HW TOP 1

Exercise 404.1(Munkres (2000) Exr 17.13).

Show that $X$ is Hausdorff iff the diagonal $\Delta =\{x\times x:x\in X\}$ is closed in $X\times X$.

Suppose $X$ is Hausdorff. Let $x_1\times x_2\in X\times X\setminus \Delta$. Then, $x_1\ne x_2$, so there exist open $U_1,U_2\subseteq X$ such that $U_1\ni x_1$, $U_2\ni x_2$, and $U_1\cap U_2=\varnothing$. Now, $U_1\times U_2$ is an open neighborhood of $x_1\times x_2$ in $X\times X$, and is disjoint from $\Delta$ since $U_1$ and $U_2$ are disjoint. Thus, $\Delta$ is closed in $X\times X$.

Conversely, suppose $\Delta$ is closed in $X\times X$. Let $x_1,x_2\in X$ be distinct. There exists a basis element $U_1\times U_2$ containing $x_1\times x_2$ disjoint from $\Delta$. It follows that, $U_1\ni x_1$, $U_2\ni x_2$, and $U_1\cap U_2=\varnothing$. Thus, $X$ is Hausdorff.

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Exercise 404.2(Munkres (2000) Exr 17.19).

If $A\subseteq X$, we define the boundary of $A$ by $\partial A=\overline{A}\cap \overline{(X\setminus A)}$.

1. Show that $A^{\circ }\cap \partial A=\varnothing$ and $\overline{A}=A^{\circ }\cup \partial A$.
2. Show that $\partial A=\varnothing ⟺A$ is clopen.
3. Show that $U$ is open $⟺$ $\partial U=\overline{U}\setminus U$.
4. If $U$ is open, is it true that $U=(\overline{U}{)}^{\circ }$?

Part 1 Suppose $x\in A^{\circ }$. There exists open $U\subseteq A$ such that $x\in U$. Since $U$ does not intersect $X\setminus A$, $x\notin \overline{X\setminus A}$, and hence $x\notin \partial A$. Next, suppose $x\in \overline{A}\setminus A^{\circ }$. Then every open neighborhood of $x$ intersects $X\setminus A$, so $x\in \overline{X\setminus A}$ by definition. Thus, $x\in \overline{A}\cap \overline{(X\setminus A)}=\partial A$.

Part 2 Using part 1, $\partial A=\varnothing ⟺\overline{A}=A^{\circ }$. Since $A^{\circ }\subseteq A\subseteq \overline{A}$, we have $A=A^{\circ }=\overline{A}$. Since $A^{\circ }$ is open and $\overline{A}$ is closed, $A$ must be clopen.

Part 3 If $U$ is open, $\overline{X\setminus U}=X\setminus U$. Thus, $\partial U=\overline{U}\cap (X\setminus U)=\overline{U}\setminus U$.

Part 4 No. Consider the set $\{a,b\}$ with the topology $\{\{a\},\{a,b\}\}$. The closure of $\{a\}$ is $\{a,b\}$, the interior of which is $\{a,b\}$.

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Exercise 404.3(Munkres (2000) Exr 18.8).

Let $Y$ be an ordered set in the order topology. Let $f,g:X\to Y$ be continuous.

1. Show that the set $\{x:f(x)⩽g(x)\}$ is closed in $X$.
2. Let $h:X\to Y$ be the function $h(x)=\min \{f(x),g(x)\}$. Show that $h$ is continuous.

$Y$ is Hausdorff under the order topology. Indeed, let $a,b\in Y$ such that $a<b$. If there exists $x$ such that $a<x<b$, then the open sets $\{y:y<x\}$ and $\{y:y>x\}$ work. Otherwise, the open sets $\{y:y<b\}$ and $\{y:y>a\}$ are disjoint and hence can be used instead.

Part 1 Consider $x\in \{x:g(x)<f(x)\}=:S$. Let $U_1,U_2\subseteq Y$ be disjoint open sets containing $g(x)$ and $f(x)$; as above, we can pick these such that $U_1<U_2$. We can find open neighborhoods $V_1,V_2\subseteq X$ of $x$ such that $g(V_1)\subseteq U_1$ and $f(V_2)\subseteq U_2$. Thus, $g(V_1\cap V_2)\subseteq U_1$ and $f(V_1\cap V_2)\subseteq U_2$, so $V_1\cap V_2\subseteq S$. It follows that $S$ is open.

Part 2 Let $A,B\subseteq X$ be the closed sets $\{x:f(x)⩽g(x)\}$ and $\{x:g(x)⩽f(x)\}$. On $A$, $h=f$, and on $B$, $h=g$. Thus, $h$ is continuous on $A$ and $B$. Clearly, $A\cup B=X$ and $f=g$ on $A\cap B$, so $h$ is well defined on $X$. By the pasting lemma, $h$ is continuous.

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Exercise 404.4(Munkres (2000) Exr 19.7).

Let ${\mathbb{R}}^{\infty }$ be the subset of ${\mathbb{R}}^{\omega }$ consisting of all sequences that are eventually zero. What is the closure of ${\mathbb{R}}^{\infty }$ in ${\mathbb{R}}^{\omega }$ in the box and product topologies?

The closure of ${\mathbb{R}}^{\infty }$ in the box topology is ${\mathbb{R}}^{\infty }$. Indeed, let $\mathbf{a}\in {\mathbb{R}}^{\omega }$ have infinitely many nonzero terms $\{a_i:i\in S\subseteq \mathbb{N}\}$. Let

$$U_i=\{\begin{array}{ll}(0,2a_i)& i\in S\\ \mathbb{R}& i\notin S.\end{array}$$

Then, ${\prod }_{i\in \mathbb{N}}{U}_i$ is an open neighborhood of $\mathbf{a}$ which does not intersect ${\mathbb{R}}^{\infty }$.

The closure of ${\mathbb{R}}^{\infty }$ in the product topology is ${\mathbb{R}}^{\omega }$. Let $\mathbf{a}\in {\mathbb{R}}^{\omega }$, and let $U={\prod }_{i\in \mathbb{N}}{U}_i$ be a basis element containing $\mathbf{a}$ (where $U_i=\mathbb{R}$ for all $i\notin S^{\prime }$, $|S^{\prime }|<\infty$). Then,

$$\mathbf{b}=\{\begin{array}{ll}\text{some member of }{U}_i& i\in S^{\prime }\\ 0& i\notin S^{\prime }\end{array}$$

is a member of ${\mathbb{R}}^{\infty }$ contained in $U$.

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Exercise 404.5(Munkres (2000) Exr 20.5).

What is the closure of ${\mathbb{R}}^{\infty }$ in ${\mathbb{R}}^{\omega }$ in the uniform topology?

If $\mathbf{a}\in {\mathbb{R}}^{\omega }$ is in the closure of ${\mathbb{R}}^{\infty }$, then every $ϵ$-ball centered at $\mathbf{a}$ must contain a sequence that is eventually zero. This is exactly the same as saying $\mathbf{a}$, seen as a sequence, converges to $0$. Thus, The closure of ${\mathbb{R}}^{\infty }$ in ${\mathbb{R}}^{\omega }$ in the uniform topology is the set of all sequences that converge to $0$.

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Exercise 404.6(Munkres (2000) Exr 20.6).

Let $\overline{\rho }$ be the uniform metric on ${\mathbb{R}}^{\omega }$. Given $\mathbf{x}=(x_1,x_2,\dots )\in {\mathbb{R}}^{\omega }$ and given $0<ϵ<1$, let

$$U(\mathbf{x},ϵ)=(x_1-ϵ,x_1+ϵ)\times \cdots \times (x_n-ϵ,x_n+ϵ)\times \dots .$$

1. Show that $U(\mathbf{x},ϵ)$ is not equal to the $ϵ$-ball $B_{\overline{\rho }}(\mathbf{x},ϵ)$.
2. Show that $U(\mathbf{x},ϵ)$ is not open in the uniform topology.
3. Show that

$$B_{\overline{\rho }}(\mathbf{x},ϵ)=\bigcup _{\delta <ϵ}U(\mathbf{x},\delta ).$$

Part 1 Consider the sequence ${\mathbf{x}}^{\prime }=\{x_n+ϵ-1/n{\}}_{n=1}^{\infty }\in U(\mathbf{x},ϵ)$. Since ${\sup }_{n\in \mathbb{N}}(ϵ-1/n)=ϵ$, ${\mathbf{x}}^{\prime }\notin B_{\overline{\rho }}(\mathbf{x},ϵ)$.

Part 2 ${\mathbf{x}}^{\prime }$ is an element of $U(\mathbf{x},ϵ)$ which does not have an open neighborhood in $U(\mathbf{x},ϵ)$. For ${ϵ}^{\prime }>0$, let $N$ be such that $1/N<{ϵ}^{\prime }$. Then, the sequence

$${\left\{\{\begin{array}{ll}{x}_n+ϵ-1/n& n<N\\ x_n+ϵ& n⩾N\end{array}\right\}}_{n=1}^{\infty }$$

is in $B_{\overline{\rho }}({\mathbf{x}}^{\prime },{ϵ}^{\prime })$ but not in $U(\mathbf{x},ϵ)$.

Part 3 Let $\mathbf{y}\in B_{\overline{\rho }}(\mathbf{x},ϵ)$. Let $\delta$ be such that $\overline{\rho }(\mathbf{x},\mathbf{y})<\delta <ϵ$. Clearly, $\mathbf{y}\in U(\mathbf{x},\delta )$. The stated equality follows.

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Exercise 404.7(Munkres (2000) Exr 21.8).

Let $X$ be a topological space and let $Y$ be a metric space. Let $f_n:X\to Y$ be a sequence of continuous functions. Let $x_n$ be a sequence of points of $X$ converging to $x$. Show that if $\{f_n\}\rightrightarrows f$, then $\{f_n(x_n)\}\to f(x)$.

Write

$$\begin{array}{rl}d(f_n(x_n),f(x))& ⩽d(f_n(x_n),f(x_n))+d(f(x_n),f(x)).\end{array}$$

Let $ϵ>0$. Let $N_1$ be such that $d(f(x),f_n(x))<ϵ$ for all $n⩾N_1$ and for all $x\in X$. By Munkres (2000) Thm 21.6, $f$ is continuous. It follows that $\{f(x_n)\}\to f(x)$. Let $N_2$ be such that $d(f(x_n),f(x))<ϵ$ for all $n⩾N_2$. Let $N=\max (N_1,N_2)$. Then,

$$\begin{array}{r}d(f_n(x_n),f(x))⩽2ϵ.\end{array}$$

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References

Munkres, J. R. (2000). Topology (2nd ed). Prentice Hall, Inc.