Point-Set Topology

If there exists a sequence ${(a_n)}_{n\ge 1}$ in $A$ with limit $x\in X$, then every neighborhood of $x$ contains all $a_n$ for $n$ sufficiently large, so $x\in \overline{A}$ by definition. Conversely, if $x\in \overline{A}$ and $\{U_n\mid n\ge 1\}$ is a basis of open neighborhoods of $x$, then we may pick $a_n\in A\cap U_n$ for each $n\ge 1$, and the sequence ${(a_n)}_{n\ge 1}$ has limit $x$. □

Let $V$ be an open set in $Y$, let $x\in f^{-1}(V)$, and let $\{U_n\mid n\ge 1\}$ be a countable basis of open neighborhoods of $x$. If $f^{-1}(V)$ does not contain an open neighborhood of $x$, then it cannot contain any $U_n$. Assume this, and for each $n\ge 1$, choose $x_n\in U_n$ such that $x_n\notin f^{-1}(V)$. Then the sequence ${(x_n)}_{n\ge 1}$ converges to $x$, and by assumption, ${(f(x_n))}_{n\ge 1}$ converges to $f(x)$. Therefore, there exists $N\ge 1$ such that $f(x_n)\in V$ for all $n\ge N$, so $x_n\in f^{-1}(V)$ for such $n$, contradicting our assumption. □

Let $𝒰$ be an open cover of $X$, and let $B=\{B_n\mid n\ge 1\}$ be a countable base of $X$. Let $S$ be the set of positive integers $n$ such that $B_n$ is contained in an element of $U$. For each $n\in S$, let $U_n\in 𝒰$ with $B_n\subseteq U_n$. We claim that $\{U_n\mid n\in S\}$ covers $X$.

If $x\in X$, then $x\in U$ for some $U\in 𝒰$. Since ${𝔹}_n$ is a base for the topology on $X$, there exists $n\ge 1$ such that $x\in B_n\subseteq U$. By definition of $S$, we then have $n\in S$, and so $x\in U_n$. Thus, we have proven the claim. □

Let $B=\{U_n\mid n\ge 1\}$ be a countable base of $X$. For each $n\ge 1$, choose $x_n\in U_n$, and set $A=\{x_n\mid n\ge 1\}$. Then for any $x\in X$ and open neighborhood $V$ of $x$, there exists $n\ge 1$ such that $x\in U_n\subseteq V$, so in particular $x_n\in V$. Thus, $x\in \overline{A}$, so $A$ is dense in $X$. □

By Propositions 4.1.9 and 4.1.11, it suffices to show that Lindeöf spaces are second-countable and separable spaces are second-countable. Let $X$ be a metric space.

Suppose first that $X$ is Lindelöf. For positive integers $n\ge 1$, let $B_n$ be a countable subcover of the open cover of $X$ by balls of radius $\frac{1}{n}$. Let $B={\bigcup }_{n=1}^{\infty }{B}_n$, which is also countable. We claim that $B$ is a base for the metric topology on $X$. Let $x\in X$ and $𝜖>0$, and consider the open ball $B(x,𝜖)$. Choose $n$ with $\frac{2}{n}<𝜖$, and let $y\in X$ be such that $x\in B(y,\frac{1}{n})\in B_n$. Then $B(y,\frac{1}{n})\subseteq B(x,𝜖)$ by the triangle inequality, hence the claim.

Next, suppose that $X$ is separable, and let $S\subseteq X$ be a countable dense subset. We claim that the countable collection

of open balls is a base for the metric topology on $X$. Let $x\in X$ and $𝜖>0$. As $S$ is dense, $B(x,\frac{𝜖}{3})$ contains a point $y\in S$, and if we take $n\ge 1$ such that $\frac{𝜖}{3}\le \frac{1}{n}\le \frac{2𝜖}{3}$, then $x\in B(y,\frac{1}{n})\subseteq B(x,𝜖)$ by the triangle inequality, hence the claim. □

Suppose $X$ is $T_1$. If $x,y\in X$ are distinct, then $X-\{y\}$ and $X-\{x\}$ are distinct open neighborhoods of $x$ and $y$, respectively, so $X$ is $T_0$. Moreover, $\overline{\{x\}}=\{x\}$ for all $x\in X$, so for $y\ne x$, the sets $\{x\}$ and $\{y\}$ are separated. That is, $X$ is symmetric.

Conversely, if $X$ is $T_0$ and $x\in X$, then for any $y\in X$, there exists an open neighborhood $U$ containing $x$ and not $y$. If $X$ is also symmetric, then since $x$ and $y$ have distinct sets of neighborhoods, $\{x\}$ and $\{y\}$ are separated. In particular, $y\notin \overline{\{x\}}$, which is to say that $\{x\}$ is closed. □

By Lemma 4.2.7, a $T_1$-space is $T_0$, and by Lemma 1.5.9, a $T_2$-space is $T_1$.

Suppose $X$ is $T_3$. Take $x,y\in X$ with $x\ne y$. By the $T_0$-axiom, there is without loss of generality an open neighborhood $W$ of $x$ than does not contain $y$. Then $W^c$ and $x$ are disjoint, hence by the regularity condition, are contained in disjoint open sets $U$ and $V$, respectively. Then $x\in U$ and $y\in V$, so $X$ is Hausdorff.

Similarly, suppose that $X$ is $T_4$. Let $x\in X$, and let $A\subset X$ be closed with $x\notin A$. Since $X$ is $T_1$, the set $\{x\}$ is closed. Therefore, by normality there exist open disjoint sets $U$ containing $\{x\}$ and $V$ containing $A$, as desired. □

Let $X$ be regular (resp., normal). Let $A$ be a singleton (resp., closed) subset of $X$, and let $W$ be an open subset containing $A$. Then $B=W^c$ is disjoint from $A$, so by regularity (resp., normality) of $X$, there exist disjoint open sets $U$ containing $A$ and $V$ containing $B$. Then $A\subseteq U\subseteq V^c\subseteq W$. Since $V^c$ is closed, $W$ contains $\overline{U}$, so $U$ is the desired open set.

Conversely, suppose that for any singleton (resp., closed) subset $A$ of $X$, every open set of $X$ containing $A$ contains the closure of an open set containing $A$. Let $B$ be a closed set disjoint from $A$, and let $U$ be an open subset of $B^c$ containing $A$ and such that $\overline{U}\subseteq B^c$. Take $V={\overline{U}}^c$ so that $B\subseteq V$, and note that $U$ and $V$ are disjoint. Thus, $X$ is regular (resp., normal). □

Let $X={\prod }_{i\in I}{X}_i$, where $\{X_i\mid i\in I\}$ is a collection of $T_3$-topological spaces. Then $X$ is Hausdorff, and in particular, its points are closed. Let $x={(x_i)}_{i\in I}\in X$, and let $U$ be an open neighborhood of $x$. Then $U$ contains a basic open neighborhood ${\prod }_{i\in I}{U}_i$, where each $U_i$ is open in $X_i$ and all but finitely many $U_i$ equal $X_i$. For every $i$, we may choose an open neighborhood $V_i$ of $x_i$ with $\overline{V_i}\subseteq U_i$ by Lemma 4.2.13. If $U_i=X_i$, we take $V_i=X_i$ as well. Then ${\prod }_{i\in I}{V}_i$ contains $x$ and has closure ${\prod }_{i\in I}\overline{V_i}\subseteq {\prod }_{i\in I}{U}_i$, as desired. □

Let $X$ be regular and second-countable. Let $B$ be a countable base for the topology on $X$. Let $A$ and $B$ be disjoint closed subsets of $X$. By regularity, for each $x\in X$ we can find an open neighborhood $U$ of $x$ with $\overline{U}$ disjoint from $B$, and contained in $U$ we can find some neighborhood in $B$ of $x$. Together, these basis elements give a countable covering $𝒰=\{U_n\mid n\ge 1\}$ of $A$ with $\overline{U_n}\cap B=\varnothing$ for each $n\ge 1$. Similarly, we can find a countable covering $𝒱=\{V_n\mid n\ge 1\}$ of $B$ with $\overline{V_n}\cap A=\varnothing$. For each $n\ge 1$, consider the open sets

If $a\in A$, then $a\in U_n$ for some $n\ge 1$, and $a\notin \overline{V_i}$ for all $i$, so $a\in U_n^{\prime }$. The open sets $U={\bigcup }_{n\ge 1}{U}_n^{\prime }$and $V={\bigcup }_{n\ge 1}{V}_n^{\prime }$ contain $A$ and $B$, respectively, and they are disjoint, since if $u\in U_n^{\prime }$ for some $n\ge 1$, then $u\notin V_i$ for $i\le n$ by definition of $U_n^{\prime }$ and $u\notin V_i^{\prime }$for $i>n$ by definition of $V_n^{\prime }$. □

Let $A$ and $B$ be disjoint, closed subsets of a metrizable space $X$, and let $d$ be a metric on $X$. For each $a\in A$, there exists ${𝜖}_a>0$ with $B(a,{𝜖}_a)\cap B=\varnothing$, and similarly, for each $b\in B$, there exists ${\delta }_b>0$ with $B(b,{\delta }_b)\cap A=\varnothing$. Set

Then $U$ and $V$ are open containing $A$ and $B$, respectively. If $x\in U\cap V$, then there exist $a\in A$ and $b\in B$ such that $d(a,x)<\frac{{𝜖}_a}{2}$ and $d(x,b)<\frac{{\delta }_b}{2}$. By the triangle inequality, we then have $d(a,b)<\frac{{𝜖}_a}{2}+\frac{{\delta }_b}{2}\le \max ({𝜖}_a,{\delta }_b)$. If $d(a,b)<{𝜖}_a$, then $b\in B(a,{𝜖}_a)$, a contradiction, and if $d(a,b)<{\delta }_b$, then $a\in B(b,{\delta }_b)$, contradiction. Thus $U$ and $V$ are disjoint. □

Let $X$ be a compact Hausdorff space, and let $A$ and $B$ be disjoint closed subsets of $X$, which are necessarily compact. Lemma 3.2.7 implies that compact Hausdorff spaces are regular. That is, for each $b\in B$, we may choose disjoint open sets $U_b$ containing $A$ and $V_b$ containing $b$. Then the sets $V_b$ cover $B$, hence have a finite subcover, say by $V_{b_1},\dots ,V_{b_n}\in B$. Then $U={\bigcap }_{i=1}^n{U}_{b_i}$ and $V={\bigcup }_{i=1}^n{V}_{b_i}$ are disjoint open sets containing $A$ and $B$, respectively. □

Let $X$ be locally compact and Hausdorff. Let $Y$ be its one-point compactification, which is compact Hausdorff and therefore normal. As $Y$ is $T_4$, it is also $T_3$, and $X$ is $T_3$ as a subspace of $Y$. □

Let $X$ be a normal space. Set $U_1=B^c$, and by normality of $X$, pick an open set $U_0$ containing $A$ with $\overline{U_0}\subseteq U_1$. We will construct open sets $U_q$ for $q\in \mathbb{Q}\cap (0,1)$ such that if $q,r\in \mathbb{Q}\cap [0,1]$ with $q<r$, then $\overline{U_q}\subseteq U_r$.

Fix any bijection $\varphi :{\mathbb{Z}}_{\ge 0}\to \mathbb{Q}\cap [0,1]$ with $\varphi (0)=0$ and $\varphi (1)=1$. For each $n\ge 2$, let $q_n$ be the largest value of $\varphi (m)$ with $m<n$ that is less than $\varphi (n)$, and let $r_n$ be the smallest value of $\varphi (m)$ with $m<n$ that is greater than $\varphi (n)$. Suppose by induction that we have constructed $U_{\varphi (m)}$ for all $m<n$. By Lemma 4.2.13, we may choose an open set $U_{\varphi (n)}$ containing $\overline{U_{q_n}}$ and such that $\overline{U_{\varphi (n)}}$ is contained in $U_{r_n}$. For negative $r\in \mathbb{Q}$, set $U_r=\varnothing$, and for $r\in \mathbb{Q}$ greater than $1$, set $U_r=X$. In this way, we have constructed open sets $U_q$ for all rational numbers $q$ such that $\overline{U_q}\subseteq U_r$ if $q<r$ are rational numbers.

We now define $f$ by setting

for $x\in X$. If $x\in A$, then $x\in U_0$, so $f(x)=0$. If $x\in B$, then $x\notin U_1$ but $x\in U_r$ for all rationals $r>1$, so $f(x)=1$. If $x\in \overline{U_q}$, then $x\in U_r$ for all rationals $r>q$, so $f(x)\le q$. Given a nonempty open interval $(a,b)$ in $ℝ$ and $x$ in its inverse image, we then have $q,r\in \mathbb{Q}$ with $a<q<f(x)<r<b$. Since $f(x)<r$, we have $x\in U_r$, and since $f(x)>q$, we have $x\notin \overline{U_q}$. That is, we have $x\in U_r-\overline{U_q}$, which is an open neighborhood with image under $f$ contained in $[q,r]\subset (a,b)$. Thus, $f$ is continuous.

As for the converse, note that if $A$ and $B$ are disjoint closed sets in $X$ and $f:X\to [0,1]$ is a continuous function with $f(A)=\{0\}$ and $f(B)=\{1\}$, then $U=f^{-1}([0,1∕2))$ and $V=f^{-1}((1∕2,1])$ are disjoint open sets containing $A$ and $B$, respectively. □

Since $X$ is second-countable and regular, it is normal. Let $\{U_n\mid n\ge 1\}$ be a countable base of open sets in $X$. If $\overline{U_m}\subseteq U_n$, then by Urysohn’s lemma, there exists a continuous function $f_{m,n}:X\to [0,1]$ with $f_{m,n}(\overline{U_m})=\{1\}$ and $f(X-U_n)=\{0\}$.

The set $A^c$ contains some neighborhood $U_n$ of $x$, which contains the closure of some neighborhood $U_m$ of $x$ by the regularity of $X$. Then $f_{m,n}(x)=1$ and $f_{m,n}(A)=\{0\}$. Since the collection of functions $f_{m,n}$ is countable, we have the lemma. □

We may take $J=\{n\mid n\ge 1\}$. Recall that ${ℝ}^J$ is metrizable by Proposition 2.2.16, so to show metrizability, it suffices to embed our second-countable regular space $X$ in ${[0,1]}^J$. If we can do this for a given $X$, then ${[0,1]}^J$ is regular Hausdorff and second-countable, and then so is $X$.

Let ${(f_n)}_{n\ge 1}$ be a sequence of functions $f_n:X\to [0,1]$ as in Lemma 4.3.5 and use them to define a function $f={(f_n)}_{n\ge 1}:X\to {[0,1]}^J$, which is continuous as each $f_n$ is continuous. Since $X$ is Hausdorff, for any $x,y\in X$ with $x\ne y$, the set ${\{x\}}^c$ is an open neighborhood of $y$, so we can for each $x\ne y$ in $X$ find an $n\ge 1$ such that $f_n(x)=0$ and $f_n(y)=1$. Thus, $f$ is injective.

It remains to show that $f$ is an open map to its image. Let $U$ be an open set in $X$, let $a={(a_n)}_{n\ge 1}\in f(U)$, and let $x\in U$ with $f(x)=a$. Choose an $m\ge 1$ such that $f_m(x)=1$ and $f_m(y)=0$ for all $y\notin U$. Let $V$ be the open set ${\pi }_m^{-1}((0,1])\cap f(X)$ in the image of $f$, where ${\pi }_m$ is the $m$th projection map from ${\pi }_m:{[0,1]}^J\to [0,1]$. Then ${\pi }_m(a)=f_m(x)=1$, so $a\in V$. For $b\in V$, we have $b=f(y)$ for some $y$ with ${\pi }_m(b)=f_m(y)>0$, so $y\in U$ and therefore $b\in f(U)$. That is, we have $a\in f(U)\subseteq V$. Thus, $f(U)$ is open. □

It is sufficient to prove this result for maps into $[-1,1]$, $[-1,1)$ and $(-1,1)$, since all intervals in $ℝ$ are homeomorphic to one of these. We next show how the result reduces to maps to $[-1,1]$. Given any $g:X\to [-1,1]$, let $B=g^{-1}(1)$ and $C=g^{-1}(-1)$. By Urysohn’s lemma, there exist continuous functions $\varphi :X\to [0,1]$ and ${\varphi }^{\prime }:X\to [0,1]$ with $\varphi (B)\subseteq \{0\}$, $\varphi (C)={\varphi }^{\prime }(C)\subseteq \{0\}$, and $\varphi (A)={\varphi }^{\prime }(A)\subseteq \{1\}$. If $g(A)\subseteq [-1,1)$, then set $g^{\prime }=\mathit{𝜙𝑔}$, and note that $g^{\prime }{|}_A=g{|}_A$. Moreover, it satisfies $g^{\prime }(x)=0$ if $x\in C$ and $g^{\prime }(x)<1$ if $x\notin C$, so $g^{\prime }(X)\subseteq [-1,1)$. Similarly, if $g(A)\subseteq (-1,1)$, then set $g^{\prime }={\varphi }^{\prime }g$. Then $g^{\prime }{|}_A=g{|}_A$, and $g^{\prime }(X)\subseteq (-1,1)$.

We next prove a weaker result. Suppose that $F:A\to [-1,1]$ is continuous. Let

By Urysohn’s lemma, there exists a continuous function

with $G(B)=\{-\frac{1}{3}\}$ and $G(C)=\{\frac{1}{3}\}$. Then

for all $a\in A$. That is, for $a\in B\cup C$, we have $|G(a)-F(a)|\le 1-\frac{1}{3}=\frac{2}{3}$, and for $x\notin B\cup C$, we have $|G(a)-F(a)|\le \frac{1}{3}-(-\frac{1}{3})=\frac{2}{3}$.

Now let $f:A\to [-1,1]$ be continuous. Suppose by induction that for some $n\ge 0$ we have found continuous functions

for $1\le i\le n$ with

for all $a\in A$. Then the argument of the previous paragraph applied to $F={(\frac{3}{2})}^n(f-{\sum }_{i=1}^n{g}_i)$ yields the next function $g_{n+1}={(\frac{2}{3})}^{n}G$ in the induction. Since ${\sum }_{n=1}^{\infty }\frac{1}{3}{(\frac{2}{3})}^{n-1}=1$, the infinite series

is well-defined, and by definition, it restricts to $f$ on $A$. Moreover, $g$ is continuous as the partial sums ${\sum }_{i=1}^n{g}_i$ converge uniformly to $g$ on $X$: that is, $|{\sum }_{i=n+1}^{\infty }{g}_n(x)|\le {(\frac{2}{3})}^n$ for all $x\in X$. □

It suffices to show the existence of $V_1$ open with $\overline{V_1}\subset U_1$ and such that $\{U_2,\dots ,U_n,V_1\}$ covers $X$, since then we can repeat with this new cover, replacing $U_2$ and so forth. Let $A=X-{\bigcup }_{i=2}^n{U}_i$. By normality of $X$, there exists an open set $V_1$ containing $A$ with $A\subset V_1\subset \overline{V_1}\subset U_1$. Then the collection $\{V_1,U_2,\dots ,U_n\}$ covers $X$. □

By Lemma 4.3.10, we can find an open cover $𝒱=\{V_i\mid 1\le i\le n\}$ with $\overline{V_i}\subseteq U_i$ for $1\le i\le n$ and an open cover $𝒲=\{W_i\mid 1\le i\le n\}$ with $\overline{W_i}\subseteq V_i$ for each $1\le i\le n$. By Urysohn’s lemma, there exist functions ${\psi }_i:X\to [0,1]$ such that ${\psi }_i(\overline{W_i})\subseteq \{1\}$ and ${\psi }_i(V_i^c)\subseteq \{0\}$. Since ${\psi }_i^{-1}(ℝ-\{0\})\subseteq V_i$, we have $\mathrm{supp}{\psi }_i\subseteq \overline{V_i}\subseteq U_i$. For each $x\in X$, we have $x\in W_i$ for some $i$, and therefore ${\sum }_j{\psi }_j(x)\ge {\psi }_i(x)=1>0$. We may then define

for $x\in X$. Then the ${\varphi }_i$ form the desired partition of unity. □

Since $X$ is a manifold, it has an open cover $𝒰$ by open sets that may be embedded in ${ℝ}^n$ for some fixed $n$. Since $X$ is compact, there exist $U_1,\dots ,U_m\in 𝒰$ that together cover $X$. Let $f_i:U_i\to {ℝ}^n$ be an open embedding. By Theorem 4.2.18, the space $X$ is normal, so there exists a partition of unity $\{{\varphi }_i\mid 1\le i\le m\}$ subordinate to $\{U_i\mid 1\le i\le m\}$. For each $i$, set $A_i=\mathrm{supp}{\varphi }_i$ and define $g_i:X\to {ℝ}^n$ by $g_i(x)={\varphi }_i(x)f_i(x)$ for $x\in U_i$ and $g_i(x)=0$ for $x\in A_i^c$. Note that $g_i$ is well-defined as ${\varphi }_i(x)=0$ for $x\in A^c$.

We set

which is continuous as a product of continuous functions. We claim that $F$ is an embedding, which will finish the proof with $N=m(1+n)$. Since $X$ is compact, it is enough to show it is injective. If $x,y\in X$ and $F(x)=F(y)$, then ${\varphi }_i(x)={\varphi }_i(y)$ and $g_i(x)=g_i(y)$ for all $1\le i\le n$. Let $i$ be such that ${\varphi }_i(x)>0$ so that $x,y\in U_i$. Then $g_i(x)={\varphi }_i(x)f_i(x)$ and $g_i(y)={\varphi }_i(x)f_i(y)$, so $f_i(x)=f_i(y)$. As $f_i:U_i\to {ℝ}^n$ is injective, we have $x=y$. □