Metric and Hilbert Spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 November 2014

Assignment 1 Solutions

1. Let $A$ and $B$ be bounded subsets of a metric space $(X,d)$ such that $A\cap B\ne \varnothing \text{.}$ Show that $$\text{diam}(A\cup B)\le \text{diam}\left(A\right)+\text{diam}\left(B\right)\text{.}$$ What can you say if $A$ and $B$ are disjoint?

   		Solution.
   	The definition of $\text{diam}\left(A\right)$ is $$\text{diam}\left(A\right)=\text{sup}\left\{d(x,y)\hspace{0.17em}\right|\hspace{0.17em}x,y\in A\}\text{.}$$ Assume $A\subseteq X$ and $B\subseteq X$ and $A\cap B=\varnothing$ and $$\text{diam}\left(A\right)<\infty \text{and}\text{diam}\left(B\right)<\infty \text{.}$$ To show: $\text{diam}(A\cup B)\le \text{diam}\left(A\right)+\text{diam}\left(B\right)\text{.}$ To show: $\text{diam}\left(A\right)+\text{diam}\left(B\right)$ is an upper bound of $\left\{d(x,y)\hspace{0.17em}\right|\hspace{0.17em}x,y\in A\cup B\}\text{.}$ To show: If $x,y\in A\cup B$ then $d(x,y)\le \text{diam}\left(A\right)+\text{diam}\left(B\right)\text{.}$ Assume $x,y\in A\cup B\text{.}$ Case 1: $x,y\in A\text{.}$ Then $$d(x,y)\le \text{diam}\left(A\right)\le \text{diam}\left(A\right)+\text{diam}\left(B\right)\text{.}$$ Case 2: $x,y\in B\text{.}$ Then $$d(x,y)\le \text{diam}\left(B\right)\le \text{diam}\left(A\right)+\text{diam}\left(B\right)\text{.}$$ Case 3: $x\in A$ and $y\in B\text{.}$ Let $z\in A\cap B\text{.}$ Then $$d(x,y)\le d(x,z)+d(z,y)\le \text{diam}\left(A\right)+\text{diam}\left(B\right)\text{.}$$ Case 4: $x\in B$ and $y\in A\text{.}$ Let $z\in A\cap B\text{.}$ Then $$d(x,y)\le d(x,z)+d(z,y)\le \text{diam}\left(A\right)+\text{diam}\left(B\right)\text{.}$$ So $\text{diam}\left(A\right)+\text{diam}\left(B\right)$ is an upper bound of $\left\{(x,y)\hspace{0.17em}\right|\hspace{0.17em}x,y\in A\cup B\}\text{.}$ So $\text{diam}(A\cup B)\le \text{diam}\left(A\right)+\text{diam}\left(B\right)\text{.}$ $\square$

2. Let $X=C[0,1]=\{f:[0,1]\to ℝ\hspace{0.17em}|\hspace{0.17em}f\hspace{0.17em}\text{is continuous}\}\text{.}$ The supremum metric $d_{\infty }:X\times X\to {ℝ}_{\ge 0}$ and the $L^1$ metric $d_1:X\times X\to {ℝ}_{\ge 0}$ are defined by $$\begin{array}{rcl}{\displaystyle d_{\infty }(f,g)}& =& {\displaystyle \text{sup}\left\{|f\left(x\right)-g\left(x\right)|\hspace{0.17em}\right|\hspace{0.17em}x\in [0,1]\}\text{and}}\\ {\displaystyle d_1(f,g)}& =& {\displaystyle {\int }_0^1\mid f\left(x\right)-g\left(x\right)\mid dx\text{.}}\end{array}$$ Consider the sequence $\{f_1,f_2,f_3,\dots \}$ in $X$ where $f_n\left(x\right)=n{x}^n(1-x)$ for $0\le x\le 1\text{.}$
   1. Determine whether $\left\{f_n\right\}$ converges in $(X,d_1)\text{.}$
   2. Determine whether $\left\{f_n\right\}$ converges in $(X,d_{\infty })\text{.}$
   (You may use any standard results about limits of real sequences.)

   		Solution.
   	The graph of $y=x-x^2$ $$\begin{array}{c} 1 1 \end{array}$$ and $y=x^{n-1}$ $$\begin{array}{c} 1 1 \end{array}$$ help to determine the graph of $y=n{x}^n(1-x)=n{x}^{n-1}(x-x^2)\text{.}$ $$\begin{array}{c} 1 1 4 \end{array}$$ Since $\underset{n\to \infty }{\text{lim}}n{x}^{n-1}=0$ for $x\in [0,1)$ and $n{x}^{n-1}=n$ for $x=1$ and $x-x^2=0$ for $x=1,$ then $\underset{n\to \infty }{\text{lim}}n{x}^{n-1}(x-x^2)=0$ for $x\in [0,1]\text{.}$ So the pointwise limit of $\left\{f_n\right\}$ is the zero function $f:[0,1]\to ℝ$ given by $f\left(x\right)=0\text{.}$ $\left\{f_n\right\}$ converges in $(X,d_1)$ if $\underset{n\to \infty }{\text{lim}}{d}_1(f_n,f)=0\text{.}$ $\left\{f_n\right\}$ converges in $(X,d_{\infty })$ if $\underset{n\to \infty }{\text{lim}}{d}_{\infty }(f_n,f)=0\text{.}$ Compute: $d_1(f_n,f)\text{.}$ $$\begin{array}{rcl}{\displaystyle d_1(f_n,f)}& =& {\displaystyle {\int }_0^1\mid f_n\left(x\right)-f\left(x\right)\mid dx}\\ & =& {\displaystyle {\int }_0^1(n{x}^n(1-x)-0)dx}\\ & =& {\displaystyle {\int }_0^1(n{x}^n-n{x}^{n+1})dx}\\ & =& {\displaystyle (\frac{n{x}^{n+1}}{n+1}-\frac{n{x}^{n+2}}{n+2}){|}_{x=0}^{x=1}}\\ & =& {\displaystyle (\frac{n}{n+1}-\frac{n}{n+2})}\\ & =& {\displaystyle \frac{n}{(n+1)(n+2)}\text{.}}\end{array}$$ So $$\begin{array}{rcl}{\displaystyle \underset{n\to \infty }{\text{lim}}{d}_1(f_n,f)}& =& {\displaystyle \underset{n\to \infty }{\text{lim}}\frac{n}{(n+1)(n+2)}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}\frac{1}{(1+\frac{1}{n})}\frac{1}{(n+2)}}\\ & =& {\displaystyle 1·0}\\ & =& {\displaystyle 0\text{.}}\end{array}$$ So $\left\{f_n\right\}$ converges in $(X,d_1)\text{.}$ Compute $d_{\infty }(f_n,f)\text{:}$ To compute $d_{\infty }(f_n,f)=\text{sup}\left\{\mid n{x}^n(1-x)-0\mid \hspace{0.17em}\right|\hspace{0.17em}x\in [0,1]\}$ find the maximum of $f_n\left(x\right)=n{x}^n(1-x)$ on the interval $[0,1]\text{.}$ This maximum occurs at $x=0$ or $x=1$ or at a critical point. Since $$\frac{d{f}_n}{dx}=\frac{dn{x}^n(1-x)}{dx}=n^2{x}^{n-1}-n(n+1)x^n=n{x}^{n-1}(n-nx-x)$$ the critical points are at $x=0$ and $x=\frac{n}{n+1}\text{.}$ Since $f_n\left(0\right)=0$ and $f_n\left(1\right)=0$ and $f_n\left(\frac{n}{n+1}\right)=n{\left(\frac{n}{n+1}\right)}^n(1-\frac{n}{n+1})$ the maximum of $f_n$ is $${\left(\frac{n}{n+1}\right)}^{n+1}={(1-\frac{1}{n+1})}^{n+1}\text{.}$$ So $$\begin{array}{rcl}{\displaystyle d_{\infty }(f_n,f)}& =& {\displaystyle \text{sup}\left\{\mid f_n\left(x\right)-f\left(x\right)\mid \hspace{0.17em}\right|\hspace{0.17em}x\in [0,1]\}}\\ & =& {\displaystyle \text{sup}\left\{n{x}^n(1-x)\hspace{0.17em}\right|\hspace{0.17em}x\in [0,1]\}}\\ & =& {\displaystyle {(1-\frac{1}{n+1})}^{n+1}}\end{array}$$ and $$\begin{array}{rcl}{\displaystyle \underset{n\to \infty }{\text{lim}}{(1-\frac{1}{n+1})}^{n+1}}& =& {\displaystyle \underset{n\to \infty }{\text{lim}}{e}^{\text{log}{(1-\frac{1}{n+1})}^{n+1}}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}{e}^{(n+1)\text{log}(1-\frac{1}{n+1})}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}{e}^{-(n+1)(\frac{1}{n+1}+\frac{1}{2}{\left(\frac{1}{n+1}\right)}^2+\frac{1}{3}{\left(\frac{1}{n+1}\right)}^3+\cdots )}}\end{array}$$ since $-\text{log}(1-x)=\int \frac{1}{1-x}dx=\int (1+x+x^2+\cdots )dx=x+\frac{1}{2}{x}^2+\frac{1}{3}{x}^3+\cdots \text{.}$ So $$\underset{n\to \infty }{\text{lim}}{d}_{\infty }(f_n,f)=\underset{n\to \infty }{\text{lim}}{(1-\frac{1}{n+1})}^{n+1}=e^{-(1+\frac{1}{2}·0+\frac{1}{3}·0^2+\cdots )}=e^{-1}\text{.}$$ So $\left\{f_n\right\}$ does not converge in $(X,d_{\infty })\text{.}$ $\square$

3. Let $X$ and $Y$ be topological spaces. Let $A\subseteq X$ and $B\subseteq Y\text{.}$ Show that $$\bar{A}\times \bar{B}=\overline{A\times B}\text{.}$$

   		Solution.
   	To show: (a) $\bar{A}\times \bar{B}\subseteq \overline{A\times B}\text{.}$ (b) $\overline{A\times B}\subseteq \bar{A}\times \bar{B}\text{.}$ (a) Assume $(x,y)\in \bar{A}\times \bar{B}\text{.}$ To show: $(x,y)\in \overline{A\times B}\text{.}$ To show: $(x,y)$ is a close point of $A\times B\text{.}$ Let $N$ be a neighbourhood of $(x,y)$ in $X\times Y\text{.}$ By the definition of the product topology on $X\times Y$ there exist $N_X,$ a neighbourhood of $x$ in $X,$ and $N_y,$ a neighbourhood of $y$ in $Y,$ such that $N_x\times N_y\subseteq N\text{.}$ Since $x\in \bar{A}$ there exists $a\in A$ with $a\in N_x\text{.}$ Since $y\in \bar{B}$ there exists $b\in B$ with $b\in N_y\text{.}$ So $(a,b)\in N_x\times N_y\subseteq N$ and $(a,b)\in A\times B\text{.}$ So $(x,y)$ is a close point of $A\times B\text{.}$ So $(x,y)\in \overline{A\times B}\text{.}$ So $\bar{A}\times \bar{B}\subseteq \overline{A\times B}\text{.}$ (b) To show: $\overline{A\times B}\subseteq \bar{A}\times \bar{B}\text{.}$ Assume $(x,y)\in \overline{A\times B}\text{.}$ To show: $(x,y)\in \bar{A}\times \bar{B}\text{.}$ To show: $x\in \bar{A}$ and $y\in \bar{B}\text{.}$ Let $N_x$ be a neighbourhood of $x\in X$ and let $N_y$ be a neighbourhood of $y\in Y\text{.}$ Then $N_x\times N_y$ is a neighbourhood of $(x,y)\in X\times Y\text{.}$ Since $(x,y)$ is a close point of $A\times B,$ there exists $(a,b)\in A\times B$ with $(a,b)\in N_x\times N_y\text{.}$ So $a\in N_x$ and $b\in N_y$ and $a\in A$ and $b\in B\text{.}$ So $x$ is a close point of $A$ and $y$ is a close point of $B\text{.}$ So $x\in \bar{A}$ and $y\in \bar{B}\text{.}$ So $(x,y)\in \bar{A}\times \bar{B}\text{.}$ $\square$

4. Let $(X,d)$ be a metric space and let $A$ be a non-empty subset of $X\text{.}$ Recall that for each $x\in X,$ the distance from $x$ to $A$ is $$d(x,A)=\text{inf}\left\{d(x,a)\hspace{0.17em}\right|\hspace{0.17em}a\in A\}\text{.}$$
   1. Prove that $\bar{A}=\{x\in X\hspace{0.17em}|\hspace{0.17em}d(x,A)=0\}\text{.}$
   2. Prove that $|d(x,A)-d(y,A)|\le d(x,y)$ for all $x,y\in X\text{.}$ [Hint: first show that $d(x,A)\le d(x,y)+d(y,A)\text{.]}$
   3. Deduce the function $f:X\to ℝ$ defined by $f\left(x\right)=d(x,A)$ is continuous.
   4. Show that if $x\notin \bar{A}$ then $U=\{y\in X\hspace{0.17em}|\hspace{0.17em}d(y,A)<d(x,A)\}$ is an open set in $X$ such that $\bar{A}\subset U$ and $x\notin U\text{.}$

   		Solution.
   	(a) To show: (aa) $\{x\in X\hspace{0.17em}|\hspace{0.17em}d(x,A)=0\}\subseteq \bar{A}\text{.}$ (ab) $\bar{A}\subseteq \{x\in X\hspace{0.17em}|\hspace{0.17em}d(x,A)=0\}\text{.}$ (aa) Assume $x\in X$ and $d(x,A)=0\text{.}$ To show: $x\in \bar{A}\text{.}$ Let $N$ be a neighbourhood of $x$ in $X\text{.}$ Then there exists $\epsilon \in {ℝ}_{>0}$ such that $B(x,\epsilon )\subseteq N\text{.}$ Since $d(x,A)=\text{inf}\left\{d(x,a)\hspace{0.17em}\right|\hspace{0.17em}a\in A\}=0,$ there exists $a\in A$ such that $d(x,a)<\epsilon \text{.}$ Then $a\in B(x,\epsilon )\subseteq N$ and $a\in A\text{.}$ So $x$ is a close point of $A\text{.}$ So $\{x\in X\hspace{0.17em}|\hspace{0.17em}d(x,A)=0\}\subseteq \bar{A}\text{.}$ (ab) To show: $\bar{A}\subseteq \{x\in X\hspace{0.17em}|\hspace{0.17em}d(x,A)=0\}\text{.}$ Let $x\in \bar{A}\text{.}$ So $x$ is a close point of $A\text{.}$ To show: $d(x,A)=0\text{.}$ Let $\epsilon \in {ℝ}_{>0}\text{.}$ Then $B(x,\epsilon )$ is a neighbourhood of $x$ in $X\text{.}$ Since $x$ is a close point of $A$ there exists $a\in A$ such that $a\in B(x,\epsilon )\text{.}$ So $d(x,a)<\epsilon \text{.}$ So $d(x,A)<\epsilon$ for all $\epsilon \in {ℝ}_{>0}\text{.}$ So $d(x,A)=0\text{.}$ So $x\in \{x\in X\hspace{0.17em}|\hspace{0.17em}d(x,A)=0\}\text{.}$ So $\bar{A}\subseteq \{x\in X\hspace{0.17em}|\hspace{0.17em}d(x,A)=0\}\text{.}$ Thus $\bar{A}=\{x\in X\hspace{0.17em}|\hspace{0.17em}d(x,A)=0\}\text{.}$ (b) Assume $x,y\in X\text{.}$ To show: (ba) $d(x,A)-d(y,A)\le d(x,y)\text{.}$ (bb) $-(d(x,A)-d(y,A))\le d(x,y)\text{.}$ (ba) Since $d(x,A)$ is a lower bound of $\left\{d(x,a)\hspace{0.17em}\right|\hspace{0.17em}a\in A\},$ if $a\in A$ then $d(x,A)\le d(x,a)\text{.}$ Using $d(x,a)\le d(x,y)+d(y,a),$ if $a\in A$ then $d(x,A)\le d(x,y)+d(y,a)\text{.}$ So $d(x,A)$ is a lower bound of $\{d(x,y)+d(y,a)\hspace{0.17em}|\hspace{0.17em}a\in A\}\text{.}$ Since $d(x,y)+d(y,A)$ is the greatest lower bound of $\{d(x,y)+d(y,a)\hspace{0.17em}|\hspace{0.17em}a\in A\}$ then $$d(x,A)\le d(x,y)+d(y,A)\text{.}$$ So $d(x,A)-d(y,A)\le d(x,y)\text{.}$ So $d(y,A)-d(x,A)\le d(y,x)=d(x,y)\text{.}$ So $-(d(x,A)-d(y,A))\le d(x,y)\text{.}$ So $d(x,A)-d(y,A)\le d(x,y)$ and $-(d(x,A)-d(y,A))\le d(x,y)\text{.}$ So $\mid d(x,A)-d(y,A)\mid \le d(x,y)\text{.}$ (c) To show: If $\epsilon \in {ℝ}_{>0}$ and $x\in X$ then there exists $\delta \in {ℝ}_{>0}$ such that if $y\in X$ and $d(x,y)<\delta$ then $d(f\left(x\right),f\left(y\right))<\epsilon \text{.}$ Assume $\epsilon \in {ℝ}_{>0}$ and $x\in X\text{.}$ To show: There exists $\delta \in {ℝ}_{>0}$ such that if $y\in X$ and $d(x,y)<\delta$ then $d(f\left(x\right),f\left(y\right))<\epsilon \text{.}$ Let $\delta =\epsilon \text{.}$ To show: If $y\in X$ and $d(x,y)<\delta$ then $d(f\left(x\right),f\left(y\right))<\epsilon \text{.}$ Assume $y\in X$ and $d(x,y)<\delta \text{.}$ To show: $d(f\left(x\right),f\left(y\right))<\epsilon \text{.}$ By part (b), $$d(f\left(x\right),f\left(y\right))=\mid d(y,A)-d(x,A)\mid \le d(x,y)<\delta =\epsilon \text{.}$$ So $f$ is continuous. (d) Assume $x\notin \bar{A}$ and let $U=\{y\in X\hspace{0.17em}|\hspace{0.17em}d(y,A)<d(x,A)\}\text{.}$ To show: (da) $x\notin U\text{.}$ (db) $U$ is open. (dc) $\bar{A}\subseteq U\text{.}$ (da) Let $D=d(x,A)\text{.}$ Since $x\notin \bar{A}$ and, by part (a), $\bar{A}=\{y\in X\hspace{0.17em}|\hspace{0.17em}d(y,A)=0\}$ then $d(x,A)\ne 0\text{.}$ So $D\ne 0\text{.}$ We know $U=\{y\in X\hspace{0.17em}|\hspace{0.17em}d(y,A)<D\}\text{.}$ Since $d(x,A)=D,$ $x\notin U\text{.}$ (db) Since $U=f^{-1}\left({ℝ}_{<D}\right)=f^{-1}\left((-\infty ,D)\right)$ and $f$ is continuous, then $U$ is open. (dc) By part (a), $$\bar{A}=\{y\in X\hspace{0.17em}|\hspace{0.17em}d(y,A)=0\}\subseteq \{y\in X\hspace{0.17em}|\hspace{0.17em}d(y,A)<D\}=U\text{.}$$ So $\bar{A}\subseteq U\text{.}$ $\square$

5. Determine whether the following sequences of functions converge uniformly.
   1. $f_n=e^{-n{x}^2},x\in [0,1]\text{;}$
   2. $g_n=e^{-x^2/n},x\in [0,1]\text{.}$
   3. $g_n=e^{-x^2/n},x\in ℝ\text{.}$

   		Solution.
   	Let $(X,\rho )$ be a metric space and let $f_n:X\to ℝ,$ $n\in {\mathbb{Z}}_{>0},$ be a sequence of functions from $X$ to $ℝ\text{.}$ Assume that $f:X\to ℝ$ defined by $$f\left(x\right)=\underset{n\to \infty }{\text{lim}}{f}_n\left(x\right)$$ is well defined. The sequence $f_1,f_2,f_3,\dots$ converges uniformly to $f$ if $$\underset{n\to \infty }{\text{lim}}\left(\text{sup}\left\{\rho (f_n\left(x\right),f\left(x\right))\hspace{0.17em}\right|\hspace{0.17em}x\in X\}\right)=0\text{.}$$ (a) Define $f_n:[0,1]\to ℝ$ by $$f_n\left(x\right)=e^{-n{x}^2}=e^{-{\left(\sqrt{n}x\right)}^2},$$ for $n\in {\mathbb{Z}}_{>0}\text{.}$ Then $$\underset{n\to \infty }{\text{lim}}{e}^{-n·0^2}=e^{-0}=1\text{and}\underset{n\to \infty }{\text{lim}}{e}^{-n·1^2}=e^{-\infty }=0$$ and if $x\in (0,1)$ then $$\underset{n\to \infty }{\text{lim}}{e}^{-n{x}^2}=e^{-\infty }=0\text{.}$$ $$\begin{array}{c} 1 \end{array}$$ Let $f:[0,1]\to ℝ$ be given by $$f\left(x\right)=\{\begin{array}{ll}1,& \text{if}\hspace{0.17em}x=0,\\ 0,& \text{if}\hspace{0.17em}x\in (0,1]\text{.}\end{array}$$ Let $n\in {\mathbb{Z}}_{>0}\text{.}$ Then $$\begin{array}{rcl}{\displaystyle \text{sup}\left\{\rho (f_n\left(x\right),f\left(x\right))\hspace{0.17em}\right|\hspace{0.17em}x\in X\}}& =& {\displaystyle \text{sup}(\left\{\mid e^{-n{x}^2}-0\mid \hspace{0.17em}\right|\hspace{0.17em}x\in (0,1]\}\cup \left\{\mid 1-1\mid \right\})}\\ & =& {\displaystyle \text{sup}\left\{e^{-n{x}^2}\hspace{0.17em}\right|\hspace{0.17em}x\in (0,1]\}}\\ & =& {\displaystyle 1\text{.}}\end{array}$$ So $$\underset{n\to \infty }{\text{lim}}\left(\text{sup}\left\{\mid e^{-n{x}^2}\mid \hspace{0.17em}\right|\hspace{0.17em}x\in [0,1]\}\right)=\underset{n\to \infty }{\text{lim}}1=1\text{.}$$ So $f_1,f_2,f_3,\dots$ is not uniformly convergent. (b) Define $g_n:[0,1]\to ℝ$ by $$g_n\left(x\right)=e^{-x^2/n}=e^{-{(x/\sqrt{n})}^2},$$ for $n\in {\mathbb{Z}}_{>0}\text{.}$ Then $$\underset{n\to \infty }{\text{lim}}{e}^{-0^2/n}=e^{-0}=1\text{and}\underset{n\to \infty }{\text{lim}}{e}^{-1^2/n}=e^{-1/\infty }=e^{-0}=1$$ and if $x\in (0,1)$ then $$\underset{n\to \infty }{\text{lim}}{e}^{-x^2/n}=e^{-0}=1\text{.}$$ $$\begin{array}{c} 1 \end{array}$$ Let $g:[0,1]\to ℝ$ by given by $g\left(x\right)=1\text{.}$ Let $n\in {\mathbb{Z}}_{>0}\text{.}$ Then $$\begin{array}{rcl}{\displaystyle \text{sup}\left\{\rho (g_n\left(x\right),g\left(x\right))\hspace{0.17em}\right|\hspace{0.17em}x\in X\}}& =& {\displaystyle \text{sup}\left\{\mid e^{-x^2/n}-1\mid \hspace{0.17em}\right|\hspace{0.17em}x\in [0,1]\}}\\ & =& {\displaystyle \mid e^{-1^2/n}-1\mid }\\ & =& {\displaystyle 1-e^{-1/n}\text{.}}\end{array}$$ So $$\begin{array}{rcl}{\displaystyle \underset{n\to \infty }{\text{lim}}\left(\text{sup}\left\{\mid e^{-x^2/n}-1\mid \hspace{0.17em}\right|\hspace{0.17em}x\in [0,1]\}\right)}& =& {\displaystyle \underset{n\to \infty }{\text{lim}}1-e^{-\frac{1}{n}}}\\ & =& {\displaystyle 1-e^{-1/\infty }}\\ & =& {\displaystyle 1-e^0}\\ & =& {\displaystyle 1-1}\\ & =& {\displaystyle 0\text{.}}\end{array}$$ So $g_1,g_2,g_3,\dots$ uniformly converges to $g\text{.}$ (c) Define $g_n:ℝ\to ℝ$ by $$g_n\left(x\right)=e^{-x^2/n}=e^{-{(x/\sqrt{n})}^2},$$ for $n\in {\mathbb{Z}}_{>0}\text{.}$ Let $x\in ℝ\text{.}$ Then $$\underset{n\to \infty }{\text{lim}}{e}^{-x^2/n}=e^{-x^2/\infty }=e^{-0}=1\text{.}$$ Let $g:ℝ\to ℝ$ be given by $g\left(x\right)=1\text{.}$ Let $n\in {\mathbb{Z}}_{>0}\text{.}$ Then $$\begin{array}{rcl}{\displaystyle \text{sup}\left\{\rho (g_n\left(x\right),g\left(x\right))\hspace{0.17em}\right|\hspace{0.17em}x\in X\}}& =& {\displaystyle \text{sup}\left\{\mid e^{-x^2/n}-1\mid \hspace{0.17em}\right|\hspace{0.17em}x\in ℝ\}}\\ & =& {\displaystyle \mid e^{-{\infty }^2/n}-1\mid }\\ & =& {\displaystyle \mid e^{-\infty }-1\mid }\\ & =& {\displaystyle \mid 0-1\mid }\\ & =& {\displaystyle 1\text{.}}\end{array}$$ So $$\underset{n\to \infty }{\text{lim}}\text{sup}\left\{\rho (g_n\left(x\right),g\left(x\right))\hspace{0.17em}\right|\hspace{0.17em}x\in X\}=\underset{n\to \infty }{\text{lim}}1=1\text{.}$$ So $g_1,g_2,\dots$ is not uniformly convergent. $\square$

6. Let $X$ be the set of all real sequences with finitely many non-zero terms with the supremum metric: if $\text{x}=\left(x_i\right)$ and $\text{y}=\left(y_i\right)$ then $d(\text{x},\text{y})=\text{sup}\left\{|x_i-y_i|\hspace{0.17em}\right|\hspace{0.17em}i\in {\mathbb{Z}}_{>0}\}\text{.}$
   For each $n\in \mathbb{N},$ let ${\text{x}}^n=(1,1/2,1/3,\dots ,1/n,0,0,\dots )\text{.}$
   1. Show that $\left\{{\text{x}}^n\right\}$ is a Cauchy sequence in $X\text{.}$
   2. Show that $\left\{{\text{x}}^n\right\}$ does not converge to a point in $X\text{.}$ (So $X$ is not complete.)

   		Solution.
   	Let $$X=\left\{(x_1,x_2,\dots )\hspace{0.17em}\right|\hspace{0.17em}{x}_i\in ℝ\hspace{0.17em}\text{and all but a finite number of}\hspace{0.17em}{x}_i\hspace{0.17em}\text{are zero}\}$$ and define $d:X\times X\to {ℝ}_{>0}$ by $$d(x,y)=\text{sup}\left\{\mid x_i-y_i\mid \hspace{0.17em}\right|\hspace{0.17em}i\in {\mathbb{Z}}_{>0}\}\text{.}$$ Let $f_n=(1,\frac{1}{2},\frac{1}{3},\dots ,\frac{1}{n},0,0,\dots ,0)\text{.}$ (a) Show that $\left\{f_n\right\}$ is a Cauchy sequence in $X\text{.}$ To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $m,n\in {\mathbb{Z}}_{>0}$ and $m>N$ and $n>N$ then $d(f_m,f_n)<\epsilon \text{.}$ Assume $\epsilon \in {ℝ}_{>0}\text{.}$ To show: There exists $N\in {\mathbb{Z}}_{>0}$ such that if $m,n\in {\mathbb{Z}}_{>0}$ and $m>N$ and $n>N$ then $d(f_m,f_n)<\epsilon \text{.}$ Let $N=\frac{1}{\epsilon }\text{.}$ Assume $m,n\in {\mathbb{Z}}_{>0},$ $m>N$ and $n>N$ and $m<n\text{.}$ To show: $d(f_n,f_m)<\epsilon \text{.}$ $$\begin{array}{rcl}{\displaystyle d(f_n,f_m)}& =& {\displaystyle \text{sup}\{1-1,\frac{1}{2}-\frac{1}{2},\dots ,\frac{1}{m}-\frac{1}{m},\frac{1}{m+1},\frac{1}{m+2},\dots ,\frac{1}{n},0-0,0-0,\dots \}}\\ & =& {\displaystyle \frac{1}{m+1}}\\ & <& {\displaystyle \frac{1}{N+1}}\\ & =& {\displaystyle \frac{1}{\frac{1}{\epsilon }+1}}\\ & =& {\displaystyle \frac{\epsilon }{\epsilon +1}}\\ & <& {\displaystyle \frac{\epsilon }{1}}\\ & =& {\displaystyle \epsilon \text{.}}\end{array}$$ So $\left\{f_n\right\}$ is a Cauchy sequence in $X\text{.}$ (b) The limit of $\{f_1,f_2,\dots \}$ is the sequence $$f=(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots )\text{.}$$ If $k\in {\mathbb{Z}}_{>0}$ the $k^{\text{th}}$ entry of $f$ is $\frac{1}{k}$ which is not equal to $0\text{.}$ So all entries of $f$ are nonzero. So $f\notin X\text{.}$ So $\{f_1,f_2,\dots \}$ is a Cauchy sequence in $X$ which does not converge to a point in $X\text{.}$ So $X$ is not complete. $\square$

7. Let $X$ be a nonempty set and let $(Y,d)$ be a complete metric space. Let $f:X\to Y$ be an injective function and define $$d_f(x,y)=d(f\left(x\right),f\left(y\right))$$ for $x,y\in X\text{.}$
   1. Explain briefly why $d_f$ is a metric on $X\text{.}$
   2. Show that $(X,d_f)$ is a complete metric space if $f\left(X\right)$ is a closed subset of $Y\text{.}$

   		Solution.
   	(a) To show: (aa) If $x,y\in X$ then $d_f(x,y)=d_f(y,x)\text{.}$ (ab) If $x\in X$ then $d_f(x,x)=0\text{.}$ (ac) If $x,y\in X$ and $d_f(x,y)=0$ then $x=y\text{.}$ (ad) If $x,y,z\in X$ then $d_f(x,y)\le d_f(x,z)+d_f(z,y)\text{.}$ (aa) Assume $x,y\in X\text{.}$ To show: $d_f(x,y)=d_f(y,x)\text{.}$ $$d_f(x,y)=d(f\left(x\right),f\left(y\right))=d(f\left(y\right),d\left(x\right))=d_f(y,x)\text{.}$$ (ab) Assume $x\in X\text{.}$ To show: $d_f(x,x)=0\text{.}$ $$d_f(x,x)=d(f\left(x\right),f\left(x\right))=0\text{.}$$ (ac) Assume $x,y\in X$ and $d_f(x,y)=0\text{.}$ To show: $x=y\text{.}$ Since $0=d_f(x,y)=d(f\left(x\right),f\left(y\right))$ and $d$ is a metric, then $f\left(x\right)=f\left(y\right)\text{.}$ Since $f:X\to Y$ is injective and $f\left(x\right)=f\left(y\right)$ then $x=y\text{.}$ (ad) Assume $x,y,z\in X\text{.}$ To show: $d_f(x,y)\le d_f(x,z)+d_f(z,y)\text{.}$ $$\begin{array}{rcl}{\displaystyle d_f(x,y)}& =& {\displaystyle d(f\left(x\right),f\left(y\right))}\\ & \le & {\displaystyle d(f\left(x\right),f\left(z\right))+d(f\left(z\right),f\left(y\right))}\\ & =& {\displaystyle d_f(x,z)+d_f(z,y)\text{.}}\end{array}$$ (b) Assume $f\left(x\right)$ is a closed subset of $Y\text{.}$ Since $f:X\to Y$ is injective and $f:X\to f(X)$ is surjective, then $f:X\to f\left(X\right)$ is bijective. Since $d_f(x,y)=d(f\left(x\right),f\left(y\right))$ for $x,y\in X$ then $$f:(X,d_f)⟶(f\left(X\right),d)$$ is an isometry. To show: $(X,d_f)$ is complete. To show: $(f\left(X\right),d)$ is complete. To show: If $z_1,z_2,\dots$ is a Cauchy sequence in $f\left(X\right)$ then $z_1,z_2,\dots$ converges with $\underset{n\to \infty }{\text{lim}}{z}_n$ in f(X). Assume $z_1,z_2,\dots$ is a Cauchy sequence in $f\left(X\right)\text{.}$ Then $z_1,z_2,\dots$ is a Cauchy sequence in $Y\text{.}$ Since $Y$ is complete $z=\underset{n\to \infty }{\text{lim}}{z}_n$ exists with $z\in Y\text{.}$ Since $z$ is a close point of $z_1,z_2,\dots ,$ then $z$ is a close point of $f\left(X\right)\text{.}$ So $z\in f\left(X\right)\text{.}$ So $f\left(X\right)$ is complete. So $(X,d_f)$ is complete. $\square$

8. Let $f:{ℝ}_{\ge 0}\to {ℝ}_{\ge 0}$ be given by $$f\left(x\right)=\frac{2}{2+x}\text{.}$$
   1. Show that $f$ defines a contraction mapping $f:{ℝ}_{\ge 0}\to {ℝ}_{\ge 0}\text{.}$
   2. Fix $x_0\ge 0$ and $x_{n+1}=f\left(x_n\right)$ for all $n\ge 0\text{.}$ Show that the sequence $\left\{x_n\right\}$ converges and find its limit with respect to the usual metric on $ℝ\text{.}$

   		Solution.
   	(a) A function $f:X\to X$ is a contraction mapping if there exists $\alpha \in (0,1)$ such that if $x,y\in X$ then $d(f\left(x\right),f\left(y\right))\le \alpha d(x,y)\text{.}$ Let $f:{ℝ}_{\ge 0}\to {ℝ}_{\ge 0}$ be given by $f\left(x\right)=\frac{2}{2+x}\text{.}$ Let $\alpha =\frac{1}{2}\text{.}$ To show: If $x,y\in {ℝ}_{>0}$ then $d(f\left(x\right),f\left(y\right))\le \alpha d(x,y)\text{.}$ Assume $x,y\in {ℝ}_{>0}\text{.}$ To show: $\mid f\left(y\right)-f\left(x\right)\mid \le \alpha \mid y-x\mid \text{.}$ $$\begin{array}{rcl}{\displaystyle \mid f\left(y\right)-f\left(x\right)\mid }& =& {\displaystyle \mid \frac{2}{2+x}-\frac{2}{2+y}\mid }\\ & =& {\displaystyle 2\mid \frac{2+y-(2+x)}{(2+x)(2+y)}\mid }\\ & =& {\displaystyle \frac{2}{(2+x)(2+y)}\mid y-x\mid }\\ & \le & {\displaystyle \frac{2}{4}\mid y-x\mid }\\ & =& {\displaystyle \frac{1}{2}\mid y-x\mid \text{.}}\end{array}$$ So $f$ is a contraction mapping. (b) The Banach fixed point theorem gives that the sequence $x_{n+1}=f\left(x_n\right)$ for $x_0\in X$ converges to the (unique) fixed point of $f\text{.}$ In our case the fixed point is $p\in {ℝ}_{>0}$ such that $$p=f\left(p\right)=\frac{2}{2+p}\text{.}$$ So $p^2+2p=2$ and $p=\frac{-2\pm \sqrt{4+4·2}}{2}=-1\pm \sqrt{3}\text{.}$ Since $p\in {ℝ}_{>0}$ then $p=-1+\sqrt{3}\text{.}$ $\square$

9. Let $X$ be a connected topological space. Let $f:X\to ℝ$ be continuous with $f\left(X\right)\subseteq \mathbb{Q}\text{.}$ Show that $f$ is a constant function.

   		Solution.
   	To show: If $f$ is not a constant function then $X$ is not connected. Assume $f$ is not a constant function. Let $a,b\in f\left(X\right)$ with $a<b\text{.}$ Let $z\in ℝ,$ $z\notin \mathbb{Q}$ with $a<z<b\text{.}$ Let $A=f^{-1}\left((-\infty ,z)\right)$ and $B=f^{-1}\left((z,\infty )\right)\text{.}$ Then $$\begin{array}{c}A\ne \varnothing \text{since}a\in A,\\ B\ne \varnothing \text{since}b\in B,\\ A\cap B=\varnothing \text{since}(-\infty ,z)\cap (z,\infty )=\varnothing \text{.}\end{array}$$ Since $z\in \mathbb{Q}$ then $z\notin f\left(X\right)$ and since $(-\infty ,z)\cup (z,\infty )=ℝ-\left\{z\right\}$ then $A\cup B=X\text{.}$ So $X$ is not connected. $\square$

10. Show that $X=\{(x,y)\in {ℝ}^2\hspace{0.17em}|\hspace{0.17em}xy=0\}$ is not homeomorphic to $ℝ\text{.}$

    		Solution.
    	Let $$\begin{array}{rcl}{\displaystyle X}& =& {\displaystyle \{(x,y)\in {ℝ}^2\hspace{0.17em}|\hspace{0.17em}xy=0\}}\\ & =& {\displaystyle \{(x,y)\in {ℝ}^2\hspace{0.17em}|\hspace{0.17em}x=0\}\cup \{(x,y)\in {ℝ}^2\hspace{0.17em}|\hspace{0.17em}y=0\}\text{.}}\end{array}\begin{array}{c} y = 0 x = 0 \end{array}$$ Assume that $f:X\to ℝ$ is a homeomorphism. Let $a=f\left((0,0)\right)\text{.}$ Then $f:X-\left\{(0,0)\right\}\to ℝ-\left\{a\right\}$ is a homeomorphism. So $X-\left\{(0,0)\right\}$ and $ℝ-\left\{a\right\}$ have the same number of connected components. Since $X-\left\{(0,0)\right\}$ has 4 connected components and $ℝ-\left\{a\right\}$ has 2 connected components, this is a contradiction. So $X$ is not homeomorphic to $ℝ\text{.}$ $\square$

Notes and References

These are a typed copy of Assignment 1 Solutions from a series of handwritten lecture notes for the class Metric and Hilbert Spaces.

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