Some exercise solutions for John M. Lee Introduction to Topological Manifolds Second Edition

2 Topological Spaces

Exercise 2.4

(a)

• Suppose $d$ and $d^{\prime }$ generate the same topology.

  For every $x\in M,r>0$.

  Because $d$ and $d^{\prime }$ generate the same topology. $B_r^{(d)}(x)$ is open in the topology induced by b, and it is also open in the topology induced by $d^{\prime }$.

  By the definition of openness in the $d^{\prime }$-topology,

  A subset $A\subseteq M$ is said to be an open subset of $M$ if it contains an open ball around each of its points.

  For every point $x\in B_r^{(d)}(x)$, there exists some $r_1>0$ such that $B_{r_1}^{(d^{\prime })}(x)\subseteq B_r^{(d)}(x)$.

  This gives the first inclusion.

  ---

  Similarly, $B_r^{(d^{\prime })}(x)$ is open in the $d^{\prime }$-topology and hence also open in the $d$-topology.

  Thus there exists some $r_2>0$ such $B_{r_2}^{(d)}(x)\subseteq B_r^{(d^{\prime })}(x)$

  This gives the second inclusion.

  ---

  Both inclusions are obtained and the inclusion property holds for every $x\in M$ and $r>0$.

• Suppose $B_{r_1}^{(d^{\prime })}(x)\subseteq B_r^{(d)}(x)$ and $B_{r_2}^{(d)}(x)\subseteq B_r^{(d^{\prime })}(x)$. Let $T_d$ denote the topology generated by $d$ and $T_{d^{\prime }}$ the topology generated by $d^{\prime }$.

  The goal is to prove that $T_d=T_{d^{\prime }}$. By the antisymmetry of $\subseteq$, the goal can be written as $T_d\subseteq T_{d^{\prime }}\wedge T_{d^{\prime }}\subseteq T_d$.

  Let $U\in T_d$ and take any $x\in U$. By the definition of open set in the metric topology, there exists $r>0$ such that $B_r^{(d)}(x)\subseteq U$.

  According to the assumption, there exists $r_1>0$ such that $B_{r_1}^{(d^{\prime })}(x)\subseteq B_r^{(d)}(x)\subseteq U$.

  Thus every point $x\in U$ also has a $d^{\prime }$-ball around it contained in $U$. Therefore $U$ is also open in the $d^{\prime }$-topology.

  Hence $U\in T_{d^{\prime }}$, proving $T_d\subseteq T_{d^{\prime }}$

  ---

  Similarly, let $V\in T_{d^{\prime }}$ and $x\in V$. Then there exists $r>0$ such that $B_r^{(d^{\prime })}(x)\subseteq V$. By the assumption, there exists $r_2>0$ such that $B_{r_2}^{(d)}(x)\subseteq B_r^{(d^{\prime })}(x)$.

  Thus $V$ is also open in the $d$-topology, proving $T_{d^{\prime }}\subseteq T_d$.

  ---

  Both inclusions hold, so $T_d=T_{d^{\prime }}$.

  $$◻$$

(b)

Let $T_d$ denote the topology generated by $d$ and $T_{d^{\prime }}$ the topology generated by $d^{\prime }$.

Let $U\in T_d$ and take any $u\in U$, there exists $r_1>0$ such that $B_{r_1}^{(d)}(u)\subseteq U$.

$$B_{r_1}^{(d)}(u)=\{y\in M:d(u,y)<r_1\}$$

Let $V\in T_{d^{\prime }}$ and take any $v\in V$, there exists $r_2>0$ such that $B_{r_2}^{(d^{\prime })}(v)\subseteq V$.

$$B_{r_2}^{(d^{\prime })}(v)=\{y\in M:d^{\prime }(v,y)<r_2\}$$

---

Now substitute $d^{\prime }(v,y)=c\cdot d(v,y)$,

$$B_{r_2}^{(d^{\prime })}(v)=\{y\in M:d^{\prime }(v,y)<r_2\}=\{y\in M:c\cdot d(v,y)<r_2\}$$

Divide both sides by $c>0$,

$$B_{r_2}^{(d^{\prime })}(v)=\{y\in M:d(v,y)<r_2/c\}=B_{r_2/c}^{(d)}(v)$$

Hence for any $v\in V$ there exists $B_{r_2/c}^{(d)}(v)\subseteq V$, proving $V\in T_d$. Namely $T_{d^{\prime }}\subseteq T_d$.

---

Similarly, substitute $d(u,y)=d^{\prime }(u,y)/c$,

$$B_{r_1}^{(d)}(u)=\{y\in M:d(u,y)<r_1\}=\{y\in M:d^{\prime }(u,y)/c<r_1\}$$

Multiply both sides by $c$,

$$B_{r_1}^{(d)}(u)=\{y\in M:d^{\prime }(u,y)<r_1\cdot c\}=B_{c\cdot r_1}^{(d^{\prime })}(u)$$

Hence for any $u\in U$ there exists $B_{c\cdot r_1}^{(d^{\prime })}(u)\subseteq U$, proving $U\in T_{d^{\prime }}$. Namely $T_d\subseteq T_{d^{\prime }}$.

Combining both inclusions, get $T_d=T_{d^{\prime }}$.

$$◻$$

(c)

Let $T_d$ denote the topology generated by the Euclidean metric and $T_{d^{\prime }}$ the topology generated by $d^{\prime }$.

---

Let $U\in T_d$ and take any $u\in U$, there exists $r_1>0$ such that $B_{r_1}^{(d)}(u)\subseteq U$.

$$B_{r_1}^{(d)}(u)=\{y\in {\mathbb{R}}^n:d(u,y)<r_1\}$$

Let $V\in T_{d^{\prime }}$ and take any $v\in V$, there exists $r_2>0$ such that $B_{r_2}^{(d^{\prime })}(v)\subseteq V$.

$$B_{r_2}^{(d^{\prime })}(v)=\{y\in {\mathbb{R}}^n:d^{\prime }(v,y)<r_2\}$$

---

To compare these two topologies, relation between the two metrics is desired.

We can prove the following inequalities.

$$d^{\prime }(x,y)\le d(x,y)\le \sqrt{n}{d}^{\prime }(x,y)$$

• $d^{\prime }(x,y)\le d(x,y)$

  By the definition of Euclidean metric,

  $$d(x,y{)}^2=\sum _{i=1}^n(x_i-y_i{)}^2$$

  Every term $(x_i-y_i{)}^2$ is nonnegative, so for any term $j$,

  $$(x_j-y_j{)}^2\le \sum _{i=1}^n(x_i-y_i{)}^2=d(x,y{)}^2$$

  Now take the square root of both sides (valid since both sides are $\ge 0$)

  $$|x_j-y_j|\le d(x,y)$$

  Because $j$ can be any coordinate, substitute by $d^{\prime }$ is valid.

  $$d^{\prime }(x,y)\le d(x,y)$$

• $d(x,y)\le \sqrt{n}{d}^{\prime }(x,y)$

  For every coordinate $i$, $|x_i-y_i|\le d^{\prime }(x,y)$ by the definition of $d^{\prime }$.

  Squaring and then summing over all coordinates gives

  $$\sum _{i=1}^n(x_i-y_i{)}^2\le n\cdot (d^{\prime }(x,y){)}^2$$

  Now take square roots (both sides are nonnegative):

  $$d(x,y)\le \sqrt{n}{d}^{\prime }(x,y)$$

---

For each $y\in B_{r_2}^{(d^{\prime })}(v)$, $d^{\prime }(v,y)<r_2$ by definition.

Then apply above inequalities,

$$d(v,y)\le \sqrt{n}{d}^{\prime }(v,y)<\sqrt{n}{r}_2$$

That is $y\in B_{\sqrt{n}{r}_2}^{(d)}(v)$. Namely $B_{r_2}^{(d^{\prime })}(v)\subseteq B_{\sqrt{n}{r}_2}^{(d)}(v)$.

Therefore, for each $v\in V$, there exists $r_2^{\prime }>0$ (take $r_2^{\prime }=\sqrt{n}{r}_2$) such that $B_{r_2^{\prime }}^{(d)}(v)\subseteq V$.

---

Similarly, if $d(u,y)<r_1$ then $d^{\prime }(u,y)\le d(u,y)<r_1$.

That is $B_{r_1}^{(d)}(u)\subseteq B_{r_1}^{(d^{\prime })}(u)$.

Therefore, for each $u\in U$, there exists $B_{r_1}^{(d^{\prime })}(u)\subseteq U$.

---

Both inclusions hold, get $T_d=T_{d^{\prime }}$.

$$◻$$

(d)

It suffices to show that if every singleton is in the topology, the topology must be the discrete topology, because arbitrary unions of singletons can generate entire power set.

For any $x\in X$ and $r>0$,

• $r\le 1$

  Since $d(x,y)<r\le 1$ can only ocurr when $x=y$ by the definition of discrete metric.

  $$B_r^{(d)}(x)=\{y\in X:d(x,y)<r\}=\{x\}$$

• $r>1$

  Then $d(x,y)<r$ holds for all $y\in X$, because $d(x,y)$ is always $1$ or $0$.

  So $B_r^{(d)}(x)=X$.

Any subset $A\subseteq X$ can be written as a union of singletons,

$$A=\bigcup _{x\in A}x$$

Since any singleton is open and the unions of open sets are still open, every subset of $X$ is open and $d$ generates the discrete topology.

$$◻$$

(e)

By Exercise 2.4 (c), discrete metric generates the discrete topology.

It suffices to show that Euclidean metric $d(x,y)=|x-y|$ also generates the discrete topology on $\mathbb{Z}$.

For $x\in \mathbb{Z}$ and $r>0$,

$$B_r^{(d)}(x)=\{y\in \mathbb{Z}:|x-y|<r\}$$

Now consider different ranges of $r$,

• $r\le 1$

  The only integer $y$ satisfying $|x-y|<1$ is $x-y$. So $B_r^{(d)}(x)=\{x\}$.

• $r>1$

  Because it has already been proved that every singleton is open, $d$ generates the discrete topology on $\mathbb{Z}$.

Therefore, both metrics generate the same (discrete) topology.

$$◻$$

Exercise 2.5

Let $\mathcal{T}$ be the topology on $X$ and ${\mathcal{T}}_Y$ be the topology on $Y$.

Three axioms of topology space must be hold.

• $\mathrm{\varnothing },Y\in {\mathcal{T}}_Y$

  Since $\mathrm{\varnothing }\in \mathcal{T}$ and $\mathrm{\varnothing }\subseteq Y$, $\mathrm{\varnothing }\in {\mathcal{T}}_Y$.

  $Y\in \mathcal{T}$ by the given assumption and $Y\subseteq Y$, so $Y\in {\mathcal{T}}_Y$.

• ${\mathcal{T}}_Y$ is closed under arbitary unions.

  Let $\{U_i{\}}_{i\in I}$ be any collection of sets in ${\mathcal{T}}_Y$. Each $U_i$ is open in $X$ and contained in $Y$ by the definition of ${\mathcal{T}}_Y$. Then

  $$\bigcup _{i\in I}{U}_i$$

  is a union of open sets in $X$, hence open in $X$, and clearly contained in $Y$ because each $U_i\in Y$.

  Thus $\bigcup _{i\in I}{U}_i\in {\mathcal{T}}_Y$.

• ${\mathcal{T}}_Y$ is closed under finite intersections.

  Let $U_1,U_2,\dots ,U_n\in {\mathcal{T}}_Y$. Each $U_i\in \mathcal{T}$ and $U_i\subseteq Y$. Then

  $$\bigcap _{i=1}^n{U}_i$$

  is a finite intersection of open sets in $X$, hence open in $X$, and also contained in $Y$. Therefore

  $$\bigcap _{i=1}^n{U}_i\in {\mathcal{T}}_Y$$

All topology axioms are satisfied, so ${\mathcal{T}}_Y=\{U\subseteq Y:U\in \mathcal{T}\}$ is a topology on $Y$.

$$◻$$

Exercise 2.6

• $\mathrm{\varnothing },Y\in \mathcal{T}$

  Since each ${\mathcal{T}}_{\alpha }$ is a topology, $\mathrm{\varnothing },X\in {\mathcal{T}}_{\alpha }$. Hence $\mathrm{\varnothing },X\in \bigcup _{\alpha \in A}{\mathcal{T}}_{\alpha }=\mathcal{T}$.

• $\mathcal{T}$ is closed under arbitary unions.

  Let $\{U_i{\}}_{i\in I}$ be any collection of sets in $\mathcal{T}$. Then $U_i\in {\mathcal{T}}_{\alpha }$ for each $\alpha \in A$. Because $T_{\alpha }$ is a topology,

  $$\bigcup _{i\in I}{U}_i\in {\mathcal{T}}_{\alpha }$$

  Since this holds for every $\alpha$,

  $$\bigcup _{i\in I}{U}_i\in \bigcup _{\alpha \in A}{\mathcal{T}}_{\alpha }=\mathcal{T}$$

  Therefore, $\mathcal{T}$ is closed under arbitary unions.

• $\mathcal{T}$ is closed under finite intersections.

  Let $U_1,U_2,\dots ,U_n\in \mathcal{T}$. Then $U_i\in {\mathcal{T}}_{\alpha }$ for every $\alpha \in A$ and $1\le i\le n$.

  Since each ${\mathcal{T}}_{\alpha }$ is a topology,

  $$\bigcap _{i=1}^n{U}_i\in {\mathcal{T}}_{\alpha }$$

  Hence

  $$\bigcap _{i=1}^n{U}_i\in \bigcap _{\alpha \in A}{\mathcal{T}}_{\alpha }=\mathcal{T}$$

  So $\mathcal{T}$ is closed under finite intersections.

All topology axioms hold. Therefore $\mathcal{T}=\bigcap _{\alpha \in A}{\mathcal{T}}_{\alpha }$ is a topology on $X$.

$$◻$$

Exercise 2.9

(a)

• Suppose $x\in \text{Int}A$.

  By definition of the interior, there exists an open set $U$ such that $x\in U\subseteq A$.

  Because $U$ is open and contains $x$, $U$ is a neighborhood $x$. Namely $x$ has a neighborhood contained in $A$.

• Suppose $x$ has a neighborhood contained in $A$.

  That means that there exists an open set $U$ such that $x\in U\subseteq A$. Then any such $U$ is part of the union that forms $\text{Int}A$ by the definition.

  Hence $x\in \text{Int}A$.

$$◻$$

(b)

• Suppose $x\in \text{Ext}A$.

  By definition, $x\in X$ and $x\notin \bar{A}$, which means that there exists a closed set $B,B\supseteq A$ and $x\notin B$.

  Then there must exists an open set $U$ containing $x$ and $U\subseteq X\setminus A$. Thus $x$ has a neighborhood $U$ contained in $X\setminus A$.

• Suppose $x$ has a neighborhood contained in $X\setminus A$.

  That is, there exists an open set $U$ with $x\in U\subseteq X\setminus A$. Then $U\cap A=\mathrm{\varnothing }$, so there exists a closed set that is a super set of $A$ and not containing $x$, resulting in $x\notin \bar{A}$.

  Therefore $x\in X\setminus \bar{A}=\text{Ext}A$.

$$◻$$

(c)

• Suppose $x\in \partial A$.

  By definition, $x\in X$, $x\notin \text{Int}A$ and $x\notin \text{Ext}A$.

  Since $x\notin \text{Int}A$, by Exercise 2.9 (a), $x$ has no neighborhood contained entirely in $A$. Therefore, every neighborhood of $x$ intersects $X\setminus A$.

  Since $x\notin \text{Ext}A$, by by Exercise 2.9 (b), $x$ has no neighborhood contained entirely in $X\setminus A$. Therefore, every neighborhood of $x$ intersects $A$.

  Combining these two facts, every neighborhood of $x$ contains at least one point of $A$ and one point of $X\setminus A$.

• Suppose every neighborhood of $x$ meets both $A$ and $X\setminus A$.

  Then no neighborhood of $x$ can be contained in $A$, which means $x\notin \mathbf{\text{Int}}A$. And no neighborhood of $x$ can be contained in $X\setminus A$, which means $x\notin \mathbf{\text{Ext}}A$.

  Thus $x\in X\setminus (\mathbf{\text{Int}}A\cup \mathbf{\text{Ext}}A)$.

$$◻$$

(d)

• Suppose $x\in \bar{A}$.

  By definition, $x$ is in every closed set $B$ that is a superset of $A$.

  Now take any neighborhood $U$ of $x$. Its complement $X\setminus U$ is closed and does not contain $x$. Assume that $U\cap A=\mathrm{\varnothing }$, then $A\subseteq X\setminus U$, and $X\setminus U$ is a closed set containing $A$ but not $x$. That would contradict $x\in \bar{A}$, since $x$ must be in every closed set containing $A$. Hence $U\cap A\ne \mathrm{\varnothing }$.

  So every neighborhood $U$ of $x$ contains at least one point of $A$.

• Suppose every neighborhood of $x$ meets $A$.

  Let $B$ be any closed set with $A\subseteq B$, then $X\setminus B$ is open and disjoint from $A$.

  Assume that $x\notin B$ for some $B$, then $x\in X\setminus B$, which is a neighborhood of $x$ disjoint from $A$. But this contradicts the assumption that every neighborhood of $x$ meets $A$. Therefore, $x\in B$ for every such closed $B$.

  That means $x$ lies in the intersection of all closed sets containing $A$. Namely, $x\in \bar{A}$.

$$◻$$

(e)

$x\in \bar{A}$ if and only if every neighborhood of $x$ meets $A$, by Exercise 2.9 (d).

$x\in \partial A$ if and only if every neighborhood of $x$ meets both $A$ and $X\setminus A$, by Exercise 2.9 (c).

$x\in \text{Int}A$ if and only if some neighborhood of $x$ lies in $A$, by Exercise 2.9 (a).

Then

$$\bar{A}=\text{Int}A\cup \{x:\text{every neighborhood meets }A\text{ but not contained in }A\}$$

and the second set equals to $\partial A$. Hence $\bar{A}=\text{Int}A\cup \partial A$.

Since $\text{Int}A\subseteq A\subseteq \bar{A}$ and $\partial A\subseteq \bar{A}$, the same argument yeilds $\bar{A}=A\cup \partial A$.

Thus both equalities hold.

$$◻$$

(f)

$\text{Int}A$ is a union of open sets, so it is open.

$\text{Ext}A$ is the complement of closed set, so it is open.

$\bar{A}$ is an intersection of closed sets. so it is closed.

$\partial A$ is the complement of open set, so it is closed.

$$◻$$

(g)

• $A$ is open $\leftrightarrow$ $\text{Int}A$

  If $A$ is open, by definition of interior, $A$ itself is open and contained in $A$. Thus $\text{Int}A=A$.

  If $\text{Int}A=A$, and $\text{Int}A$ is open by Exercise 2.9 (f), then $A$ is open.

• $A$ is open $\leftrightarrow$ $A$ contains none of its boundary points

  If $A$ is open, then $\text{Int}A=A$, so

  $$\begin{gathered} \partial A=X\setminus (\text{Int}A\cup \text{Ext}A)=X\setminus (A\cup \text{Ext}A) \\ \partial A\cap A=\mathrm{\varnothing } \end{gathered}$$

  By Exercise 2.9 (e), $\bar{A}=A\cup \partial A$. Since $\partial A\cap A=\mathrm{\varnothing }$, $\partial A=\bar{A}\setminus A$. Hence no point of $A$ lies in $\partial A$.

  Conversely, if $A$ contains no boundary points, then $A\cap \partial A=\mathrm{\varnothing }$. Since $\bar{A}=A\cup \partial A=\text{Int}A\cup \partial A$, $A\subseteq \text{Int}A$. But alwyas $\text{Int}A\subseteq A$ by Exercise 2.9 (e), so $A=\text{Int}A$, and thus $A$ is open.

• $A$ is open $\leftrightarrow$ Every point of $A$ has a neighborhood contained in $A$

  If $A$ is open, it is itself a neighborhood of each of its points. So each $x\in A$ has a neighborhood $A$ contained in $A$.

  If every point of $A$ has a neighborhood $U_x\subseteq A$, then $A=\bigcup _{x\in A}{U}_x$, a union of open sets, hence open.

$$◻$$

(h)

• $A$ is closed $\leftrightarrow A=\bar{A}$

  If $A$ is closed, $A$ itself is a closed set containing $A$, and its intersection with any larger closed set containing $A$ is still A. So $A=\bar{A}$.

  If $A=\bar{A}$, since $\bar{A}$ is always closed by Exercise 2.9 (f), $A$ is closed.

• $A$ is closed $\leftrightarrow$ $A$ contains all its boundary points

  If $A$ is closed, then $\bar{A}=A$, hence $\bar{A}=A\cup \partial A=\bar{A}\cup \partial A$, which implies that $\partial A\subseteq A$. So $A$ contains all its boundary points.

  If $A$ contains all its boundary points, then $\partial A\subseteq A$, hence $\bar{A}=A\cup \partial A=A$. So $A$ is closed.

• $A$ is closed $\leftrightarrow$ Every point of $X\setminus A$ has a neighborhood contained in $X\setminus A$

  If $A$ is closed, then $X\setminus A$ is open. So for each $x\in X\setminus A$, the set $X\setminus A$ itself is an open neighborhood of $x$ and the set itself contained in $X\setminus A$. Hence every point of $X\setminus A$ has a neighborhood contained in $X\setminus A$.

  If every $x\in X\setminus A$ has a neighborhood (open set $U_x$) contained in $X\setminus A$, then

  $$X\setminus A=\bigcup _{x\in X\setminus A}{U}_x$$

  is open because the union of open sets is open, implying $A$ is closed.

$$◻$$

Exercise 2.10

• Suppose $A$ is closed.

  Let $x$ be a limit point of $A$. Assume that $x\notin A$ for some $x$. Then $x\in X\setminus A$, which is open. So there exists a neighborhood $U\subseteq X\setminus A$ of $x$. But then $U\cap A=\mathrm{\varnothing }$, contradicting the definition of limit point. Hence $x\in A$.

• Suppose $A$ contains all of its limit points.

  Let $x\in X\setminus A$. Since $x\notin A$ and $x$ is not a limit point of $A$, there exists an open neighborhood $U$ of $x$ such that $U\cap A=\mathrm{\varnothing }$. Then $U\subseteq X\setminus A$, so every point of $X\setminus A$ has an open neighborhood contained in it. Hence $X\setminus A$ is open, and $A$ is closed.

$$◻$$

Exercise 2.11

• Suppose a subset $A\subseteq X$ is dense.

  That is $\bar{A}=X$. Let $U\subseteq X$ be a nonempty open set. For any $x\in U$, since $x\in \bar{A}$, every neighborhood of $x$ intersects $A$ by Exercise 2.8 (d). In particular, $U$ itself is a neighborhood of $x$. Hence $U\cap A\ne \mathrm{\varnothing }$. So $U$ contains a point of $A$.

• Suppose every nonempty open subset of $X$ contains a point of $A$.

  Let $x\in X$. Every neighborhood $U$ of $x$ is open and nonempty, so $U$ contains a point of $A$. Thus every neighborhood of $x$ intersects $A$, meaning $x\in \bar{A}$. Thus $\bar{A}=X$.

$$◻$$

Exercise 2.12

Let ${\mathcal{T}}_d$ be the topology induced by the metric $d$ in a metric space.

• Suppose $x_i\to x$ in the metric space.

  For any open neighborhood $U$ of $x$ in the topology ${\mathcal{T}}_d$. By definition of ${\mathcal{T}}_d$, there exists $\epsilon >0$ such that the open ball

  $$B_{\epsilon }(x)=\{y\in X:d(x,y)<\epsilon \}$$

  is contained in $U$. By metric convergence, there exists $N$ such that $i\ge N⟹d(x_i,x)<\epsilon$.

  Hence $x_i\in B_{\epsilon }(x)\subseteq U$ for all $i\ge N$. Thus $x_i\to x$ in the topology sense.

• Suppose $x_i\to x$ in the topology space.

  Fix $\epsilon >0$. The ball $B_{\epsilon }(x)$ is open in ${\mathcal{T}}_d$, hence a neighborhood of $x$. By topological convergence, there exists $N$ such that $i\ge N⟹x_i\in B_{\epsilon }(x_i)$, namely $d(x_i,x)<\epsilon$. Thus $x_i\to x$ in the metric sense.

$$◻$$

Exercise 2.13

• Suppose $x_i\to x$ in $X$.

  Then $\{x\}$ is an open neighborhood of $x$ since $X$ is a discrete topology space. By the definition of convergence, there exists $N$ such that $i\ge N⟹x_i\in \{x\}$. That is $x_i=x$ for all $i\ge N$. Hence the sequence is eventually constant.

• Suppose $(x_i)$ is eventually constant.

  That is there exists $N$ such that $x_i=x$ for all $i\ge N$. Then for any open set $U$ containing $x$, $x_i=x\in U$ holds for all $i\ge N$. Therefore $x_i\to x$.

$$◻$$

Exercise 2.14

Let $U$ be any open neighborhood of $x$. Since $x_i\to x$, by the definition of convergence, there exists $N$ such that $i\ge N⟹x_i\in U$. Each $x_i\in A$ by hypothesis, so $x_i\in A\cap U$ for all $i\ge N$. Thus every neighborhood of $x$ meets $A$. By Exercise 2.9 (d), it follows that $x\in \bar{A}$.

$$◻$$

Exercise 2.16

Let $f:X\to Y$ be a map between topological spaces.

• Suppose $f$ is continuous.

  For any closed set $C\subseteq Y$. Then $Y\setminus C$ is open in $Y$. By continuity of $f$, the preimage $f^{-1}(Y\setminus C)$ is open in $X$.

  By the preimage law

  $$f^{-1}(Y\setminus C)=X\setminus f^{-1}(C)$$

  Since $X\setminus f^{-1}(C)$ is open, $f^{-1}(C)$ must be closed in $X$. Hence preimages of closed sets are closed.

• Suppose the preimage of every closed subset is closed.

  That is, for every closed set $C\subseteq Y$, the set $f^{-1}(C)$ is closed in $X$. Then $Y\setminus C$ is open in $Y$. Similarly

  $$f^{-1}(Y\setminus C)=X\setminus f^{-1}(C)$$

  Because $f^{-1}(C)$ is closed, its complement $X\setminus f^{-1}(C)$ is open. Thus preimages of open sets are open, proving $f$ is continuous.

$$◻$$

Exercise 2.18

(a)

Let $f(x)=c$ for some fixed $x\in Y$, and $U$ is any open set in $Y$. Then

$$f^{-1}(U)=\{\begin{array}{ll}X,& c\in U\\ \mathrm{\varnothing },& c\notin U\end{array}$$

Both $X$ and $\text{empty}$ are open in $X$. Hence $f^{-1}(U)$ is open for every open set $U\subseteq Y$, proving $f$ is continuous.

(b)

For any open set $U\subseteq X$,

$${\text{Id}}_X^{-1}(U)=\{x\in X:{\text{Id}}_X(x)\in U\}=U$$

Since $U$ is open and its preimage $U$ is open, ${\text{Id}}_X$ is continuous.

(c)

Let $U\subseteq Y$ be open, and define $f{|}_U:U\to Y$ by $f{|}_U(x)=f(x)$.

Let $V\subseteq Y$ be open. By definition of preimage and restrictions:

$$(f{|}_U{)}^{-1}(V)=\{x\in U:f(x)\in V\}=U\cap f^{-1}(V)$$

Since $f$ is continuous, $f^{-1}(V)$ is open in $X$. Hence the intersection $U\cap f^{-1}(V)$ of two open sets is open in $U$ (by the subspace topology definition). Thus $f{|}_U$ is continuous.

$$◻$$

Exercise 2.20

• Reflexivity

  For every topology space $X$, the identity map

  $${\text{id}}_X:X\to X,{\text{id}}_X(x)=x$$

  is bijective and continuous, and its inverse is itself (also continuous). Hence $X$ is homeomorphic to itself.

• Symmetry

  Suppose $f:X\to Y$ is a homeomorphism. Then $f$ is bijective, continuous, and $f^{-1}:X\to Y$ is continuous. Therefore $f^{-1}$ is a homeomorphism from $Y$ to $X$. Hence if $X$ is homeomorphic to $Y$, then $Y$ is homeomorphic to $X$.

• Transitivity

  Suppose $f:X\to Y$ and $g:Y\to Z$ are homeomorphisms. Then both $f$ and $g$ are bijective and continuous, with continuous inverses. The composition

  $$g\circ f:X\to Z$$

  is bijective (since a composition of bijections is bijective), continuous (composition of continuous maps is continuous), and has continuous inverse

  $$(g\circ f{)}^{-1}=f^{-1}\circ g^{-1}$$

  Thus $g\circ f$ is a homeomorphism from $X$ to $Z$. Hence if $X$ is homeomorphic to $Y$ and $Y$ to $Z$, then $X$ is homeomorphic to $Z$.

$$◻$$

Exercise 2.21

• Suppose $f$ is a homeomorphism.

  Because $f$ is continuous, for any open set $V\in {\mathcal{T}}_2$, $f^{-1}(V)\in {\mathcal{T}}_1$.

  Because $f^{-1}$ is continuous, for any $U\in {\mathcal{T}}_1$, $f(U)\in {\mathcal{T}}_2$.

  Thus $f(U)\in {\mathcal{T}}_2$ if $U\in {\mathcal{T}}_1$ and $f^{-1}(V)\in {\mathcal{T}}_1$ if $V\in {\mathcal{T}}_2$. Therefore $f({\mathcal{T}}_1)={\mathcal{T}}_2$.

• Suppose $f({\mathcal{T}}_1)={\mathcal{T}}_2$.

  • Continuity of $f$.

    Let $V\in {\mathcal{T}}_2$. Since $f({\mathcal{T}}_1)={\mathcal{T}}_2$, there exists $U\in {\mathcal{T}}_1$ such that $V=f(U)$. Then $f^{-1}(V)=f^{-1}(f(U))=U\in {\mathcal{T}}_1$. Thus $f$ is continuous.

  • Continuity of $f^{-1}$.

    Let $U\in {\mathcal{T}}_1$. Then $f(U)\in {\mathcal{T}}_2$. Since $(f^{-1}{)}^{-1}(U)=f(U)\in {\mathcal{T}}_2$, $f^{-1}$ is continuous.

  Therefore $f$ is a homeomorphism.

$$◻$$

Exercise 2.22

Since $f$ is a homeomorphism, both $f$ and $f^{-1}$ are continuous. Because $f^{-1}$ is continuous, the preimage under $f^{-1}$ of any open set in $X$ is open in $Y$. That is $(f^{-1}{)}^{-1}(U)=f(U)$, which means $f(U)$ is open in $Y$.

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The next goal is to prove that $f{|}_U:U\to f(U)$ is a homeomorphism.

• Bijectivity

  Since $f$ is bijective, the restriction $f{|}_U$ is bijective from $U$ onto $f(U)$.

• Continuouity

  From Exercise 2.18 (c), the restriction of a continuous map to an open subset of its domain is continuous. Thus $f{|}_U$ is continuous.

• Inverse Continuouity

  The inverse of $f{|}_U$ is the restriction of $f^{-1}$ to $f(U)$. That is $(f{|}_U{)}^{-1}=f^{-1}{|}_{f(u)}$. Since $f^{-1}$ is continuous and $f(U)$ is open in $Y$, apply Exercise 2.18 (c) again shows that the restriction $f^{-1}{|}_{f(u)}$ is continuous.

Therefore, $f{|}_U$ is bijective, continuous, and has a continuous inverse.

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Exercise 2.23

By the definition of continuouity,

$${\text{id}}_X\text{ is continuous}\leftrightarrow {\text{id}}_X^{-1}(U)\in {\mathcal{T}}_1\text{ for all }U\in {\mathcal{T}}_2.$$

Because ${\text{id}}_X^{-1}(U)=U$, this becomes,

$$U\in {\mathcal{T}}_1\text{ for all }U\in {\mathcal{T}}_2$$

Namely ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$, which is the definition of ${\mathcal{T}}_1$ being finer than ${\mathcal{T}}_2$.

Thus

$${\text{id}}_X\text{ is continuous}\leftrightarrow {\mathcal{T}}_1\text{ is finer than }{\mathcal{T}}_2.$$

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By the definition of homeomorphism,

${\text{id}}_X$ is a homeomorphism if and only if ${\text{id}}_X$ is continuous, and its inverse is continuous.The second condition means that

$${\text{id}}_X^{-1}:(X,{\mathcal{T}}_2)\to (X,{\mathcal{T}}_1)\text{ is continuous }\leftrightarrow {\mathcal{T}}_2\text{ is finer than }{\mathcal{T}}_1$$

Combine with previous part,

$${\text{id}}_X\text{ is continuous }\leftrightarrow {\mathcal{T}}_1\text{ is finer than }{\mathcal{T}}_2.$$

Hence both ${\mathcal{T}}_1\subseteq {\mathcal{T}}_2$ and ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$ must hold.

Thus,

$${\text{id}}_X\text{ is a homeomorphism }\leftrightarrow {\mathcal{T}}_1={\mathcal{T}}_2.$$$$◻$$

Exercise 2.33

Let $\mathcal{T}=\{\mathrm{\varnothing },Y\}$ be the trivial topology of $Y$, $(y_n)$ be any sequence of $Y$, and let $y\in Y$ be any point.

In the trivial topology, the only open set containing $y$ is $Y$ it self. Since $Y$ also contains every term of the sequence, by definition, $y$ is the limit point of sequence $(y_n)$.

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Exercise 2.35

Let $p,q\in X,p\ne q$. By assumption, there is a continuous function $f_p:X\to \mathbb{R}$ such that $f_p^{-1}(0)=p$. Since $p\ne q$, $f_p(q)\ne 0$.

Let

$$\epsilon =\frac{|f_p(q)|}{2}>0$$

Define the open sets,

$$U=f_p^{-1}(-\epsilon ,\epsilon ),V=f_p^{-1}(\mathbb{R}\setminus [-\epsilon ,\epsilon ])$$

Both sets are open because $f_p$ is continuous.

Then

• $p\in U$ because $f_p(p)=0\in (-\epsilon ,\epsilon )$.

• $q\in V$ because $|f_p(q)|>\epsilon$.

• $U\cap V=\mathrm{\varnothing }$ by construction.

Thus every pair $p\ne q$ is contained in disjoint open neighborhood, proving $X$ is Hausdorff.

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Exercise 2.40

• Suppose $U\in \mathcal{T}$,

  Let $p\in U$. Since $\mathcal{B}$ is a basis for $\mathcal{T}$, every open set (including $U$) is a union of basis elements. Therefore, there exists some basis element $B\in \mathcal{B}$ such that $p\in B\subseteq U$.

• Suppose for each $p\in U$, there exists $B_p\in \mathcal{B}$ such that $p\in B_p\subseteq U$.

  Then $U=\bigcup _{p\in U}{B}_p$. Each $B_p$ is open because basis elements are open. A union of open sets is open, so $U$ is open.

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Exercise 2.42 (a)

• Every cube ${\mathbf{C}}_s(x)$ is open.

  By definition of open cube, it can be rewritten as

  $${\mathbf{C}}_s(x)=\prod _{i=1}^n(x_i-s/2,x_i+s/2)$$

  a Cartesian product of open intervals. Since products of open intervals are exactly the usual Euclidean open boxes, each ${\mathbf{C}}_s(s)$ is an open subset of ${\mathbb{R}}^n$.

• Every Euclidean open set is a union of cubes ${\mathbf{C}}_s(x)$.

  Let $U\subseteq {\mathbb{R}}^n$ be an open set. Fix $p\in U$, there exists an open ball $B_r(p)\subseteq U,r>0$ since $U$ is open. Consider the cube with side length $s=\frac{2r}{\sqrt{n}}$. Then for any $y\in {\mathbf{C}}_s(p)$,

  $$|y_i-p_i|<s/2$$$$\parallel y-p{\parallel }_2<\sqrt{\sum _{i=1}^n(s/2{)}^2}=\sqrt{n}\frac{s}{2}=r$$

  Thus ${\mathbf{C}}_s(p)\subseteq B_r(p)\subseteq U$. Therefore every point of $U$ is contained in some cube entirely contained in $U$. Hence any Euclidean open set is a union of members of ${\mathcal{B}}_1$.

Exercise 2.51

Let $\mathcal{B}=\{B_n:n\in \mathbb{N}\}$ be a countable basis for the topology on $X$.

Define

$$D=\{x_n:B_n\ne \mathrm{\varnothing },x_n\in B_n\}$$

Since $D$ is constructed by choosing one $x_n$ from each $B_n$, $D$ is countable.

Let $U\subseteq X$ be any nonempty open set. Since $\mathcal{B}$ is a basis, there exists some $B_k\in \mathcal{B}$ such that $B_k\subseteq U$ and $B_k\ne \mathrm{\varnothing }$. Then $x_k\in D\cap B_k$ and $D\cap B_k\subseteq D\cap U$. Hence every nonempty open set intersects $D$, by Exercise 2.11, $D$ is dense.

Thus $D$ is a countable dense subset of $X$.

Exercise 2.54

• Suppose $X$ is a $0$-manifold.

  For each $x\in X$, there exists an open $U_x\subseteq X$ and a homeomorphism ${\phi }_x:U_x\to \{0\}$. Since ${\phi }_x$ is a homeomorphism and $\{0\}$ is open in itself, $U_x={\phi }_x^{-1}(\{0\})$ is open. But $U_x$ contains exactly one point because $\phi$ is injective. Hence $U_x=\{x\}$ is open. So $X$ is discrete.

  ---

  A $0$-manifold is second countable by definition. Therefore a discrete second-countable space must be countable, because a discrete space has the basis $\{\{x\}:x\in X\}$. A basis must be countable, so $X$ must be countable.

  Thus $X$ is a countable discrete space.

• Suppose $X$ is a discrete countable space.

  In a discrete space, every point $x\in X$ has an open neighborhood $U_x=\{x\}$, which is open. Then the map $\{x\}\to \{0\},x\mapsto 0$ is a homeomorphism onto $\{0\}$. So each point has a neighborhood homeomorphic to an open subset of ${\mathbb{R}}^0$. And a countable discrete space is second countable and Hausdorff.

  Thus $X$ satisfies the definition of a $0$-manifold.

$$◻$$