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MATH 447: Real Variables - Homework #4

Jerich Lee

December 22, 2024

Problem.
[13.3] Let $B$ be the set of all bounded sequences $\mathbf{x} = (x_{1}, x_{2}, \ldots{})$ and define $d(\mathbf{x}, \mathbf{y})=\sup{} \{ \vert x_{j}-y_{j} \vert :j=1,2, \ldots{} \} $.
  1. Show $d$ is a metric for $B$.
Solution. We will show that $d$ is a metric for $B$ by showing that it satisfies the symmetry, non-degeneracy, and triangle inequality with respect to $B$, i.e. the set of all bounded sequences.
Proof.
  1. Symmetry: We wish to show that $d(\mathbf{x}, \mathbf{y})=d(\mathbf{y}, \mathbf{x})$. $$\begin{align} d(\mathbf{x}, \mathbf{y}) = \sup\left\{ \left\vert x_{j}-y_{j} \right\vert : j=1,2, \ldots \right\} \\[10pt] d(\mathbf{x}, \mathbf{y}) = \sup\left\{ \left\vert y_{j}-x_{j} \right\vert : j=1,2, \ldots \right\} \\[10pt] \end{align}$$ From the properties of absolute value, we know that $\left\vert a-b \right\vert = \left\vert b-a \right\vert $. Setting $a=x_{j}$ and $b=y_{j} $, we can state: $$\begin{align} \left\vert x_{j} -y_{j} \right\vert =\left\vert y_{j} -x_{j} \right\vert \end{align}$$
  2. Non-degeneracy We wish to show that $d(\mathbf{x},\mathbf{y} )=\sup \left\{ \left\vert x_{j}-y_{j} \right\vert : j=1,2, \ldots \right\}=0 $ iff $\mathbf{x} =\mathbf{y} $. $$\begin{align} \left\vert x_{j} -y_{j} \right\vert =0 \iff x_{j} =y_{j} \\[10pt] x_{j} -y_{j} =0 \end{align}$$
  3. Triangle Inequality We wish to show that $d(\mathbf{x}, \mathbf{z})\leq d(\mathbf{x},\mathbf{y})+d(\mathbf{y}, \mathbf{z})$. $$\begin{align} \text{RHS} = d(\mathbf{x}, \mathbf{z})+d(\mathbf{y}, \mathbf{zy})\\[10pt] = \sup\left\{ \left\vert x_{j}-y_{j} \right\vert : j=1,2, \ldots \right\}+ \sup\left\{ \left\vert y_{j}-z_{j} \right\vert : j=1,2, \ldots \right\} \\[10pt] \geq \sup\left\{ \left\vert x_{j}-y_{j} \right\vert +\left\vert y_{j} -z_{j} \right\vert : j=1,2, \ldots \right\}\\[10pt] \geq \sup \left\{ (x_{j} -y_{j} )+(y_{j} -z_{j} ) \right\} \\[10pt] =\sup \left\{ \left\vert x_{j} -z_{j} \right\vert \right\} =\text{LHS} \end{align}$$
Lines 8 and 9 can be shown in Homework 3 Problem 4.
∎

Problem.
[13.4] Prove (iii) and (iv) in Discussion 13.7
  1. The union of \emph{any} collection of open sets is open.
  2. The intersection of \emph{finitely many} open sets is again an open set.
Solution.
  1. Proof. Suppose we have a collection of open sets, $\{ U_k\} $. Then if $x\in{} \{ U_k\} $, then $x\in{} \cup_k{} \{ U_k \} $. Because $\{ U_k \} $ is a collection of open sets, $\cup_k{} \{ U_k \} $ is also itself an open set.
    ∎

  2. Proof. Suppose we have a collection of finitely open sets, $\{ U_k \}$. Then, for each element $x$ in the intersection $\cap_k{} U_k$, there exists a neighborhood $N_k$ about the element $x$ for each open set $U_{ k=1,2, \ldots{} k} $ such that $N_k\subset{} U_k, k=1,2, \ldots{} k $. For each $N_k$, there exists a radius $r_k$. We can choose the smallest radius out of all neighborhoods $N_k$ such that $r_{\mathop{\min}}=\mathop{\min} \{ N_1,N_{2}, \ldots{} N_k \} $. Then, this neighborhood is a subset of all open sets $U_k$, $N_{k_{\mathop{\min}(r) } }\subset{} \{U_k \} $. Therefore, the intersection $\cap_k{} U_k$ is open for a finite collection of open sets.
    ∎

Problem.
[13.10] Show that the interior of each of the following sets is the empty set.
  1. $\left\{ \frac{1}{n}:n\in\mathbb{{N}} \right\} $
Solution.
Proof. The definition of the interior point of a set $E$ is a point that contains at least one neighborhood such that $N\subset{} E$. The interior of a set $E$ is a set that contains only interior points. A set $E$ is called open if the set $E$ is equal to its interior. Therefore, to show that the interior of the set $E=\{ \frac{1}{n}:n\in{} \mathbb{{N}} \} $ is the empty set, we must show that there exists a point in $E$ such that its neighborhood contains elements that are included in the set $\{ \frac{1}{n}:n\in{}\mathbb{{N}} \}$. We will show that for any point in $E$, there is no neighborhood of $s_{o} $ such that $B_{r}^{o}(s_{o} )\subset{} E $ for all $r$. We will first show that between any two rational numbers, there exists an irrational number. $$\begin{align} 0<\frac{1}{\sqrt{2} }<1 \\[10pt] r_{1} + 0 < r_{1} + \frac{1}{\sqrt{2} }(r_{1} -r_{2} )<r_{1} +(r_{2} -r_{1} )\\[10pt] r_{1} <r_{1} +\frac{1}{\sqrt{2} }(r_{2} -r_{1} )<r_{2} \end{align}$$ Suppose that the radius of any neighborhood about any point in $E$ is a rational number. Then, by above, there exists members of $B_{r}^{o}(s_{o} )$ that are irrational, and $\notin{} E $, as $E$ contains only rational numbers. Now suppose that we set our radius $r$ to be an irrational number. We know that between any rational and irrational number exists an irrational number. Let $A$ be a rational number, and $C$ to be an irrational number. Then, $B=\frac{A+C}{2}$ is irrational, and $A<B<C$. Therefore, we have shown that there is no value of radius $r$ that satisfies $B_{r}^{o}(s_{o} )\subset{} E$ for any rational point $s_{o} $ in $E$.
∎

Problem.
[13.12] Let $(S,d)$ be any metric space.
  1. Show that the finite union of compact sets in $S$ is compact.
Solution.
Proof. To show that the finite union of compact sets in $S$ is compact, we will use the property of compactness of each set $U_k$ in the finite union $\cup_k{} \{ U_k\} $. If a set $U_k$ is compact, then there exists a finite subcover for every open cover of $U_k$. Then, there exists $\cup_k{} U_k$ for finitely many $k$.
∎

Problem.
Show that a sequence in a metric space $(S,d)$ cannot have more than one limit.
Solution. We will show that if a sequence $s_{n} $ converges to $L_{1} $ and $L_{2} $, $\forall{} \varepsilon{} >0$, $\exists{} N_{1}\ s.t. \ m,n >N \implies{} \vert s_{n} -L_{1} \vert <\frac{\varepsilon}{2}$ and $\exists{} N_{2} \ s.t. \ n>N_{2} \implies{} \vert s_{n} -L_{2} \vert <\frac{\varepsilon}{2}$. Choose $N_{\text{max} }=\text{max}(N_{1},N_{2} ) $. Then, $n>N_{\text{max} } \implies{} $ $$\begin{align} d\left( s_{n} -L_{1} + (L_{2}-s_{n}) \right) < d\left( \frac{\varepsilon}{2} \right) + d\left( \frac{\varepsilon}{2} \right)< \varepsilon \\[10pt] = d\left( \frac{\varepsilon}{2} \right) <\varepsilon \end{align}$$ This implies that $L_{1} =L_{2} $, so there cannot be more than one limit in $(S,d)$.
Problem.
  1. Suppose a sequence $s_{n} $ in a metric space $(S,d)$ converges to $s\in S$. Prove that any subsequence of $s_{n} $ converges to $s$ as well.
Solution. Let $s_{n} $ be a sequence in $S$. By the given, $\forall{} \varepsilon{} >0,\exists{} N\ s.t. \ n>N\implies{} d(s_{n}-s)$. Let $s_{n_{k}} $ be a subsequence of $s_{n} $. We know that $n_{k} >k$ for all $k$. Now let $s=\lim_{n \to \infty} s_{n} $ and let $\varepsilon{} >0$. There exists $N$ such that $n>N$ implies $d(s_{n} -s)<\varepsilon{} $. Now $k>N$ implies $n_{k} >N$, implying $d(s_{n_{k} }-s )<\varepsilon{} $. Therefore, $$\begin{align} \lim_{k \to \infty} s_{n_{k} }=s \end{align}$$
Problem.
Suppose $(S,d)$ is a complete metric space, and $E \subset{} S$. We can view $E$ as a metric space, equipped with the metric inherited from $S$. Prove that $E$ is complete iff it is a closed subset of $S$.
Solution. $\implies{} :$ Suppose that $E$ is not complete. Then, $E$ does not contain its limit points. By definition, a closed set is a set that contains all of its limit points. Therefore, $E$ cannot be closed. $\impliedby{} :$ Suppose that $E$ is not a closed subset of $S$. Then, $E$ does not contain all of its limit points. Suppose that $E$ is complete. Then, every Cauchy sequence in $E$ converges to some limit point in $E$.
Problem.
Suppose $s_{n} $ is a Cauchy sequence in a metric space $(S,d)$ which has a convergent subsequence. Is it true that the sequence $s_{n} $ itself converges?
Solution. We know that the following holds: $$\begin{align} \forall \varepsilon >0, \exists N\ s.t. \ m,n>N\implies d(s_{m} -s_{n} )<\varepsilon \end{align}$$ Want: If $s_{n} $ has a convergent subsequence, then $L\in{} S$, $d(s_{n_{k} }-L )<\varepsilon{}\implies{} d(s_{n} -L)<\varepsilon{} $. Choose $N$ such that $n>N\implies{} d(s_{n_{k} } -L)<\frac{\varepsilon}{2}$. Choose $N_{1} $ such that $d(s_{n_{k+1} }-s_{n_{k} } )<\frac{\varepsilon}{2}$ Then we get: $$\begin{align} d(s_{n_{k+1} }-s_{n_{k} }+s_{n_{k} }-L )<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}<\varepsilon \\[10pt] d(s_{n_{k+1} }-L )<\varepsilon \end{align}$$
Problem.
Is the metric space $(B,d)$ defined in Problem 13.3(a) complete?
Solution. If $B$ is the set of all bounded sequences $\mathbf{x} = (x_{1}, x_{2}, \ldots{}, x_n )$, endowed with the distance function $d(\mathbf{x}, \mathbf{y})=\sup{} \{ \vert x_i - y_{i} \vert : i=1,2, \ldots{} n \}$, then the metric space $(B,d)$ is complete.
Proof. We wish to find a proof that $(B, d)$ such that $B$ is the set of all bounded sequences, and $d$ is the distance defined as: $$\begin{align} d(\mathbf{x}, \mathbf{y})=\sup \left\{ \left\vert x_i - y_{i} \right\vert : i=1,2, \ldots n \right\} \end{align}$$ The idea is to treat the set $B$ of sequences as a set of infinite dimensional vectors in $\mathbb{{R}} $. We will utilize and prove the following lemma:
Lemma.
Consider a sequence $(\mathbf{x}^{k})_k$ in $\mathbb{{R}}^n$, with $\mathbf{x}^{k}=(x_i^{k})^{n}_{i=1}$
  1. $(\textbf{x}^{(k)})_{k}$ is Cauchy iff $(\textbf{x}^{k})_{k}$ is Cauchy for $1 \leq i \leq n $
  2. $(\textbf{x}^{(k)})_{k}$ converges to $\mathbf{x} = (x_{i})^{n}_{i=1}$ iff $\lim_{k \to \infty} x_{i}^{(k)}$ for $1 \leq i \leq n $
$\implies{} :$ To prove part one of the lemma above, we will suppose that $(<strong>x</strong>^{k})_{k}$ is Cauchy. Fix $i$. Want: $(x_{i}^{(k)})_{k}$ is Cauchy. For $\varepsilon{} > 0$ we need to find $N\ s.t. \ \vert x_{i}^{(k)} - x_{i}^{(m)} \vert <\varepsilon{} $ for $k, m>N$. We need to find an $N\ s.t. \ d(\mathbf{x}^{k}, \mathbf{x}^{m})<\varepsilon{}$ for all $k,m>N$. $\varepsilon{} > d(\mathbf{x}^{(k)}, \mathbf{x}^{(m)})=\sup{} \{ \vert x_{i}- y_{i}\vert : i= 1,2, \ldots{} , n \} \geq{} \vert x_{i}-y_{i} \vert $. Therefore, this $N$ works in the forward direction. $\impliedby{} :$ Suppose $(x_{i}^{(k)})_{k}$ is Cauchy $\forall{} \ i$. We want to find: $(\mathbf{x}^{(k)})_{k}$ is Cauchy. We fix $\varepsilon{} >0$. We need to find $N\ s.t. \ d(\mathbf{x}^{k}, \mathbf{x}^{m})<\varepsilon{} $ for $k, m>N$. For $1\leq{} i\leq{} n$ find $N_{i} \in{} \mathbb{{N}} \ s.t. \ \vert x_{i}^{(k)}- x_{i}^{(m)} \vert < \varepsilon{}$ for $k,m>N_{i} $. Then, we know that: $$\begin{align} x_{i}^{(k)} <\varepsilon + x_{i}^{(m)} \\[10pt] x_{i}^{(m)} < \varepsilon + x_{i}^{(k)} \end{align}$$ Therefore, by the properties of the least upper bound, we have the following: $$\begin{align} \sup x_{i}^{(k)} \leq \varepsilon + x_{i}^{(m)}\\[10pt] \sup x_{i}^{(m)} \leq \varepsilon + x_{i}^{(k)}\\[10pt] \end{align}$$ We can prove a quick fact proposition about differences of supremums of sequences:
Proposition.
$$\begin{align} \sup (s_{n} -t_{n} )\leq \sup s_{n} - t_{n} \\[10pt] \end{align}$$
Proof. $$\begin{align} s_{n} \leq \sup s_{n} \\[10pt] t_{n} \leq \sup t_{n} \\[10pt] s_{n} - t_{n} \leq \sup s_{n} -\sup t_{n} \\[10pt] \sup (s_{n} -t_{n} ) \leq \sup s_{n} -\sup t_{n} \end{align}$$
∎

Therefore, we can state the following: $$\begin{align} \sup (x_{i}^{(k)} -x_{i}^{(m)})\leq \sup x_{i}^{(k)}-\sup x_{i}^{(m)}\\[10pt] \leq (x_{i}^{(m)} +\varepsilon )-(x_{i}^{(k)} +\varepsilon ) \end{align}$$ We know from the implication that the following is true: $$\begin{align} \left\vert x_{i}^{(k)}- x_{i}^{(m)} \right\vert< \varepsilon \\[10pt] \sup (x_{i}^{(k)}-x_{i}^{(m)} )<\varepsilon \end{align}$$ Therefore, we can say the following: $$\begin{align} \sup \left( \left\vert x_{i}^{(k)}-x_{i}^{(m)} \right\vert\right) <\varepsilon \end{align}$$ To show the second part of Lemma 1, we can use similar reasoning as the first part of Lemma 1 to show that the limit exists $B$.
∎