$$ and both $$n_k, n \geq M_1$$, then \begin{align*} d(p_n, p) &\leq d(p_n, p_{n_k}) + d(p_{n_k}, p) \\ &< \frac{\varepsilon}5 + \frac{\varepsilon}5 < \varepsilon\tag*{\square} \end{align*}

## Rudin Chapter 3 Q21

For each $$n\in\mathbb{N}$$ we choose a point $$p_n \in E_n$$. We first claim that $$(p_n)_{n\in\mathbb{N}}$$ is Cauchy. For any $$\varepsilon>0$$, since $$\lim_{n\to\infty}\operatorname{diam}E_n = 0$$, find large enough $$N\in\mathbb{N}$$ such that $$\operatorname{diam}E_N < \varepsilon$$, then since we have a descending chain of sets, for any $$m,n> N$$, $$p_m, p_n\in E_N$$, so $$d(p_m, p_n) \leq \operatorname{diam}E_N < \varepsilon$$. Since our metric space $$X$$ is complete, let $$p\in X$$ such that $$(p_n)_{n\in\mathbb{N}} \to p$$. The immediate observation we can make here is that $$p \in \bigcap E_n$$. This is because for each $$n\in\mathbb{N}$$, $$E_n$$ is closed and $$p_{n+1}, p_{n+2}, \dots$$ is a sequence of points converging at $$p$$, so $$p$$ is a limit point of $$E_n$$ and $$p\in E_n$$.

Now we show that no other point $$p'$$ can exist in the intersection. Suppose $$p' \ne p$$ lies in the intersection, then we have a contradiction as by choosing a large enough $$M$$ we can have $$E_M$$ such that $$\operatorname{diam}E_M < d(p',p)$$, a contradiction. $$\square$$

## Q

Use the definition of Cauchy sequences to prove that if $$(x_n)_{n\in\mathbb{N}}$$ and$$(y_n)_{n\in\mathbb{N}}$$ are two Cauchy sequences of real numbers, then $$(x_ny_n)_{n\in\mathbb{N}}$$ is also a Cauchy sequence.

We first note that the Cauchy sequences are convergent, thus bounded. Let $$X>0$$ be a bound for $$(x_n)_{n\in\mathbb{N}}$$ and $$Y>0$$ a bound for $$(y_n)_{n\in\mathbb{N}}$$, so for any $$n\in\mathbb{N}$$ we have $\left\lvert x_n \right\rvert < X, \left\lvert y_n \right\rvert < Y.$ To show that $$(x_ny_n)_{n\in\mathbb{N}}$$ is a Cauchy sequence, first fix any $$\varepsilon>0$$, we want to find $$N\in\mathbb{N}$$ such that $m,n>N \implies \left\lvert x_ny_n - x_my_m \right\rvert < \varepsilon. \tag{5.1}$ By Cauchy criterion applied to $$(x_n)_{n\in\mathbb{N}}$$ and $$(y_n)_{n\in\mathbb{N}}$$, we get existence of $$N_1, N_2\in\mathbb{N}$$ such that \begin{align*} m,n>N_1 &\implies \left\lvert x_n - x_m \right\rvert < \frac{\varepsilon}{2X} \\ m,n>N_2 &\implies \left\lvert y_n - y_m \right\rvert < \frac{\varepsilon}{2Y}. \end{align*} Choose $$N = \max(N_1, N_2)$$, then for any $$m,n>N$$, \begin{align*} \left\lvert x_ny_n - x_my_m \right\rvert &= \left\lvert (x_ny_n - x_ny_m) + (x_ny_m - x_my_m) \right\rvert \\ &\leq \left\lvert x_n(y_n - y_m) \right\rvert + \left\lvert y_m(x_n - x_m) \right\rvert \\ &= \left\lvert x_n \right\rvert\left\lvert y_n - y_m \right\rvert + \left\lvert y_m \right\rvert\left\lvert x_n - x_m \right\rvert \\ &< X\frac{\varepsilon}{2X} + Y\frac{\varepsilon}{2Y} \\ &= \varepsilon\tag*{\square} \end{align*}

# Homework set 6

## Rudin Chapter 3 Q2

Observe that \begin{align*} \sqrt{n^2 + n} - n &= \frac{ n^2 + n - n^2 }{ \sqrt{n^2 + n} + n } \\ &= \frac{ 1 }{ \frac{\sqrt{n^2 + n}}n + 1 } \\ &= \frac{ 1 }{ \sqrt{1 + \frac1n} + 1 } \\ \end{align*} so the limit as $$n\to\infty$$ is $$\frac12$$. $$\square$$

## Question A

Define $S := {\left\{ f(x,y): x\in X, y\in Y \right\}}.$ We show that $\inf_{x\in X} \inf_{y\in Y} f(x,y) = \inf S.$ We can see that $S_1 := \inf_{x\in X} (\inf_{y\in Y} f(x,y)) \geq \inf S.$ To show the reverse inequality, we show that for all $$x\in X, y\in Y$$, $$f(x,y) \geq S_1$$, this is true because $f(x,y) \geq \inf_{y'\in Y} f(x,y') \geq \inf_{x'\in X} \inf_{y'\in Y} f(x',y') = S_1.$ A similar argument proves $$S = S_2 := \inf_{y\in Y}\inf_{x\in X} f(x,y)$$. $$\square$$

In general we cannot interchange the order of $$\liminf_{m\to\infty}$$ and $$\liminf_{n\to\infty}$$. If the limit exists, then $$\liminf$$ and limit are equal, so a counterexample similar to the tutorial question, like $$\left(1/n\right)^{1/m}$$ can be applied.

## Question B

Given a real sequence $$(x_n)_{n\in\mathbb{N}}$$. Suppose $$x = \liminf_{n\to\infty} x_n$$. By definition, $$x = \inf E$$ where $$E$$ is the set of subsequential limits. Since $$E$$ is closed we see that $$x$$ is also a subsequential limit of $$(x_n)_{n\in\mathbb{N}}$$. Now let $$\varepsilon>0$$, we want to show that there exists $$N$$ such that for all $$n>N$$, $$x_n > x - \varepsilon$$. Suppose not, then we have a number $$\varepsilon_0 > 0$$ such that for all $$N$$, there exists $$n>N$$ such that $$x_n \leq x - \varepsilon_0$$. This property will allow us to create a subsequence bounded above by $$x -\varepsilon_0$$, which contradicts the definition.

Conversely suppose $$x$$ is a subsequential limit of $$(x_n)_{n\in\mathbb{N}}$$ and it has the property $\forall\varepsilon>0.~ \exists N\in\mathbb{N}.~ \forall n > N.~ x_n > x - \varepsilon.$ We want to show that $$x = \inf E$$. Since $$x\in E$$ already we just have to show that there cannot be an $$x' < x$$ with $$x' \in E$$. Suppose not, so we have a subsequence $$n_1 < n_2 < \dots$$ such that $$(x_{n_i})_{i\in\mathbb{N}} \to x'$$. Because it converges, we can just consider the tail of the subsequence, so without loss of generality $$\left\lvert x_{n_1} - x' \right\rvert < r$$ where $$r = (x - x')/2$$. Now we see that the property of $$x$$ is violated, consider the number $$r$$, for all $$N > n_1$$, we can find the next term in the subsequence $$x_{n_k}$$ where $$n_k > N$$ and $\left\lvert x_{n_k} - x' \right\rvert < r$ so in particular $x_{n_k} - x' < (x - x')/2$ so $$x_{n_k} < (x + x')/2 = x - r$$ which is a contradiction. $$\square$$

## Question C

Suppose $$(x_n)_{n\in\mathbb{N}}\to x$$, where $$x \in \overline{\mathbb{R}}$$. Case $$x = \infty$$, We want to show that $$(y_n) \to \infty$$. Fix any real $$M>0$$, there is a $$N\in\mathbb{N}$$ such that $n > N \implies x_n \geq 4M.$ Now we consider the sequence $z_n = \frac\alpha{n} \text{ where } \alpha = x_1 + \dots + x_N$ which clearly converges to $$0$$. By convergence let $$L\in\mathbb{N}$$ such that $n > L \implies \left\lvert z_n \right\rvert < M$

So for all sufficiently large $$n$$ ($$n > \max(2N, L)$$), we have \begin{align*} y_n &= \frac{x_1 + x_2 + \dots + x_n}{n} \\ &= \frac{x_1 + \dots + x_N}{n} + \frac{x_{N+1} + \dots + x_n}{n} \\ &\geq \frac{x_1 + \dots + x_N}{n} + \frac{(n-N)(4M)}{n} \\ &\geq \frac{x_1 + \dots + x_N}{n} + \frac{n(4M)}{2} \\ &= \frac{x_1 + \dots + x_N}{n} + 2nM \\ &> -M + 2nM = (2n-1)M \geq M \end{align*} The case where $$x = -\infty$$ is symmetric but similar.

Case $$x\in\mathbb{R}$$, we show $$y_n \to x$$. Fix any $$\varepsilon>0$$ and by convergence of $$x_n$$, we can find $$N\in\mathbb{N}$$ such that $n>N \implies \left\lvert x_n - x \right\rvert < \varepsilon/2.$

Again we consider the sequence $z'_n = \frac{\alpha'}{n} \text{ where } \alpha' = (x_1 - x) + (x_2 - x) + \dots + (x_N - x).$ Similarly let $$L'\in\mathbb{N}$$ such that $n > L' \implies \left\lvert z'_n \right\rvert < \varepsilon.$

Then for any sufficiently large $$n$$ ($$n > \max(N,L)$$) \begin{align*} \left\lvert y_n - x \right\rvert &= \left\lvert \frac{x_1 + \dots + x_N + x_{N+1} + \dots + x_n}{n} - \frac{\overbrace{x + \dots + x}^{n \text{copies}}}{n} \right\rvert \\ &= \left\lvert \frac{(x_1-x) + \dots + (x_N-x) + (x_{N+1}-x) + \dots + (x_n-x)}n \right\rvert \\ &< \frac{\varepsilon}2 + \left\lvert \frac{(x_{N+1}-x) + \dots + (x_n-x)}n \right\rvert \\ &\leq \frac{\varepsilon}2 + \left\lvert \frac{(x_{N+1}-x) + \dots + (x_n-x)}n \right\rvert \\ &< \frac{\varepsilon}2 + \frac{(n-N)(\varepsilon/2)}{n} \\ &\leq \frac{\varepsilon}2 + \frac{n\varepsilon}{2n} \\ &\leq \frac{\varepsilon}2 + \frac{\varepsilon}2 = \varepsilon \tag*{\square} \end{align*}

As for the counter example, consider $$x_n = (-1)^n$$ which is not convergent, but the Cesaro average of this sequence converges to 0.