## Topology

### James R. Munkres

#### Second Edition (2000)



\documentclass[12pt]{article}

\begin{document}
\title{Host}

\author{Sándor Szabó (Dept of Analysis)}

Chapter 2

\maketitle

Section 12 Topological Spaces

Section 13 Basis for a Topology

\textbf{Exercises}

\textbf{1}. Let $X$ be a topological space; let $A$ be a subset
of $X$. Suppose that for each $x\in A$ there is an open set $U$
containing $x$ such that $U\subset A$. Show that $A$ is open in
$X$.

\textcolor{green}{Solution}. $A=\cup_{\alpha}\{a_{\alpha}\}$. Then
$\forall\alpha\exists U_{\alpha}:a_{\alpha}\in U_{\alpha}\subseteq A$. Therefore
$A=\cup_{\alpha}\{a_{\alpha}\}\subseteq\cup_{\alpha}U_{\alpha}\subseteq A$,
that is, $A=\cup_{\alpha}U_{\alpha}$.

\textbf{2}. Consider the nine topologies on the set $X=[a,b,c]$ indicated
in Example 1 of Sec. 12. Compare them; that is, for each pair of topologies,
determine whether they are comparable, and if so, which is the finer.

\textbf{3}. Show that the collection $\tau_{c}$ given in Example
4 of Section 12 is a topology on the set $X$. Is the collection$\tau_{\infty}:=\{U\,|\, X-U\,is\, infinite\, or\, empty\, or\, all\, of\, X\}$
a topology on $X$?

\textcolor{green}{Solution}. $\tau_{c}$.

$\emptyset,X\in\tau_{c}$. If $U_{\alpha}\in\tau_{c}$ then $X-\bigcup_{\alpha}U_{\alpha}=\bigcap_{\alpha}(X-U_{\alpha})$
is countable and $X-\bigcap_{i=1}^{n}U_{i}=\bigcup_{i=1}^{n}(X-U_{i})$
is also countable.

$\tau_{\infty}$.

Assume $X=A\cup^{\star}B\cup^{\star}\{x\}$, $card\, A,\,card\, B=\infty$.
Then $A,B\in\tau_{\infty}$, but $X-A\cup B=\{x\}$ is finite.

\textbf{4}. (a) If $\{\tau_{\alpha}\}$ is a family of topologies
on $X$, show that $\cap\tau_{\alpha}$ is a topology on $X$. Is
$\cup\tau_{\alpha}$ a topology on $X$?

(b) Let $\{\tau_{\alpha}\}$ is a family of topologies on $X$. Show
that there is a unique smallest topology on $X$ containing all the
collections $\tau_{\alpha}$, and a unique largest topology contained
in all $\tau_{\alpha}$.

(c) If $X:=[a,b,c]$, let $\tau_{1}:=\{\emptyset,X,\{a\},\{a,b\}\}\qquad\tau_{2}:=\{\emptyset,X,\{a\},\{b,c\}\}.$
Find the smallest topology containing $\tau_{1}$ and $\tau_{2}$,
and the largest topology contained in $\tau_{1}$ and $\tau_{2}$.

\textcolor{green}{Solution}. (a) $\cap\tau_{\alpha}=:\tau$.

$\emptyset,X\in\tau$. If $U_{\beta}\in\tau$ then $U_{\beta}\in\tau_{\alpha}$
where $\alpha$ is arbitrary but fixed. Since $\tau_{\alpha}$ is
topology, thus $\cup_{\beta}U_{\beta}\in\tau_{\alpha}$ and $\cap_{i=1}^{n}U_{\beta_{i}}\in\tau_{\alpha}$.
Since $\alpha$ is arbitrary, $\cup_{\beta}U_{\beta}\in\tau$ and
$\cap_{i=1}^{n}U_{\beta_{i}}\in\tau$.

$\cup\tau_{\alpha}$.

Let $A,B\neq\emptyset$, $A\cap B=\emptyset$,
$X:=A\cup B$, $(A,\tau_{A})$, $(B,\tau_{B})$ arbitrary topological
spaces. Denote $\tau_{1}:=\tau_{A}\cup\{X\}$, $\tau_{2}:=\tau_{B}\cup\{X\}$.
If $\emptyset,A\neq U_{1}\in\tau_{A}$, $\emptyset,B\neq U_{2}\in\tau_{B}$
then $U_{1}\cup U_{2}\notin\tau_{1}\cup\tau_{2}$.

(b) $\cup\tau_{\alpha}:=A\subseteq P(X)$.

If $A=P(X)$ then $\tau=P(X)$ is the smallest topology containing
$A$. If $A\neq P(X)$, then $\tau:=\bigcap_{\sigma}\sigma,$
where $\sigma$ topology on $X$ containing $A$. Since $A\subset P(X)$
the intersection is not empty. $\tau$ is the unique smallest topology
on $X$ containing all the collections $\tau_{\alpha}$.

The unique largest topology contained in all $\tau_{\alpha}$ is $\tau:=\cap\tau_{\alpha}$.

\textbf{5}. Show that if $\mathcal{A}$ is a basis for a topology
on $X$, then the topology generated by $\mathcal{A}$ equals the
intersection of all topologies on $X$ that contain $\mathcal{A}$.
Prove the same if $\mathcal{A}$ is a subbasis.

\textcolor{green}{Solution}. See \textbf{4}. (b).

\textbf{6}. Show that the topologies of $\mathbf{R}_{\ell}$ and $\mathbf{R}_{K}$
are not comparable.

\textcolor{green}{Solution}. Let $\tau_{\ell}$ and $\tau_{K}$ be
the corresponding topologies. $B:=(-1,1)-K$. Then $0\in B\in\tau_{K}$,
but there is no $[a,b)$ such that $0\in[a,b)\subset B$.

On the other hand, given the basis element $[a,b)$ of $\tau_{\ell}$,
there is no open interval $(c,d)$ nor the set of the form $(c,d)-K$
that contains $a$ and lies in $[a,b)$. Therefore $\tau_{\ell}$
and $\tau_{K}$ are not comparable.

\textbf{7}. Consider the following topologies on $\mathbf{R}$:

$\tau_{1}:=$ the standard topology,

$\tau_{2}:=$ the topology of $\mathbf{R}_{K}$,

$\tau_{3}:=$ the finite complement topology,

$\tau_{4}:=$ the upper limit topology, having all sets $(a,b]$ as
basis,

$\tau_{5}:=$ the topology having all sets $(-\infty,a)=\{x\,|\, x\lt a\}$
as basis.

Determine, for each of these topologies, which of the others it contains.

\textbf{8}. (a) Apply Lemma 13.2 to show that the countable collection$\mathcal{B}:=\{(a,b)\,|\, a\lt b,\, a\, and\, b\, rational\}$
is a basis that generates the standard topology on $\mathbf{R}$.

(b) Show that the collection $\mathcal{C}:=\{[a,b)\,|\, a\lt b,\, a\,and\, b\,rational\}$
is a basis that generates a topology different from the lower limit
topology on $\mathbf{R}$.

\textcolor{green}{Solution}. (b) Denote $\tau$ the generated topology.
Then $\tau\subset\tau_{l}$. Since $[\sqrt{2},2)\notin\tau$ $\tau_{l}$
is strictly finer.

Section 14 The Order Topology

Section 15 The Product Topology on $X\times Y$

Section 16 The Subspace Topology

\textbf{Exercises}

\textbf{1}. Show that if $Y$ is a subspace of $X$, and $A$ is a
subset of $Y$, then the topology $A$ inherits as a subspace of $Y$
is the same as the topology it inherits as a subspace of $X$.

\textcolor{green}{Solution}. Denote $\tau$ the topology on $X$,
$\tau_{Y}$ and $\tau_{A}$ the corresponding subspace topology. Then
\begin{eqnarray*}
\tau_{Y} & = & \{Y\cap U\,|\, U\in\tau\},\\
\tau_{A} & = & \{A\cap V\,|\, V\in\tau\}.\end{eqnarray*}
Then we have$A\cap V=(A\cap Y)\cap V=A\cap(Y\cap V)=A\cap W,$
where $V\in\tau$, $W\in\tau_{Y}$.

\textbf{2}. If $\tau$ and $\tau'$ are topologies on $X$ and $\tau'$
is strictly finer than $\tau$, what can you say about the corresponding
subspace topologies on the subset $Y$ of $X$?

\textcolor{green}{Solution}. The $\tau_{Y}$ can be the same as $\tau'_{Y}$.
Let $A,B\neq\emptyset$, $A\cap B=\emptyset$, $X:=A\cup B$, $(A,\tau_{A})$
arbitrary topological spaces. Denote $\tau_{1}:=\tau_{A}\cup\{X\}$.
We give $\tau_{2}$ by basis, $\tau_{A}\cup\{X\}\cup\{B\}$. Then
$\tau_{1},\tau_{2}$ are topologies on $X$, $\tau_{2}$ is strictly
finer than $\tau_{1}$, because $\tau_{1}\subset\tau_{2}$, $B\in\tau_{2}$,
but $B\notin\tau_{1}$, and $\tau_{1A},\tau_{2A}\equiv\tau_{A}$.

\textbf{3}. Consider the set $Y:=[-1,1]$ as a subspace of $\mathbf{R}$.
Which of the following sets are open in $Y$? Which are open in $\mathbf{R}$?
\begin{eqnarray*}
A: & = & \left\{ x\,|\,\frac{1}{2}\lt |x|\lt 1\right\} ,\\
B: & = & \left\{ x\,|\,\frac{1}{2}\lt |x|\leq1\right\} ,\\
C: & = & \left\{ x\,|\,\frac{1}{2}\leq|x|\lt 1\right\} ,\\
D: & = & \left\{ x\,|\,\frac{1}{2}\leq|x|\leq1\right\} ,\\
E: & = & \left\{ x\,|\,0\lt |x|\lt 1\,and\,\frac{1}{x}\notin\mathbf{Z}_{+}\right\} .\end{eqnarray*}

\textcolor{green}{Solution}. In $\mathbf{R}$: $A=\left(-1,-\frac{1}{2}\right)\cup\left(\frac{1}{2},1\right)$.
$E=(-1,0)\cup_{n=1}^{\infty}\left(\frac{1}{n+1},\frac{1}{n}\right)$.

In $Y$: $A=A\cap Y$, $B=\left(\frac{1}{2},2\right)\cap Y$, $E=E\cap Y$.

\textbf{4}. A map $f:X\to Y$ is said to be \textbf{open map} if for
every open set $U$ of $X$, the set $f(U)$ is open in $Y$. Show
that $\pi_{1}:X\times Y\to X$ and $\pi_{2}:X\times Y\to Y$ are open
maps.

\textcolor{green}{Solution}. If $G\in X\times Y$ is open then $G=\cup_{\alpha}U_{\alpha}\times V_{\alpha}$,
where $U_{\alpha}\in X$ and $V_{\alpha}\in Y$ are open sets. Thus
$\pi_{1}(G)=\cup_{\alpha}U_{\alpha}$ is open, similarly for $\pi_{2}$.

\textbf{5}. Let $X$ and $X'$ denote a single set in the topologies
$\tau$ and $\tau'$, respectively; let $Y$ and $Y'$ denote a single
set in the topologies $\mathcal{U}$ and $\mathcal{U}'$, respectively.
Assume these sets are nonempty.

(a) Show that if $\tau'\supset\tau$ and $\mathcal{U}'\supset\mathcal{U}$,
then the product topology on $X'\times Y'$ is finer than the product
topology on $X\times Y$.

\textbf{6}. Show that the countable collection$\{(a,b)\times(c,d)\,|\, a\lt b\,and\,c\lt d,\, and\,a,b,c,d\,are\, rational\}$
is a basis for $\mathbf{R}^{2}$.

\textbf{7}. Let $X$ be an ordered set. If $Y$ is a proper subset
of $X$ that is convex in $X$, does it follow that $Y$ is an interval
or a ray in $X$?

\textcolor{green}{Solution}. Not. Let $\omega:=(X,\preceq)$ without
right endpoints, $\omega^{\star}:=(X,\succeq)$, Then $\omega\times\omega^{\star}\times\omega\times\omega^{\star}$
with the dictionary order is an ordered set. The subset $\omega^{\star}\times\omega$
is convex, but neither an interval nor a ray, because it has no endpoints
at all.

\textbf{8}. If $L$ is a straight line in the plane, describe the
topology $L$ inherits as a subspace of $\mathbf{R}_{\ell}\times\mathbf{R}$
and as a subspace of $\mathbf{R}_{\ell}\times\mathbf{R}_{\ell}$.
In each case it is a familiar topology.
\end{document}

\textbf{9}. Show that the dictionary order topology on the set $\mathbf{R}\times\mathbf{R}$
is the same as the product topology $\mathbf{R}_{d}\times\mathbf{R}$,
where $\mathbf{R}_{d}$ denotes $\mathbf{R}$ in the discrete topology.
Compare this topology with the standard topology on $\mathbf{R}^{2}$.

\textcolor{green}{Solution}. The order topology on $\mathbf{R}\times\mathbf{R}$
has as basis the collection of all open intervals of the form $\{a\}\times(b,d)$,
$b\lt d$ (see Section 14, Example 2), which is the same as the product
topology on $\mathbf{R}_{d}\times\mathbf{R}$. The standard topology
on $\mathbf{R}^{2}$ has as basis the collection of all open rectangles
of the form $(e,f)\times(g,h)$. Obviously$(e,f)\times(g,h)=\cup_{a\in(e,f)}(\{a\}\times(g,h)).$
It implies, the order topology is strictly finer than the standard
topology.

\textbf{10}. Let $I:=[0,1]$. Compare the product topology on $I\times I$,
the dictionary order topology on $I\times I$, and the topology $I\times I$
inherits as a subspace of $\mathbf{R}\times\mathbf{R}$ in the dictionary
order topology.

Section 17 Closed Sets and Limit Points

\textbf{Exercises}

\textbf{1}. Let $\mathcal{C}$ be a collection of subsets of the set
$X$. Suppose that $\emptyset$ and $X$ are in $\mathcal{C}$, and
that finite unions and arbitrary intersections of elements of $\mathcal{C}$
are in $\mathcal{C}$. Show that the collection$\tau:=\{X-C\,|\, C\in\mathcal{C}\}$
is a topology on $X$.

\textcolor{green}{Solution}. $\emptyset,X\in\tau$.

For union:$\cup_{\alpha}(X-C_{\alpha})=X-\cap_{\alpha}C_{\alpha}.$

For intersection:$\bigcap_{i=1}^{n}(X-C_{\alpha})=X-\bigcup_{i=1}^{n}C_{\alpha}.$

\textbf{2}. Show that if $A$ is closed in $Y$ and $Y$ is closed
in $X$, then $A$ is closed in $X$.

\textcolor{green}{Solution}. $A=Y\cap C$, where $C$ is closed in
$X$. Since $Y$is closed, $A$ is closed in $X$.

\textbf{3}. Show that if $A$ is closed in $X$ and $B$ is closed
in $Y$, then $A\times B$ is closed in $X\times Y$.

\textcolor{green}{Solution}. $X\times Y-A\times B=((X-A)\times Y)\cup((Y-B)\times X)$.

\textbf{4}. Show that if $U$ is open in $X$ and $A$ is closed in
$X$, then $U-A$ is open in $X$, and $A-U$ is closed in $X$.

\textcolor{green}{Solution}. $U-A=U\cap(X-A)$, and, $A-U=A\cap(X-U)$.

\textbf{5}. Let $X$ be an ordered set in the order topology. Show
that $\overline{(a,b)}\subset[a,b]$. Under what conditions does the
equality hold?

\textcolor{green}{Solution}. Obviously $(a,b)\subset[a,b]$, and $[a,b]$
is closed. By Exe. 6 (a) we obtain $\overline{(a,b)}\subset[a,b]$.
If $(a,b)$, $[a,b)$, $(a,b]$ are not closed then we have equality.

\textbf{6}. Let $A,B$, and $A_{\alpha}$ denote subsets of a space
$X$. Prove the following:

(a) If $A\subset B$, then $\overline{A}\subset\overline{B}$.

(b) $\overline{A\cup B}=\overline{A}\cup\overline{B}$.

(c) $\overline{\cup A_{\alpha}}\supset\cup\overline{A_{\alpha}}$;
give an example where equality fails.

\textcolor{green}{Solution}. (a) $A\subset B$ $\Longrightarrow$
$A\subset\overline{B}$ $\Longrightarrow$ $\overline{A}\subset\overline{B}$.

(b) $\overline{A\cup B}\subset\overline{A}\cup\overline{B}$:

$A\subset\overline{A}$, $B\subset\overline{B}$ $\Longrightarrow$
$A\cup B\subset\overline{A}\cup\overline{B}$ $\Longrightarrow$ $\overline{A\cup B}\subset\overline{A}\cup\overline{B}$.

$\overline{A\cup B}\supset\overline{A}\cup\overline{B}$:

$A\cup B\supset A,B$ $\Longrightarrow$ $\overline{A\cup B}\supset A,B$
$\overline{A\cup B}\supset\overline{A},\overline{B}$ $\Longrightarrow$
$\overline{A\cup B}\supset\overline{A}\cup\overline{B}$.

(c) $\cup A_{\alpha}\supset A_{\beta}$ $\Longrightarrow$ $\overline{\cup A_{\alpha}}\supset A_{\beta}$
$\Longrightarrow$ $\overline{\cup A_{\alpha}}\supset\overline{A_{\beta}}$
$\Longrightarrow$ $\overline{\cup A_{\alpha}}\supset\cup\overline{A_{\beta}}$.

If $A_{\alpha}:=\{\alpha\}$, $\alpha\in\mathbf{Q}$ then $\cup\overline{A_{\alpha}}=\mathbf{Q}$,
$\overline{\cup A_{\alpha}}=\mathbf{R}$.

\textbf{7}. Criticize the following proof'' that $\overline{\cup A_{\alpha}}\subset\cup\overline{A_{\alpha}}$:
if $\{A_{\alpha}\}$ is a collection of sets in $X$ and if $x\in\overline{\cup A_{\alpha}}$,
then every neighborhood $U$ of $x$ intersects $\cup A_{\alpha}$.
Thus $U$ must intersect some $A_{\alpha}$, so that $x$ must belong
to the closure of some $A_{\alpha}$. Therefore, $x\in\cup\overline{A_{\alpha}}$.

\textcolor{green}{Solution}. Wrong. When $U$ varies it not necessarily
intersects the same $A_{\alpha}$.

\textbf{8}. Let $A,B$, and $A_{\alpha}$ denote subsets of a space
$X$. Determine whether the following equations hold; if an equality
fails, determine whether one of the inclusions $\supset$ or $\subset$
holds.

(a) $\overline{A\cap B}=\overline{A}\cap\overline{B}$.

(b) $\overline{\cap A_{\alpha}}=\cap\overline{A_{\alpha}}$.

(c) $\overline{A-B}=\overline{A}-\overline{B}$.

\textcolor{green}{Solution}. (a) $A\cap B\subset A,B$ thus
$A\cap B\subset\bar{A},\bar{B}$ therefore
$A\cap B\subset\bar{A}\cap\bar{B}$.
Thus $\overline{A\cap B}\subset\bar{A}\cap\bar{B}$.
Generally we can not have equality.
$A=(0,1)$, $B=(1,2)$. Then $A\cap B=\emptyset$,
$\bar{A}\cap\bar{B}=\{1\}$.

(b) $\overline{\cap A_{\alpha}}\subset\cap\overline{A_{\alpha}}$.

(c) $\overline{A-B}\cup\bar{B}=\overline{(A-B)\cup B}=\overline{A\cup B}\supset\bar{A}$.
$\Longrightarrow$ $\overline{A-B}\supset\bar{A}-\bar{B}$.

$A:=(0,2)$, $B:=(1,2)$. $\overline{A-B}=[0,1]$, $\bar{A}-\bar{B}=[0,1)$.

\textbf{9}. Let $A\subset X$ and $B\subset Y$. Show that in the
space $X\times Y$,$\overline{A\times B}=\overline{A}\times\overline{B}.$

\textcolor{green}{Solution}. $A\times B\subset\overline{A}\times\overline{B}$,
since $\overline{A}\times\overline{B}$ is closed (see Exe. 3) we
obtain $\overline{A\times B}\subset\overline{A}\times\overline{B}$.

Conversely, let $(x,y)\in\overline{A}\times\overline{B}$. Then any
neighborhood of $(x,y)$ contains a product neighborhood $U\times V$
where $U$ is a neighborhood of $x$, and $V$ is a neighborhood of
$y$. Then $(U\times V)\cap(A\times B)=(U\cap A)\times(V\cap B)\neq\emptyset$
because $U\cap A\neq\emptyset$ since $x\in\overline{A}$ and, similarly
$V\cap B\neq\emptyset$ since $y\in\overline{B}$.Hence, any neighborhood
of $(x,y)$ intersects $A\times B$, thus $(x,y)\in\overline{A\times B}$.

\textbf{10}. Show that every order topology is Hausdorff.

\textcolor{green}{Solution}. Let $a\prec b$. If there exists $c$
such that $a\prec c\prec b$ then $(-\infty,c)$ and $(c,\infty)$
the two disjoint neighborhoods. If there is no such $c$, then $(-\infty,b)$
and $(a,\infty)$ are the two disjoint neighborhoods.

\textbf{11}. Show that the product of two Hausdorff spaces is Hausdorff.

\textcolor{green}{Solution}. $(X,\tau_{1})$ and $(Y,\tau_{2})$ are
Hausdorff spaces. $(x_{1},y_{1}),\,(x_{2},y_{2})\in(X\times Y,\tau_{1}\times\tau_{2})$
are two different points. If $x_{1}=x_{2}$ then $y_{1}\neq y_{2}$
and there exist $V_{1},V_{2}$ disjoint neighborhoods of $y_{1},y_{2}$
respectively. Then $X\times V_{1}$ and $X\times V_{2}$ are disjoint
neighborhoods of $(x_{1},y_{1}),\,(x_{2},y_{2})$. If $x_{1}\neq x_{2}$
but $y_{1}=y_{2}$ we can argue similarly. If $x_{1}\neq x_{2}$ and
$y_{1}\neq y_{2}$ then there exist $U_{1},U_{2}$ disjoint neighborhoods
of $x_{1},x_{2}$ and $V_{1},V_{2}$ disjoint neighborhoods of $y_{1},y_{2}$.
Then $U_{1}\times V_{1}$ and $U_{2}\times V_{2}$ are disjoint neighborhoods
of $(x_{1},y_{1}),\,(x_{2},y_{2})$.

\textbf{12}. Show that that a subspace of a Hausdorff space is Hausdorff.

\textcolor{green}{Solution}. Let $x\neq y\in A\subset X$. Then there
exist $x\in U$ and $y\in V$ disjoint neighborhoods. Then $U\cap A$
and $V\cap A$ are disjoint neighborhoods in the subspace topology.

\textbf{13}. Show that $X$ is Hausdorff if and only if the \textbf{diagonal}
$\Delta:=\{x\times x\,|\, x\in X\}$ is closed in $X\times X$.

\textcolor{green}{Solution}. If: We show that the complement of the
diagonal is open. Let $(x,y)\in X\times X-\Delta$. Since $x\neq y$
there exist $x\in U$, $y\in V$ disjoint neighborhoods, so $(x,y)\in U\times V$.
Since $U,V$ are disjoints, $U\times V\subset X\times X-\Delta$.

Only if: If $(x,y)\in X\times X-\Delta$ then there exists a basis
element of the form $U\times V\subset X\times X-\Delta$, $U,V$ are
open sets, such that $x\in U$, $y\in V$. Then $U,V$ are disjoint,
because $z\in U\cap V$ gives $(z,z)\in U\times V$ which is impossible.

\textbf{14}. In the finite complement topology on $\mathbf{R}$, to
what point or points does the sequence $x_{n}=1/n$ converge?

\textcolor{green}{Solution}. To every points.

\textbf{15}. Show the $T_{1}$ axiom is equivalent to the condition
that for each pair of points of $X$, each has a neighborhood not
containing the other.

\textcolor{green}{Solution}. $T_{1}\Longrightarrow$.

Let $x\neq y$. Then $x\in X-\{y\}$ and $y\in X-\{x\}$ are the required
neigborhoods.

$\Longrightarrow T_{1}$. We show that $\{x\}$ is closed. Let $y\in X-\{x\}$.
Then there exists $y\in U_{y}$ neighborhood such that $x\notin U_{y}$.
Then $X-\{x\}=\cup_{y\in X-\{x\}}U_{y}$ is open.

\textbf{16}. Consider the five topologies on $\mathbf{R}$ given in
Exercise 7 of Section 13.

(a) Determine the closure of the set $K=\{1/n\,|\, n\in\mathbf{Z}_{+}\}$
under each of these topologies.

(b) Which of these topologies satisfy the Hausdorff axiom? the $T_{1}$
axiom?

\textbf{17}. Consider the lower limit topology on $\mathbf{R}$ and
the topology given by the basis $\mathcal{C}$ of Exercise 8 of Section
13. Determine the closure of the intervals $A:=(0,\sqrt{2})$ and
$B:=(\sqrt{2},3)$ in these two topologies.

\textbf{18}. Determine the closures of the following subsets of the
ordered square \begin{eqnarray*}
A: & = & \left\{\frac{1}{n}\times 0\,|\, n\in\mathbf{Z}_{+}\right\},\\
B: & = & \left\{\left(1-\frac{1}{n}\right)\times\frac{1}{2}\,|\, n\in\mathbf{Z}_{+}\right\},\\
C: & = & \left\{x\times 0\,|\,0\lt x\lt 1\right\},\\
D: & = & \left\{x\times\frac{1}{2}\,|\,0\lt x\lt 1\right\},\\
E: & = & \left\{\frac{1}{2}\times y\,|\,0\lt y\lt 1\right\}.\end{eqnarray*}

\textbf{19}. If $A\subset X$, we define the \textbf{boundary} of
$A$ by the equation$Bd\,A=\overline{A}\cap\overline{(X-A)}.$

(a) Show that $Int\,A$ and $Bd\,A$ are disjoint, and
$\overline{A}=Int\,A\cup Bd\,A$.

(b) Show that $Bd\,A=\emptyset$ $\Longleftrightarrow$ $A$
is both open and closed.

(c) Show that $U$ is open $\Longleftrightarrow$ $Bd\,U=\overline{U}-U$.

(d) If $U$ is open, is it true that $U=Int\,(\overline{U})$?

\textcolor{green}{Solution}. First of all, the meaning of the definition,
$x\in Bd\,A$ if and only if every neighborhood of $x$ intersects
$A$ and $X-A$.

(a) If $x\in Int\,A\cap Bd\,A$, then there exists an $x\in U$
neighborhood such that $U\subset Int\,A\subset A$, thus $U\cap(X-A)=\emptyset$
which is impossible.

$A\subset Int\,A\cup Bd\,A$. Indeed, if $x\in A-Int\,A$
then every neighborhood of $x$ intersects $A$ ($x\in A$) and intersects
$X-A$ (in opposite case we have $x\in Int\,A$), therefore $x\in Bd\,A$.

$Int\,A\cup Bd\,A$ is closed. If $x\notin Int\,A\cup Bd\,A$
then $x\notin Int\,A$ so every $U$ neighborhood of $x$ intersects
$X-A$. Since $x\notin Bd\,A$ there exists a $V$ neighborhood
of $x$ such that $V\cap A=\emptyset$ or $V\cap(X-A)=\emptyset$.
The last one is impossible, hence there exists a $x\in V$ neighborhood
such that $V\cap A=\emptyset$. Consequently, $X-V\supset\bar{A}$.
Obviously $Int\,A, Bd\,A\subset\overline{A}$. Thus $x$
has a $V$ neighborhood which is disjoint from $Int\,A\cup Bd\,A$.

Thus $A\subset Int\,A\cup Bd\,A\subset\overline{A}$ and
the middle set is closed. By the definition of closure we obtain the
statement.

(b) $\Longrightarrow:$

Using (a) we obtain $\overline{A}= Int\,A$. Since $Int\,A\subset A\subset\overline{A}$
we have $Int\,A=A=\overline{A}$.

$\Longleftarrow:$

$A=\overline{A}=Int\,A\cup Bd\,A=A\cup Bd\,A$. Since
$A=Int\,A$ and $Bd\,A$ are disjoint we get $Bd\,A=\emptyset$.

(c) $\Longrightarrow:$

Using (a) $\overline{U}=U\cup Bd\,U$. Since $U, Bd\,U$
are disjoint we obtain $Bd\,U=\overline{U}-U$.

$\Longleftarrow:$

Using (a) $Bd\,U=\overline{U}-U=(Bd\,U\cup Int\,U)-U=Bd\,U-U$.
It implies $Bd\,U\cap U=\emptyset$. Thus we can write $\emptyset=(\overline{U}-Int\, U)\cap U=U-(U\cap Int\,U)$
so $U=Int\,U$.

(d) $U=Int U\subset Int \overline{U}$.
$U:=\mathbf{R}-\{0\}$. Then $Int \overline{U}=\mathbf{R}$.

\textbf{20}. Find the boundary and the interior of each of the following
subsets of $\mathbf{R}^{2}$:

(a) $A=\{x\times y\,|\, y=0\}$

(b) $B=\{x\times y\,|\, x \gt 0\,and\,y\neq 0\}$

(c) $C=A\cup B$

(d) $D=\{x\times y\,|\, x\,is\,rational\}$

(e) $E=\{x\times y\,|\,0 \lt x^{2}-y^{2}\leq 1\}$

(f) $F=\{x\times y\,|\, x\neq 0\,and\,y\leq 1/x\}$

\textbf{21*}. (Kuratowski) Consider the collection of all subsets
$A$ of the topological space $X$. The operations of closure $A\to\overline{A}$
and complementation $A\to X-A$ are functions from this collection
to itself.

(a) Show that starting with a given set $A$, one can form no more
than $14$ distinct sets by applying these two operations successively.

(b) Find a subset $A$ of $\mathbf{R}$ (in its usual topology) for
which the maximum $14$ is obtained.

Section 18 Continuous Functions

\textbf{Exercises}

\textbf{1}. Prove that for functions $f:\mathbf{R}\to\mathbf{R}$,
the $\varepsilon-\delta$ definition of continuity implies the open
set definition.

\textbf{2}. Suppose that $f:X\to Y$ is continuous. If $x$ is a limit
point of the subset $A$ of $X$, is it necessarily true that $f(x)$
is a limit point of $f(A)$?

\textcolor{green}{Solution}. Not. Let $Y$ be the discrete topological
space, $y\in Y$, $f(x):=y$ for every $x$. Then $f(A)$ has no limit
point at all.

We say a $f:X\to Y$ function is \textbf{locally constant }in $x$
if there exists an $x\in U$ neighborhood such that $f$ is constant
on $U$. Let $A\subset X$, $x\in\overline{A}$. We say $f:A\to Y$
is locally constant in $x$ on the subspace $A$, if there exists
an $x\in U$ neighborhood such that $f$ is constant on $U\cap A$.

If we assume that $f$ is not locally constant in $x$ on the subspace
$A$, then $f(x)$ is a limit point of $f(A)$.

Indeed, let $f(x)\in V$, $V$ is open. Then $x\in f^{-1}(V)$, $f^{-1}(V)$
is open. Since $x$ is a limit point and $f$ is not locally constant,
there exists $x\neq y\in A\cap f^{-1}(V)$ such that $f(y)\neq f(x)$
and $f(y)\in f(A)\cap V$.

\textbf{3}. Let $X$ and $X'$ denote a single set in the two topologies
$\tau$ and $\tau'$, respectively. Let $i:X'\to X$ be the identity
function.

(a) Show that $i$ is continuous $\Longleftrightarrow$ $\tau'$ is
finer that $\tau$.

(b) Show that $i$ is a homeomorphism $\Longleftrightarrow$ $\tau'=\tau$.

\textcolor{green}{Solution}. (a) $i$ is continuous $\Longleftrightarrow$
$i^{-1}(\tau)=\tau\subset\tau'$.

\textbf{4}. Given $x_{0}\in X$ and $y_{0}\in Y$, show that the maps
$f:X\to X\times Y$ and $g:Y\to X\times Y$ defined by$f(x)=x\times y_{0}\quad and\quad g(y)=x_{0}\times y$
are imbeddings.

\textcolor{green}{Solution}. $f':X\to X\times y_{0}$, $f'(x)=x\times y_{0}$
and $g':Y\to x_{0}\times Y$, $g'(y)=x_{0}\times y$ are bijections.
Let $x\times y_{0}\in U\times V$, $U,V$ are open. Then $f'^{-1}(U\times V)=U$
is open, and $f'(U)=U\times y_{0}$ which is open in the $X\times y_{0}$
subspace topology, hence $f'$ is homeomorphism. Similarly for $g'$.

\textbf{5}. Show that the subspace $(a,b)$ of $\mathbf{R}$ is homeomorphic
with $(0,1)$ and the subspace $[a,b]$ of $\mathbf{R}$ is homeomorphic
with $[0,1]$.

\textcolor{green}{Solution}. $f(x):=\frac{1}{b-a}(x-a)$.

\textbf{6}. Find a function $f:\mathbf{R}\to\mathbf{R}$ that is continuous
at precisely one point.

\textcolor{green}{Solution}. The function $f$, $f(x):=x$ if $x\in\mathbf{Q}$,
$f(x):=-x$ if $x\in\mathbf{R}-\mathbf{Q}$ is continuous only at $0$.

\textbf{7}. (a) Suppose that $f:\mathbf{R}\to\mathbf{R}$ is continuous
from the right'', that is,$\lim_{x\to a+}f(x)=f(a),$

for each $a\in\mathbf{R}$. Show that $f$ is continuous when considered
as a function from $\mathbf{R}_{\ell}$ to $\mathbf{R}$.

(b) Can you conjecture what functions $f:\mathbf{R}\to\mathbf{R}$
are continuous when considered as maps from $\mathbf{R}$ to $\mathbf{R}_{\ell}$?
As maps from $\mathbf{R}_{\ell}$ to $\mathbf{R}_{\ell}$? We shall

\textcolor{green}{Solution}. (a) Let $f(a)\in V$, $V$ is open. Then
there exists $[a,b)$ such that $f([a,b))\subset V$ because of the
right limit. It implies $[a,b)\subset f^{-1}(V)$. Since $[a,b)$
is open $\mathbf{R}_{\ell}$, we obtain the statement.

(b) Only constant functions are continuous as maps from $\mathbf{R}$
to $\mathbf{R}_{\ell}$. Let $x$ be arbitrary fixed. Since $[f(x),f(x)+\varepsilon)$
($\varepsilon\gt 0$) is open there exists $x\in U$, $U$ is open such
that $f(x)\leq f(y)\lt f(x)+\varepsilon$. Hence $x$ is a local minimum
point.

A map $f:\mathbf{R}_{\ell}\to\mathbf{R}_{\ell}$ is continuous if
and only if for any $x$ and $\varepsilon$ there exists $\delta \gt 0$
such that $f([x,x+\delta))\subset[f(x),f(x)+\varepsilon)$. Hence,
$f:\mathbf{R}_{\ell}\to\mathbf{R}_{\ell}$ is continuous if and only
if $f:\mathbf{R}\to\mathbf{R}$ is right continuous and for every $x\in\mathbf{R}$
we have there exists $\delta \gt 0$ such that
$\forall y\in[x,x+\varepsilon)$ $f(x)\leq f(y)$.

\textbf{8}. Let $Y$ be an ordered set in the order topology. Let
$f,g:X\to Y$ be continuous.

(a) Show that the set $\{x\,|\, f(x)\preceq g(x)\}$ is closed in $X$.

(b) Let $h:X\to Y$ be the function$h(x):=\min\{f(x),g(x)\}.$

Show that $h$ is continuous. [Hint: Use the pasting lemma.]

\textcolor{green}{Solution}. (a) We show the set $A:=\{x\,|\, f(x)\succ g(x)\}$
is open in $X$. Let $x\in A$. Since order topology is $T_{2}$ there
exist $f(x)\in U$, $g(x)\in V$, $U\succ V$ disjoint open sets.
Then $x\in f^{-1}(U)\cap g^{-1}(V)\subset A$ and $f^{-1}(U)\cap g^{-1}(V)$
is open.

(b) Let $A:=\{x\,|\, f(x)\succeq g(x)\}$, $B:=\{x\,|\, f(x)=g(x)\}$,
$C:=\{x\,|\, f(x)\preceq g(x)\}$. Then $X=A\cup B\cup C$, $A,B,C$
are closed and $B=A\cap C$. By pasting lemma $h$ is continuous.
The same is true for max.

\textbf{9}. Let $\{A_{\alpha}\}$ be a collection of subsets of $X$;
let $X=\cup_{\alpha}A_{\alpha}$. Let $f:X\to Y$; suppose that $f\left|A_{\alpha}\right.$
is continuous for each $\alpha$.

(a) Show that if the collection $\{A_{\alpha}\}$ is finite and each
set is closed, then $f$ is continuous.

(b) Find an example where the collection $\{A_{\alpha}\}$ is countable
and each $A_{\alpha}$ is closed, but $f$ is not continuous.

(c) An indexed family of sets $\{A_{\alpha}\}$ is said to be \textbf{locally
finite} if each point $x$ of $X$ has a neighborhood that intersects
$A_{\alpha}$ for only finitely many values of $\alpha$. Show that
if the family $\{A_{\alpha}\}$ is locally finite and each $A_{\alpha}$
is closed, then $f$ is continuous.

\textcolor{green}{Solution}. (a) By pasting lemma and Theorem 18.2
(f).

(b) Let $A_{n}:=\frac{1}{2^{n}}$, $(n=0,\ldots)$, $A_{\omega}:=0$,
$f(A_{n}):=2^{n}$, $f(A_{\omega}):=0$. Then $f:X\to\mathbf{R}$
is not continuous at $0$.

(c) Let $f(x)\in V$ open, we find $x\in W\subset f^{-1}(V)$ open.
We can write $f^{-1}(Y-V)=\cup_{\alpha}\left[\left(f\left|_{A_{\alpha}}\right.\right)^{-1}(Y-V)\right]$.
Let $x\in U$ open. Denote $F_{x}:=\cup_{\alpha}\left[U\cap\left(f\left|_{A_{\alpha}}\right.\right)^{-1}(Y-V)\right]$.
The sets $\left(f\left|_{A_{\alpha}}\right.\right)^{-1}(Y-V)$ are
closed, and by the local finiteness $F_{x}$ is a union of finite
many closed sets, so it is closed. $W:=U-F_{x}$.

\textbf{10}. Let $f:A\to B$ and $g:C\to D$ be continuous functions.
Let us define a map $f\times g:A\times C\to B\times D$ by the equation$(f\times g)(a\times c):=f(a)\times g(c).$
Show that $f\times g$ is continuous.

\textcolor{green}{Solution}. Let $(f_{j}:X_{j}\to Y_{j})_{j\in J}$
be an indexed family of continuous maps. Define $\mathbf{f}=(f_{j}):\prod X_{j}\to\prod Y_{j}$
to be the map that takes $\mathbf{x}=(x_{j})\in\prod X_{j}$ to $\mathbf{f}(\mathbf{x})=(f_{j}(x_{j}))\in\prod Y_{j}$.
The commutative diagram $\begin{array}{ccc} \prod X_{j} & \mathbf{f}{\longrightarrow} & \prod Y_{j}\\ \pi_{k}\downarrow & & \downarrow\pi_{k}\\ X_{k} & f_{k}{\longrightarrow} & Y_{k}\end{array}$
shows that $\pi_{k}\circ\mathbf{f}=f_{k}\circ\pi_{k}$ is continuous
for all $k\in J$. By Theorem 19.6 we obtain the statement.

\textbf{11}. Let $F:X\times Y\to Z$. We say that $F$ is \textbf{continuous
in each variable separately} if for each $y_{0}$ in $Y$, the map
$h:X\to Z$ defined by $h(x):=F(x\times y_{0})$ is continuous, and
for each $x_{0}$ in $X$, the map $k:Y\to Z$ defined by $k(y):=F(x_{0}\times y)$
is continuous. Show that if $F$ is continuous, then $F$ is continuous
in each variable separately.

\textcolor{green}{Solution}. Let $F:\prod X_{j}\to Y$ be continuous,
$x_{j}\in X_{j}$ are given. Then $i_{j}:X_{j}\to\prod X_{j}$, $i_{j}(x)=(x_{1},\ldots,x,\ldots,x_{k},\ldots)$
are imbeddings. Then $F\circ i_{j}$ are continuous.

\textbf{12}. Let $F:\mathbf{R}\times\mathbf{R}\to\mathbf{R}$ be defined
by the equation$F(x\times y):=\left\{ \begin{array}{cc} \frac{xy}{x^{2}+y^{2}} & if\, x\times y\neq0\times0,\\ 0 & if\, x\times y=0\times0.\end{array}\right.$

(a) Show that $F$ is continuous in each variable separately.

(b) Compute the function $g:\mathbf{R}\to\mathbf{R}$ defined by $g(x):=F(x\times x)$.

(c) Show that $F$ is not continuous.

\textbf{13}. Let $A\subset X$; let $f:A\to Y$ be continuous; let
$Y$ be Hausdorff. Show that if $f$ may be extended to a continuous
function $g:\overline{A}\to Y$, then $g$ is uniquely determined
by $f$.

\textcolor{green}{Solution}. We may assume that $\overline{A}\neq\{a\}$.
Assume there are two different extensions, $g_{1}$ and $g_{2}$.
Denote $G=(g_{1},g_{2})$. Then $G:\overline{A}\to Y\times Y$ is
continuous. Since $g_{1}\neq g_{2}$ there exists $x\in\overline{A}$
such that $g_{1}(x)\neq g_{2}(x)$. Since $Y$ is Hausdorff so $Y\times Y$.
Then $\Delta$ is closed, $(g_{1}(x),g_{2}(x))\notin\Delta$, thus
there exist $g_{i}(x)\in U_{i}$ open sets, such that $U_{1}\times U_{2}\cap\Delta=\emptyset$.
Because of the continuity of $g_{i}$ the sets $g_{i}^{-1}(U_{i})$
are open. Obviously $x\in g_{1}^{-1}(U_{1})\cap g_{2}^{-1}(U_{2})$.
It implies there exists $y\in A$ such that $y\in g_{1}^{-1}(U_{1})\cap g_{2}^{-1}(U_{2})$.
Hence $(g_{1}(y),g_{2}(y))\in U_{1}\times U_{2}$. However $g_{i}(y)=f(y)$
which is impossible.

Section 19 The Product Topology

\textbf{5}. \textcolor{green}{Solution}. Since in the box topology
$\pi_{\beta}$ is continuous, $f_{\beta}=\pi_{\beta}\circ f$ is also
continuous.

\textbf{6}. \textcolor{green}{Solution}. If $\lim\mathbf{x}_{n}=\mathbf{x}$
then continuity of $\pi_{\alpha}$ gives $\lim\pi_{\alpha}(\mathbf{x}_{n})=\pi_{\alpha}(\mathbf{x})$.
It is true for product and box topology.

If $\lim\pi_{\alpha}(\mathbf{x}_{n})=y_{\alpha}$ in the product topology
then we show that $\lim\mathbf{x}_{n}=\mathbf{x}$ where $\mathbf{x}=\cap_{\alpha}\pi_{\alpha}^{-1}(\{y_{\alpha}\})$.
Let $\mathbf{x}\in U$ basis element. We can write $U$ as the product
$U=\prod_{\alpha\in J}U_{\alpha}$, where $U_{\alpha}$ denotes the
entire space $X_{\alpha}$ if $\alpha\neq\beta_{1},\ldots,\beta_{k}$
and $U_{\alpha}$ is a basis element in $X_{\alpha}$ if $\alpha=\beta_{1},\ldots,\beta_{k}$.
It implies $y_{\alpha}\in U_{\alpha}$. Since $\lim\pi_{\beta_{j}}(\mathbf{x}_{n})=y_{\beta_{j}}$
($j=1,\ldots,k$), therefore there exists $n_{0}$ such that $\pi_{\beta_{j}}(\mathbf{x}_{n})\in U_{\beta_{j}}$
for $n\gt n_{0}$, $j=1,\ldots,k$. Thus we have obtained $\mathbf{x}_{n}\in U$
for $n\gt n_{0}$. We proved until that $\mathbf{x}\in\lim\mathbf{x}_{n}$.
Assume $\mathbf{x}\neq\mathbf{x}^{\star}\in\lim\mathbf{x}_{n}$. Then
$\pi_{\alpha}(\mathbf{x})=\lim\pi_{\alpha}(\mathbf{x}_{n})=\pi_{\alpha}(\mathbf{x}^{\star})$
for all $\alpha$ which is contradiction.

In the case of box topology we show a counter example.

Let $J:=\mathbf{N}$, $\mathbf{x}_{n}:=\left(\frac{1}{n},\ldots,\frac{1}{n}\right)$,
$\mathbf{x}:=(0,\ldots,0)$. Then $\lim\pi_{k}(\mathbf{x}_{n})=\lim\frac{1}{n}=0=\pi_{k}(\mathbf{x})$,
however $\lim\mathbf{x}_{n}\neq\mathbf{x}$, because $U=\prod_{k=1}^{\infty}\left(-\frac{1}{k},\frac{1}{k}\right)$
is a neighborhood of $\mathbf{x}$. (So $\mathbf{x}_{n}$ is divergent.)

\textbf{7}. \textcolor{green}{Solution}. In product topology $\overline{\mathbf{R}^{\infty}}=\mathbf{R}^{\omega}$.
Indeed, let $\mathbf{r}\in\mathbf{R}^{\omega}$. Let $\mathbf{r}\in U$
basis element. Then $U=\prod_{\alpha=1}^{\infty}U_{\alpha}$ where
$U_{\alpha}$ denotes the entire space $\mathbf{R}$ if $\alpha\neq\beta_{1},\ldots,\beta_{n}$,
$U_{\beta_{j}}=(a_{j},b_{j})$. Now pick up $\mathbf{s}\in\mathbf{R}^{\infty}$,
$\mathbf{s}:=(s_{1},\ldots)$, where $s_{j}=0$ if $j\neq\beta_{1},\ldots,\beta_{n}$,
$s_{j}\in(a_{j},b_{j})$, $j=\beta_{1},\ldots,\beta_{n}$. Then $\mathbf{s}\in U$.

In box topology $\overline{\mathbf{R}^{\infty}}=\mathbf{R}^{\infty}$.
We show that $\mathbf{R}^{\omega}-\mathbf{R}^{\infty}$ is open. Let
$\mathbf{r}\in\mathbf{R}^{\omega}-\mathbf{R}^{\infty}$. Choose $U_{i}=\mathbf{R}$
if $r_{i}=0$, $U_{i}=\mathbf{R}-\{0\}$ if $r_{i}\neq0$. Then $U:=\prod_{i=1}^{\infty}U_{i}$
is open and if $y\in U$ then $y_{i}\neq0$ whenever $r_{i}\neq0$,
so $y\notin\mathbf{R}^{\infty}$.

\textbf{8}. \textcolor{green}{Solution}. $h$ is bijection. Obviously
$h^{-1}((y_{1},y_{2},\ldots))=\left(\frac{y_{1}-b_{1}}{a_{1}},\frac{y_{2}-b_{2}}{a_{2}},\ldots\right)$.
Since $h^{-1}$ is also linear in coordinates it is enough to investigate
$h$. It maps open intervals onto open intervals, so $h$ is a homeomorphism
in both topology.

\textbf{10}. \textcolor{green}{Solution}. (a) If $A$ has discrete
topology then functions $f_{\alpha}$ are continuous. The topology
$\tau$ is the intersection of all topologies on $A$ for which the
functions $f_{\alpha}$ are continuous, see Exe. 13.4.

(b), (c) It is the same as the product topology using functions $f_{\alpha}$
instead of $\pi_{\alpha}$.

(d) Consider first a single map $g:A\to X$, and give $A$ the initial
topology so that the open sets in $A$ are the sets of the form $g^{-1}(U)$
where $U$ is open in $X$. Then $g:A\to g(A)$ is continuous by Theorem
18.2 (d) and open because $g(g^{-1}(U))=g(A)\cap U$.

Now, the initial topology for the set maps $\{f_{\alpha}\,|\,\alpha\in J\}$
is the same as the initial topology for the single map $\mathbf{f}:=(f_{\alpha}):A\to\prod_{\alpha\in J}X_{\alpha}$.
As just observed, $\mathbf{f}:A\to f(A)$ is continuous and open.\\

Section 20 The Metric Topology

\textbf{1}. \textcolor{green}{Solution}. On a finite dimensional normed
space any two norms are equivalent.

\textbf{2}. \textcolor{green}{Solution}. In Exe. 16.9 we have seen
that the dictionary order topology on $\mathbf{R}\times\mathbf{R}$
is the same as the product topology on $\mathbf{R}_{d}\times\mathbf{R}$,
where $\mathbf{R}_{d}$ denotes $\mathbf{R}$ in the discrete topology.
The discrete topology is metrizable with the discrete metric and the
product of two metric spaces is also metrizable.

\textbf{3}. \textcolor{green}{Solution}. (a) Let $(x_{1},x_{2})\in X\times X$.
Now $|d(x_{1},x_{2})-d(y_{1},y_{2})|\leq|d(x_{1},x_{2})-d(y_{1},x_{2})|+|d(y_{1},x_{2})-d(y_{1},y_{2})|\leq d(x_{1},y_{1})+d(x_{2},y_{2})$.

(b) Denote $\tau_{m}$ the topology induced by $d$. Assume $\tau$
is a topology on $X$ such that $d:(X\times X,\tau\times\tau)\to\mathbf{R}$
is continuous. We have to show that $\tau_{m}\subset\tau$. Let $U\in\tau_{m}$
and $x\in U$. Then there exists $\varepsilon\gt 0$ such that $B_{m}(x,\varepsilon)\subseteq U$.
Since $d(x,x)=0$ and $d$ is continuous, $d^{-1}(-\varepsilon,\varepsilon)$
is a neighborhood of $(x,x)$ in $(X\times X,\tau\times\tau)$. Hence
there are neighborhoods $x\in V_{1}\in\tau$ and $x\in V_{2}\in\tau$
such that $V_{1}\times V_{2}$ is a neighborhood of $(x,x)$ and $V_{1}\times V_{2}\subset d^{-1}(-\varepsilon,\varepsilon)$.
It means if we choose the point $x\in V_{1}$ and the arbitrary point
$y\in V_{2}$ then $d(x,y)\lt\varepsilon$. Since $x\in V_{2}$ we obtain
$V_{2}\subseteq B_{m}(x,\varepsilon)$, thus $x\in V_{2}\subseteq U$
which implies that $U$ is open in $\tau$.

\textbf{4}. \textcolor{green}{Solution}. (a) We know by Theorem 20.4
$\tau_{\pi}\subset\tau_{u}\subset\tau_{b}$. Since the coordinate functions
are linear it is enough to investigate the continuity at $t=0$.

$f$ is continuous in $\tau_{\pi}$, but not in $\tau_{u},\tau_{b}$;

$g$ is continuous in $\tau_{\pi}$, but not in $\tau_{u},\tau_{b}$;

$h$ is continuous in $\tau_{\pi}$, but not in $\tau_{u},\tau_{b}$.

(b) The spaces $\tau_{\pi},\tau_{u},\tau_{b}$ are Hausdorff spaces
so the limit is unique.

$\mathbf{w}$ converges to $0$ in $\tau_{\pi}$, but not in $\tau_{u},\tau_{b}$;

$\mathbf{x}$ converges to $0$ in $\tau_{\pi}$, but not in $\tau_{u},\tau_{b}$;

$\mathbf{y}$ converges to $0$ in $\tau_{\pi}$, but not in $\tau_{u},\tau_{b}$;

$\mathbf{z}$ converges to $0$ in $\tau_{\pi},\tau_{u},\tau_{b}$.

\textbf{5}. \textcolor{green}{Solution}. $\overline{\mathbf{R}^{\infty}}=c_{0}$,
the sequences converge to $0$. Obviously $\mathbf{R}^{\infty}\subset c_{0}$.
We show that $c_{0}$ is closed in $(\mathbf{R}^{\omega},\tau_{u})$,
or, equivalently, $\emptyset\neq\mathbf{R}^{\omega}-c_{0}$ is open.
Let $\mathbf{r}=(r_{n})\in\mathbf{R}^{\omega}-c_{0}$. Since $\lim\, r_{n}\neq0$
there exists $1\gt\varepsilon\gt 0$ such that $|r_{n}|\gt\varepsilon$ for
infinitely many $n$. Then $B_{d}\left(\mathbf{r},\frac{\varepsilon}{2}\right)\subset\mathbf{R}^{\omega}-c_{0}$.

Now we show that if $\mathbf{x}=(x_{n})\in c_{0}$ then for every
$\epsilon\gt 0$ there exists $\mathbf{s}\in\mathbf{R}^{\infty}$ such
that $d_{U}(\mathbf{x},\mathbf{s})\lt\varepsilon$. We may assume that
$1\gt\varepsilon$. Since $\lim\, x_{n}=0$, therefore there exists
$n_{0}$ such that for $n\gt n_{0}(\varepsilon)$ we have $\left|x_{n}\right|\lt\varepsilon$.
Choose $s_{n}:=x_{n}$ if $n\leq n_{0}(\varepsilon)$, $s_{n}:=0$
if $n\gt n_{0}(\varepsilon)$. Then $\mathbf{s}\in\mathbf{R}^{\infty}$
and $d_{u}(\mathbf{x},\mathbf{s})\lt\varepsilon$.

\textbf{6}. \textcolor{green}{Solution}. (a) Obviously $B_{\overline{\varrho}}(\mathbf{x},\varepsilon)\subseteq U(\mathbf{x},\varepsilon)$.
Let $\mathbf{y}:=\left(x_{1}+\left(1-\frac{1}{2}\right)\varepsilon\right)\times\cdots\times\left(x_{n}+\left(1-\frac{1}{2^{n}}\right)\varepsilon\right)\times\cdots$.
Then $\mathbf{y}\in U(\mathbf{x},\varepsilon)$, but $\mathbf{y}\notin B_{\overline{\varrho}}(\mathbf{x},\varepsilon)$
because $\overline{\varrho}(\mathbf{x},\mathbf{y})=\varepsilon$.

(b) We show that $\mathbf{y}$ is a boundary point of $U(\mathbf{x},\varepsilon)$.
Let $\varepsilon\gt\delta\gt 0$ be arbitrary fixed. Then $\mathbf{z}\in B_{\overline{\varrho}}(\mathbf{y},\delta)\cap U(\mathbf{x},\varepsilon)$
where $\mathbf{z}=(z_{n})$, $z_{n}=y_{n}$ if $\frac{1}{2^{n}}\varepsilon\geq\frac{\delta}{2}$,
and $z_{n}=x_{n}+\varepsilon-\frac{\delta}{2}$ if $\frac{1}{2^{n}}\varepsilon\lt\frac{\delta}{2}$.
Similarly, $\mathbf{w}\in B_{\overline{\varrho}}(\mathbf{y},\delta)\cap(\mathbf{R}^{\omega}-U(\mathbf{x},\varepsilon))$
where $\mathbf{w}=(w_{n})$, $w_{n}=y_{n}$ if $\frac{1}{2^{n}}\varepsilon\geq\frac{\delta}{4}$,
and $w_{n}=x_{n}+\varepsilon+\frac{\delta}{4}$ if $\frac{1}{2^{n}}\varepsilon\lt\frac{\delta}{4}$.

(c) If $\delta\lt\varepsilon$ then $U(\mathbf{x},\delta)\subset B_{\overline{\varrho}}(\mathbf{x},\varepsilon)$,
consequently $\cup_{\delta\lt\varepsilon}U(\mathbf{x},\delta)\subseteq B_{\overline{\varrho}}(\mathbf{x},\varepsilon)$.
On the other hand, if $\mathbf{y}\in B_{\overline{\varrho}}(\mathbf{x},\varepsilon)$
then $\tau:=\sup\{\left|x_{n}-y_{n}\right|\}\lt\varepsilon$ and $\mathbf{y}\in U(\mathbf{x},\tau)$.

\textbf{7}. \textcolor{green}{Solution}. See Exe. 19.8.

\textbf{11}. \textcolor{green}{Solution}. Obviously $d'\lt 1$. Furthermore
$d'(x,y)=d'(y,x)$ and $d'(x,y)=0$ $\Longleftrightarrow$ $x=y$.
Now investigate the triangle inequality. We have$d'(x,z)+d'(z,y)-d'(x,y)=\frac{d(x,y)d(y,z)+d(x,z)+d(z,y)-d(x,y)}{(1+d(x,y))(1+d(x,z))}\gt 0.$
\\

Section 21 The Metric Topology II

\textbf{1}. \textcolor{green}{Solution}. Obviously $d|A\times A$
is a metric on $A$. We show that the topology on $A$ induced by
$d|A\times A$ is the same as the subspace topology on $A$ it inherits
from the space $X$. It is enough to compare their basis. If they
are not empty, the basis of $A$ inherited from $X$ has the form
$A\cap B_{d}(x,\varepsilon)$, where $x\in X$ and $\varepsilon\gt 0$
are arbitrary. Since $\emptyset\neq A\cap B_{d}(x,\varepsilon)$,
we can choose $a\in A\cap B_{d}(x,\varepsilon)$. In this case $d|A\times A(a,x):=\delta\lt\varepsilon$.
Thus $B_{d|A\times A}(a,\varepsilon-\delta)\subseteq A\cap B_{d}(x,\varepsilon)$.
In the other direction $B_{d|A\times A}(a,\varepsilon)\supseteq A\cap B_{d}(a,\varepsilon)$
(in fact we have equality).

\textbf{2}. \textcolor{green}{Solution}. $f$ is a continuous open
bijection between $X$ and $f(X)$.

\textbf{3}. \textcolor{green}{Solution}. (a) See page 122, the proof
that $\varrho$ is a metric.

(b) See the proof of Theorem 20.5.
\\

Section 22 The Quotient Topology

\textbf{2}. \textcolor{green}{Solution}. (a) Since $p$ is continuous
we have to show that $p$ is surjective and if $p^{-1}(U)$ is open
in $X$ than $U$ is open in $Y$. The assumption $p\circ f=id_{Y}$
implies the surjectivity. Furthermore using the continuity of $f$
we obtain $(f^{-1}\circ p^{-1})(U)=id_{Y}(U)=U$ is open.

(b) A retraction is a continuous surjective map. Assume $r^{-1}(U)$
is open in $X$. Since $U\subseteq A$ therefore $r^{-1}(U)=U$.

\textbf{3}. \textcolor{green}{Solution}. Obviously $q$ is surjective
and since $\pi_{1}$ is continuous so is. Let $B:=\mathbf{R}\times\{0\}\subseteq A$.
Then $q|B:B\to B$ is a homeomorphism. Suppose $V\subseteq\mathbf{R}$
and $U:=q^{-1}(V)$ is open in $A$. Then $U\cap B$ is open in $B$,
so $q(U\cap B)=V$ is open. Hence $q$ is a quotient map.

$q$ is not open, because $B(1,1)\cap A$ is open in the subspace
topology, however $q$ maps it onto $[0,1)$ which is not open. $q$
is not closed, because $C:=\{x\times y\,|\, xy=1\}$ is closed in
$A$, but its image $(0,\infty)$ is not closed.

\textbf{4}. \textcolor{green}{Solution}. (a) Define $g:\mathbf{R}^{2}\to\mathbf{R}$
by $g(x\times y):=x+y^{2}$. Then $g$ is a continuous surjection.
By Corollary 22.3 $g$ induces a bijective continuous map $f:X^{\star}\to\mathbf{R}$.
Define $h:\mathbf{R}\to\mathbf{R}^{2}$ by $h(x):=x\times0$. If $p:\mathbf{R}^{2}\to X^{\star}$
is the quotient map, then $f^{-1}=p\circ h$, which is continuous.
Therefore $f$ is a homeomorphism and $X^{\star}$ is homeomorphic
to $\mathbf{R}$.

(b) Define $g:\mathbf{R}^{2}\to[0,\infty)$ by $g(x\times y):=\sqrt{x^{2}+y^{2}}$.
Then $g$ is a continuous surjection. By Corollary 22.3 $g$ induces
a bijective continuous map $f:X^{\star}\to[0,\infty)$. Define $h:[0,\infty)\to\mathbf{R}^{2}$
by $h(x):=x^{2}\times0$. If $p:\mathbf{R}^{2}\to X^{\star}$ is the
quotient map, then $f^{-1}=p\circ h$, which is continuous. Therefore
$f$ is a homeomorphism and $X^{\star}$ is homeomorphic to $[0,\infty)$.

\textbf{5}. \textcolor{green}{Solution}. If $U\subseteq A$ is open
in $A$ then $U=A\cap V$, where $V$ is open in $X$. Since $p$
is open map the set $q(U)=p(U)=p(A)\cap p(V)$ is open in $p(A)$
in the subspace topology. If we consider $q$ as a map from $A$ to
$Y$, and $A$ is open, then $p(A)$ is open and $q(U)$ is also open
in the topology of $X$.

\textbf{6}. \textcolor{green}{Solution}. If $U$ is open in $\mathbf{R}$
then $U-K$ is open in $\mathbf{R}_{K}$ since $U=\cup_{\alpha\in J}(a_{\alpha},b_{\alpha})$
implies $U-K=\cup_{\alpha\in J}((a_{\alpha},b_{\alpha})-K)$. In particular,
$\mathbf{R}-K$ is open in $\mathbf{R}_{K}$, so $K$ is closed in
$\mathbf{R}_{K}$.

(a) A quotient space is $T_{1}$ if and only if each element of the
partition are closed subsets (see page 141, bottom lines). Now the
quotient space consists of the singletons $\{x\}$ for $x\notin K,$
which are closed in $\mathbf{R}$ and consequently in $\mathbf{R}_{K}$,
and $K$ itself, which is closed in $\mathbf{R}_{K}$. Hence $Y$
is $T_{1}$.

Now $p(0)$ and $p(K)$ are distinct points of $Y$. Let $V_{1}$
and $V_{2}$ be neighborhoods of $p(0)$ and $p(K)$, respectively.
Then $U_{1}=p^{-1}(V_{1})$ and $U_{2}=p^{-1}(V_{2})$ are open sets
in $\mathbf{R}_{K}$ containing $0$ and $K$, respectively. It means,
there exist $a\lt 0\lt b$ such that $(a,b)-K\subseteq U_{1}$. Take $n\in\mathbf{Z}_{+}$
such that $\frac{1}{n}\lt b$. Then $\frac{1}{n}\in U_{2}$, so there
exist $c\lt\frac{1}{n}\ltd$ with $(c,d)\subseteq U_{2}$. (Every basic
neighborhood of a point of $K$ is an interval.) We may assume that
$\frac{1}{n+1}\leq c$. Then $\left(c,\frac{1}{n}\right)\subseteq U_{1}\cap U_{2}$,
so $V_{1}\cap V_{2}$ is non-empty. Hence $Y$ is not Hausdorff.

(b) The diagonal $\Delta_{2}$ in $Y\times Y$ is not closed, because
$Y$ is not Hausdorff (see Exe. 17.13). Denote $\Delta_{1}$ the diagonal
in $\mathbf{R}^{2}$ which is closed in the usual topology, and hence
in $\mathbf{R}_{K}\times\mathbf{R}_{K}$. Since $K$ is closed in
$\mathbf{R}_{K}$, $K\times K$ is closed in $\mathbf{R}_{K}\times\mathbf{R}_{K}$.
Now $(p\times p)^{-1}(\Delta_{2})=\Delta_{1}\cup K\times K$ which
is closed, so $p\times p$ is not a quotient map.
\\

Section 22 Topological Groups

\textbf{2}. \textcolor{green}{Solution}. (e) Denote $G:=GL(n)$. Then
$G$ is a $T_{1}$ space. The formula for multiplying two matrices
and the formula for inverting a matrix employ only continuous functions
of the entries of the matrices.

\textbf{3}. \textcolor{green}{Solution}. Any subspace of a $T_{1}$
space is a $T_{1}$ space. The mapping $\varphi(x\times y):=xy$ of
$H\times H$ onto $H$ and the mapping $\psi(x):=x^{-1}$ of $H$
onto $H$ are continuous, since they are restrictions of the corresponding
mappings of $G\times G$ and $G$. Hence $H$ is a topological group.

Let $A$ and $B$ be subsets of a topological group $G$. Then we
have

(i) $\overline{A}\,\overline{B}\subseteq\overline{AB}$;

(ii) $\left(\overline{A}\right)^{-1}=\overline{\left(A^{-1}\right)}$.

Indeed, $\varphi(\overline{A}\times\overline{B})=\varphi(\overline{A\times B})\subseteq\overline{\varphi(A\times B)}$,
and $\psi$ is a homeomorphism of $G$.

Using these we obtain $\left(\overline{H}\right)^{2}\subseteq\overline{H^{2}}\subseteq\overline{H}$
and $\left(\overline{H}\right)^{-1}=\overline{\left(H^{-1}\right)}\subseteq\overline{H}$.

\textbf{4}. \textcolor{green}{Solution}. $f_{\alpha}$ and $\left(f_{\alpha}\right)^{-1}=f_{\alpha^{-1}}$
are continuous bijections, similar for $g_{\alpha}$.

\textbf{5}. \textcolor{green}{Solution}. (a) Left multiplication with
$\alpha$ induces a commutative diagram\begin{eqnarray*}
G & f_{\alpha}\longrightarrow & G\\
p\downarrow &  & \downarrow p\\
G/H & f_{\alpha}/H\longrightarrow & G/H\end{eqnarray*}
which shows that $f_{\alpha}/H$ is a homeomorphism.

(b) The one-point sets in $G/H$ are $\alpha H=f_{\alpha}(H)$. If
$H$ is closed then $\alpha H$ is closed because $f_{\alpha}$ is
homeomorphism.

(c) If $U$ is open in $G$ then $UH=\cup_{h\in H}Uh=\cup_{h\in H}g_{h}(U)$
is open because $g_{h}$ is a homeomorphism.

(d) By (b) $G/H$ is $T_{1}$. If $H$ is a normal subgroup then $G/H$
is a group. The map $\varphi:G\times G\to G$, $\varphi(x\times y)=xy^{-1}$
is continuous and it induces a commutative diagram \begin{eqnarray*}
G\times G & \varphi\longrightarrow & G\\
p\times p\downarrow &  & \downarrow p\\
G/H\times G/H & \varphi/H\longrightarrow & G/H\end{eqnarray*}
where $p\times p$ is a quotient map since $p$ is an open map and
the product of open maps are open map.

\textbf{6}. \textcolor{green}{Solution}. $mod\,1$. $[0,1)$.

\textbf{7}. \textcolor{green}{Solution}. (a) By continuity of the
map $x\times y\to xy$ there is a neighborhood $W$ of $e$ such that
$WW\subseteq U$. Now $V:=W\cap W^{-1}\subseteq W$ is a symmetric
neighborhood of $e$. It is clear that $VV\subseteq U$.

(b) If $x\in G$ and $e\in U$ is a neighborhood then $Vx$ is a neighborhood
of $x$. Now, let $x\neq y$. Since $xy^{-1}$ is closed there is
a neighborhood of $U$ of $e$ not containing $xy^{-1}$. Then there
exists a symmetric neighborhood $V$ of $e$ such that $VV\subseteq U$.
Then $Vx\cap Vy=\emptyset$. Indeed, assume that $v_{1}x=v_{2}y$.
Then $xy^{-1}=v_{1}^{-1}v_{2}\in V^{-1}V=VV\subseteq U$ which is

(c) Since $e\notin xA^{-1}$ which is a closed set, there is a neighborhood
$U$ of $e$ disjoint from $xA^{-1}$. Then there exists a symmetric
neighborhood $V$ of $e$ such that $VV\subseteq U$. Then $Vx\cap VA=\emptyset$.
Indeed, assume that $v_{1}x=v_{2}a$. Then $xa^{-1}=v_{1}^{-1}v_{2}\in V^{-1}V=VV\subseteq U$
which is a contradiction because $xa^{-1}\in xA^{-1}$.

(d) The point $xH$ is closed in $G/H$ (see 5 (b)). Let $x\in G$
be a point outside a saturated closed subset $A\subset G$. Then by
(c) there exists a symmetric neighborhood $V$ of $e$ such that $VV\subseteq U$
and $Vx\cap VA=\emptyset$. Then also $VxH\cap VAH=\emptyset$. Since
$p:G\to G/H$ is open (see 5 (c)), $p(x)\in p(Vx)$ and $p(A)\in p(VA)$
are disjoint open sets in $G/H$.


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