Metrization Theorems and Paracompactness

Munkres Topology

Theorem 1: Let $X$ be a metrizable space. If $\mathcal{A}$ is an open covering of $X$, there is an open covering $\mathcal{E}$ of $X$ refining $\mathcal{A}$ that is countably locally finite.

Proof) We use the motif of imposing an order on coverings and subtracting sets to reduce the covering to something more tractable. By the well-ordering theorem, we can impose a well-ordering $<$ on $\mathcal{A}$. For any open set $U$ and $n\in \mathbb{N}$, let $S_n(U)=\{x|B(x,1/n)\subset U\}$; note that $S_n(U)$ is closed. Now, take $T_n(U)=S_n(U)-{\bigcup }_{V<U}V$. For any $U<V$, $d(T_n(U),T_n(V))\ge 1/n$ as $d(T_n(U),T_n(V))\ge d(T_n(U),U^C)\ge d(S_n(U),U^C)\ge 1/n$. We just have to make this an open covering now; taking $E_n(U)=B(T_n(U),1/(3n))$ gives $d(E_n(U),E_n(V))\ge 1/(3n)$ for any $U$, $V$. Thus, $\{E_n(U)\}$ is a locally finite collection as for any $x\in X$, $B(x,1/(12n))$ intersects at most one element of $\{E_n(U)\}$, and $\{E_n(U){\}}_{n\in \mathbb{N}}$ is countably locally finite. As this collection is clearly a refinement of $\mathcal{A}$, it suffices to show this covers $X$. For any $x\in X$, as $\mathcal{A}$ is well-ordered there is a least element $U\in \mathcal{A}$ that contains $x$. Take $n$ such that $B(x,1/(3n))\subset U$; then $x\in E_n(U)$, completing our proof.

Recall a subset $A\subset X$ is a $G_{\delta }$ set if it is the intersection of a countable collection of open sets.

Theorem 2: Let $X$ be regular with a countably locally finite basis $\mathcal{B}$. Then, $X$ is normal and all closed sets are $G_{\delta }$.

Proof) Let $\mathcal{B}={\bigcup }_{n\in \mathbb{N}}{\mathcal{B}}_n$ where ${\mathcal{B}}_n$ is locally finite. First, we show for any open set $U\subset X$ there is a countable collection $U_n$ of open sets such that $U={\bigcup }_n{U}_n={\bigcup }_n\overline{U_n}$. This is a straightforward explicit construction; let $U_n$ be the union of elements of $B\in {\mathcal{B}}_n$ such that $\overline{B}\subset U$. By local finiteness, $\overline{U_n}=\bigcup \overline{B}\subset U$, and thus ${\bigcup }_n{U}_n\subset {\bigcup }_n\overline{U_n}\subset U$. It suffices to show ${\bigcup }_n{U}_n=U$. This holds because $\mathcal{B}$ is a basis and, by regularity, there is some $B\in \mathcal{B}$ for any $x\in U$ such that $x\subset B$ and $\overline{B}\subset U$. It naturally follows that all closed sets are $G_{\delta }$; for any closed subset $V\subset X$, $(X-V)$ is open and thus there is a countable collection $U_n$ such that $X-V={\bigcup }_n{U}_n={\bigcup }_n\overline{U_n}$. Then, $V={\bigcap }_n(X-\overline{U_n})$, which is the intersection of a countable collection of open sets.

We now have to show normality; consider any two disjoint closed sets $A$ and $B$. By our previous reasoning there exists a countable collection of sets $A_i$ such that $\bigcup A_i=\bigcup \overline{A_i}=X-A$, and thus $A_i$ is a countable open cover of $B$ such that the closures are disjoint from $A$. Similarly, we can take a countable open cover $B_i$ of $A$ such that the closures are disjoint from $B$. We use a strategy similar to Theorem 1 to extract separations of $A$ and $B$; let $C_i=A_i-{\bigcup }_{j\le i}\overline{B_j}$ and let $D_i=B_i-{\bigcup }_{j\le i}\overline{A_j}$. Clearly, $C_i$ is an open cover of $B$ such that $\overline{C_i}$ is disjoint from $A$, and vice versa. Taking $C={\bigcup }_i{C}_i$ and $D={\bigcup }_i{D}_i$ gives us our desired separation.

Theorem 3: Let $X$ be normal, and let $A$ be a closed $G_{\delta }$ set in $X$. There is a continuous function $X\to [0,1]$ such that $f(x)=0$ for $x\in A$ and $f(x)\ne 0$ for $x\notin A$.

Proof) Let $A={\bigcap }_n{A}_n$ where $A_n$ is open. By the Urysohn Lemma, we can take a continuous function $f_n:X\to [0,1]$ such that $f_n(A)=0$ and $f_n(X-A_n)=1$. Let $f={\sum }_{n\in \mathbb{N}}{f}_n/2^n$; this is a uniformly convergent function series, and thus $f$ is continuous. Clearly, $f(x)=0$ for $x\in A$ and $f(x)>0$ for $x\notin A$, as desired.

Theorem 4 (Nagata-Smirnov Metrization Theorem): A space $X$ is metrizable if and only if $X$ is regular and has a countably locally finite basis.

Proof) The forward direction is quite straightforward. If $X$ is metrizable it is clearly regular; let $\mathcal{B}$ be a basis for $X$. Then, by Theorem 1, there is an open refinement $\mathcal{E}$ of $\mathcal{B}$ that is a countably locally finite covering. This $\mathcal{E}$ must generate the topology of $X$, and thus $\mathcal{E}$ is the desired countably locally finite basis.

We now show the reverse direction, in which we replicate the structure of the Urysohn metrization theorem with more technical precision. Specifically, we aim to find an embedding of $X$ into $[0,1{]}^J$ with the uniform topology. Let $\mathcal{B}={\bigcup }_{n\in \mathbb{N}}{\mathcal{B}}_n$ be the countably locally finite basis of $X$. By Theorem 2, $X$ is normal, and by Theorem 3, for any $n\in \mathbb{N}$ and $B\in {\mathcal{B}}_n$ we can find a continuous function $f:X\to [0,1]$ such that $f(x)=0$ for $x\notin B$ and $0<f(x)\le 1/n$ for $x\in B$. Consider the collection of all such functions $\{f_{n,B}{\}}_J$ and the map $g:X\to [0,1{]}^J$ such that $x\mapsto (f_{n,B}(x){)}_J$. We claim this is our desired imbedding.

We first work in the product topology $[0,1{]}^J$ and show that $g$ is an embedding. $g$ is continuous as $f_{n,B}$ is continuous for all $(n,B)\in J$, and is injective as $X$ is $T1$. It suffices to show open sets in $X$ are mapped to open sets in $g(X)\subset [0,1{]}^J$. Consider some open set $U\subset X$ and $x\in X$; there is some $(n,B)$ such that $x\subset B$ and $\overline{B}\subset U$. Then, $f_{n,B}>0$ in $B\subset U$ only. Therefore, $g(X)\cap {\pi }_{(n,B)\in J}^{-1}((0,1])$ is an open set containing $g(x)$ that is a subset of $g(U)$, implying $g(U)$ is open. Thus, $g$ is an embedding into the product topology $[0,1{]}^J$.

We are not done; we do not know the cardinality of $J$, and thus cannot conclude the product topology $[0,1{]}^J$ is metrizable. We must go a step further and show $g$ also embeds into the uniform topology $[0,1{]}^J$; note that in $[0,1{]}^J$ the uniform metric $\rho$ is equivalent to $\rho (x,y)=\sup \{|x_i-y_i|{\}}_{i\in J}$. From our previous discussion, we know $g$ is injective and maps open sets into open sets (as the uniform topology is finer than the product topology). We only have to additionally show that $g$ is continuous. It suffices to show that for any $ϵ>0$ and $x\in X$, there is some neighborhood $U$ of $x$ such that for any $u\in U$, $\rho (g(u),g(x))<ϵ$.

Take some $N$ such that $1/N<ϵ$; we only have to concern ourselves with the behavior of $f_{n,B}$ for $n<N$. For each such $n$, ${\mathcal{B}}_n$ is locally finite and thus there is a neighborhood $U_n$ of $x$ such that $U_n$ intersects finitely many elements of ${\mathcal{B}}_n$. Furthermore, for every $B\in {\mathcal{B}}_n$ such that $B\cap U_n\ne \varnothing$, by continuity there is a neighborhood $V_{n,B}\subset U_n$ such that $|f_{n,B}(V_{n,B})-f_{n,B}(x)|<ϵ$. As there are only finitely many such $B$s for each $n$ and there are finitely many $n$s less than $N$, the total intersection

$$V=\bigcap _{n<N,B\cap U_n\ne \varnothing ,B\in {\mathcal{B}}_n}{V}_{n,B}$$

is a neighborhood of $x$ such that for any $v\in V$ $\rho (g(v),g(x))<ϵ$, as desired.

Recall that a space $X$ is paracompact if every open cover has a locally finite open refinement that also covers $X$. We show some interesting properties of paracompactness, particularly relating to partitions of unity, then prove the Smirnov metrization Theorem.

Theorem 5: Let $X$ be regular. The following conditions are equivalent: Any open covering $\mathcal{A}$ of $X$ has a refinement that is: (1) An open covering of $X$ and countably locally finite. (2) A covering of $X$ and locally finite. (3) A closed covering of $X$ and locally finite. (4) An open covering of $X$ and locally finite (i.e. $X$ is paracompact).

Proof) $(4)\Rightarrow (1)$ is trivial. Assume $(1)$ holds and let $\mathcal{B}=\bigcup {\mathcal{B}}_n$ be the refinement of $\mathcal{A}$ that is a countably locally finite open cover of $X$. We reuse the motif of shrinking sets until we get our desired structure. Let $U_n={\bigcup }_{B\in {\mathcal{B}}_n}B$ and for any $B\in {\mathcal{B}}_N$ let $S_N(B)=B-{\bigcup }_{n<N}{U}_n$. Clearly, the collection of sets $S_N(B)$ for all $N\in \mathbb{N}$ and $B\in {\mathcal{B}}_N$ is a refinement of $\mathcal{A}$ and covers $X$. It suffices to show it is locally finite; for any $x\in X$, let $n$ be the smallest number such that $x\in B$ for some $B\in {\mathcal{B}}_n$. For $i\le n$, take $V_i$ such that $V_i$ is a neighborhood of $x$ that intersects finitely many elements of ${\mathcal{B}}_i$; the intersection $V=V_1\cap V_2\cap \dots \cap V_n\cap B$ is a neighborhood of $x$ that intersects finitely many elements of $\{S_N(B)\}$.

$(2)\Rightarrow (3)$ is also quite straightforward. Let $\mathcal{B}$ be the collection of open sets $U$ such that $\overline{U}\subset A$ for some $A\in \mathcal{A}$; by regularity, $\mathcal{B}$ is an open cover of $X$ and is also a refinement of $\mathcal{A}$. By $(2)$, there is a refinement of $\mathcal{B}$ that is a locally finite covering of $X$; call this $\mathcal{C}$. Taking the set $\mathcal{D}=\{\overline{C}|C\in \mathcal{C}\}$, we get a closed refinement of $\mathcal{A}$ that is a locally finite covering of $X$, as desired.

$(3)\Rightarrow (4)$ is by far the trickiest part of this proof. We usually obtained desired properties such as local finiteness by shrinking covers in some way; to show $(3)\Rightarrow (4)$, we have to expand each set of the locally finite closed covering to obtain a locally finite open covering. This expansion must be done cautiously, however, as we must preserve local finiteness.

Let $\mathcal{B}$ be the closed refinement of $\mathcal{A}$ that is a locally finite covering. Our goal is to expand the elements of $\mathcal{B}$ to obtain our desired refinement. For any $x\in X$, there is a neighborhood $U_x$ that intersects only finitely many elements of $\mathcal{B}$; this constitutes an open covering of $X$. Take a closed refinement $\mathcal{C}$ of this collection that is locally finite; any $C\in \mathcal{C}$ only intersects finitely many elements of $\mathcal{B}$. Consider the collection $\mathcal{D}$ that we obtain by substituting each $B\in \mathcal{B}$ with the open set $X-{\bigcup }_{C\in {\mathcal{U}}_B}C$, where ${\mathcal{U}}_B$ is the subcollection of $\mathcal{C}$ consisting of elements of $C$ that are disjoint from $B$; this is guaranteed to be open by the local finiteness of $\mathcal{C}$. $\mathcal{D}$ is clearly an open covering of $X$, and is also locally finite by the following argument: every $C\in \mathcal{C}$ clearly only intersects finitely many elements of $\mathcal{D}$, and for every $x\in X$ we can take a neighborhood $W$ of $x$ that intersects finitely many elements of $\mathcal{C}$. We don’t however, have the guarantee that $\mathcal{D}$ is a refinement of $A$; this is an easy fix; for any $B\in \mathcal{B}$, let $F(B)$ denote its corresponding expanded open set in $\mathcal{D}$, and let $E(B)$ be some element of $\mathcal{A}$ that contains $B$. Let $\mathcal{E}$ be the collection of sets $F(B)\cap E(B)$; this is an open cover of $X$ as $B\subset F(B)\cap E(B)$, and is locally finite as $F(B)$ is locally finite.

We now expand the definition of partitions of unity to be indexed by any set $J$ instead of a finite subset of $\mathbb{N}$. Let $\{U_{\alpha }{\}}_{\alpha \in J}$ be an open cover of $X$. An indexed family of continuous functions ${\varphi }_{\alpha }:X\to [0,1]$ is a partition of unity dominated by $\{U_{\alpha }\}$ if: (1) $\text{support}({\varphi }_{\alpha })\subset U_{\alpha }$ (2) The indexed family $\{\text{support}({\varphi }_{\alpha }){\}}_{\alpha \in J}$ is locally finite (3) $\sum {\varphi }_{\alpha }(x)=1$ for all $x\in X$. The last condition becomes quite tractable with the observation that for every $x$ there is a neighborhood $W$ on which only finite many ${\varphi }_{\alpha }$s achieve nonzero values, and thus the limit becomes a finite sum of continuous functions.

Theorem 6: Let $X$ be a paracompact Hausdorff space. Let $\{U_{\alpha }{\}}_{\alpha \in J}$ be a open cover of $X$. There exists a locally finite indexed family $\{V_{\alpha }{\}}_{\alpha \in J}$ that is an open cover of $X$ such that $\overline{V_{\alpha }}\subset U_{\alpha }$ for all $\alpha \in J$.

Proof) Essentially, we hope to shrink the open cover and obtain local finiteness while conserving the size of the open cover. Let $\mathcal{A}$ be the collection of open sets $W$ such that $\overline{W}\subset U_{\alpha }$ for some $\alpha \in J$. As $X$ is paracompact and Hausdorff, it is normal, which forces $\mathcal{A}$ to be an open cover. $\mathcal{A}$ is an open cover of $X$, and by Theorem 5, there is a refinement $\mathcal{B}$ of $\mathcal{A}$ that is a locally finite open cover. For any $\alpha \in J$, let $V_{\alpha }$ be the union of elements in $\mathcal{B}$ that are subsets of some $W$ such that $\overline{W}\subset U_{\alpha }$. As $\mathcal{B}$ is locally finite, the collection $\{V_{\alpha }{\}}_{\alpha \in J}$ must also be locally finite. Furthermore, This is clearly a open cover of $X$ with the same index set $J$, and by local finiteness ${\overline{V}}_{\alpha }\subset U_{\alpha }$.

Theorem 7: Let $X$ be a paracompact Hausdorff space, and let $\{U_{\alpha }{\}}_{\alpha \in J}$ be an indexed open covering of $X$. There exists a partition of unity on $X$ dominated by $\{U_{\alpha }\}$.

Proof) This proof is similar to the partition of unity proof on manifolds. We apply Theorem 6 twice, obtaining a locally finite open cover $\{V_{\alpha }\}$ such that $\overline{V_{\alpha }}\subset U_{\alpha }$ and another locally finite open cover $\{W_{\alpha }\}$ such that $\overline{W_{\alpha }}\subset V_{\alpha }$. By normality, we can take a continuous ${\psi }_{\alpha }:X\to [0,1]$ such that ${\psi }_{\alpha }(\overline{W_{\alpha }})=1$ and ${\psi }_{\alpha }(X-V_{\alpha })=0$; then, $\text{support}({\psi }_{\alpha })\subset U_{\alpha }$, and the indexed family $\{\text{support}({\psi }_{\alpha }){\}}_{\alpha \in J}$ is locally finite as $\{\overline{V_{\alpha }}\}$ is locally finite.

Take $\psi (x)={\sum }_{\alpha \in J}{\psi }_{\alpha }(x)$. For any $x$, there is a neighborhood $W_x$ such that $W_x$ only intersects finitely many sets $\text{support}({\psi }_{\alpha })$. On $W_x$, $\psi (x)$ is precisely the sum of finitely many continuous functions and is thus continuous, and the sets $W_x$ form an open cover of $X$; thus, $\psi (x)$ is continuous. Furthermore, $\psi (x)>0$ for all $x\in X$, and thus taking ${\varphi }_{\alpha }={\psi }_{\alpha }/\psi$ gives us our desired partition of unity.

The following theorem demonstrates how partitions of unity can be used to “patch together” locally defined functions.

Theorem 8: Let $X$ be a paracompact Hausdorff space, and let $\mathcal{C}$ be a locally finite collection of subsets of $X$; associate each $C\in \mathcal{C}$ with a positive number ${ϵ}_C$. There exists a continuous function $f:X\to \mathbb{R}$ such that $f(x)>0$ for all $x\in X$ and $f(x)\le {ϵ}_C$ for all $x\in C$ for each $C\in \mathcal{C}$.

Proof) For every $x\in X$, there is some neighborhood $U_x$ such that $U_x$ intersects finitely many elements of $\mathcal{C}$. Thus, there is a open covering $\mathcal{B}$ of $X$ such that each element intersects finitely many elements of $\mathcal{C}$. Let $\{{\varphi }_{\beta }{\}}_{\beta \in \mathcal{B}}$ be the partition of unity dominated by $\mathcal{B}$. We define $f$ as follows: each $B\in \mathcal{B}$ intersects finitely many sets in $\mathcal{C}$; call these sets $C_1,C_2,\dots ,C_n$, and let ${\delta }_B=\min ({ϵ}_{C_1},{ϵ}_{C_2},\dots ,{ϵ}_{C_n})$ (set ${\delta }_B=1$ if $B$ doesn’t intersect any sets in $\mathcal{C}$). Then, let $f(x)={\sum }_{\beta \in B}{\varphi }_{\beta }{\delta }_{\beta }$; clearly, $f(x)>0$ for all $x\in X$. For some $x\in C$, for any $B\in \mathcal{B}$ such that $x\in B$, ${\delta }_B\le {ϵ}_C$. Therefore, $f(x)\le {ϵ}_C$, as desired. Continuity follows from the observation that for every $x\in X$ there is some neighborhood on which $f$ is the finite sum of continuous functions, as the collection $\{\text{support}({\varphi }_{\beta })\}$ is locally finite.

Theorem 9 (Smirnov Metrization Theorem): Note that a space $X$ is locally metrizable if every $x\in X$ has a neighborhood that is metrizable in the subspace topology. $X$ is metrizable if and only if it is a paracompact Hausdorff space that is locally metrizable.

Proof) By Theorems 1 and 5, if $X$ is metrizable any open covering has a refinement that is a countably locally finite open cover, and thus $X$ is paracompact. Furthermore, $X$ is clearly locally metrizable and Hausdorff; the forward direction is thus true.

For the reverse direction, it suffices to show that a locally metrizable paracompact Hausdorff space must have a countably locally finite basis. Let $\mathcal{A}$ be a locally finite open covering of $X$ such that every element $A\in \mathcal{A}$ is metrizable. For any $n\in \mathbb{N}$, let ${\mathcal{B}}_n$ be the collection of open $1/n$-balls taken within the metrics of $A\in \mathcal{A}$, and let ${\mathcal{C}}_n$ be the refinement of ${\mathcal{B}}_n$ such that ${\mathcal{C}}_n$ is a locally finite open covering of $X$. It suffices to show that the collection $\bigcup {\mathcal{C}}_n$ is a basis of $X$; take any open set $U$ and $x\in U$. $x$ is in finitely many sets $A\in \mathcal{A}$; denote these as $A_1,A_2,\dots ,A_n$. For each $A_i$, there is some ${ϵ}_i$ such that $B(x,{ϵ}_i)\subset A_i\cap U$; let $ϵ=\min \{{ϵ}_i\}$. Taking $N$ such that $1/N<ϵ/2$, there is some $C\in {\mathcal{C}}_N$ such that $x\in C$. $C$ is contained in some ball of radius $1/N$ in one of the metrics of $A_1,A_2,\dots ,A_n$, and in any case $C\subset U$.