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.