CW implies normal

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., CW-space) must also satisfy the second topological space property (i.e., normal space)
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This article involves a proof using cellular induction, viz, it inductive construction on the $n$-skeleton of a cellular space

Statement

Every CW-space (viz every space which can be given a CW-complex structure) is normal, viz it is Hausdorff and any two disjoint closed sets can be separated by disjoint open sets. (note that Hausdorffness follows from one of the definitions, so we only prove normality).

Proof

Goal of the proof

Let $X$ be a CW-space. Equip $X$ with a CW-complex structure and let $X^n$ denote the $n$-skeleton with respect to that structure.

Let $A,B\subset X$ be closed subsets. The goal is to construct a function $f:X\to [0,1]$ such that $f(A)=0$ and $f(B)=1$ (this will show that $A$ and $B$ are separated by disjoint open sets).

To construct this $f$, we construct a family of functions $f_n$ on the $n$-skeletons, such that $f_n$ restricted to $X^m$ is $f_m$ for $m\le n$, and such that $f_n(A\cap X^n)=0$ and $f_n(B\cap X^n)=1$.

Proof details

Suppose we have constructed $f_{n-1}:X^{n-1}\to I$ which takes the value $0$ on $A\cap X^{n-1}$ and $1$ on $B\cap X^{n-1}$. We need to use this to define $f_n$ on $X^n$, which extends $f_{n-1}$.

Now we extend the function separately on the interior of each disc. Note that the choice of how we extend the function on the interior of one disc, does not affect the choice on the interior of any other disc.

For a disc corresponding to a cell $e_i$ attached via a map ${\mathrm{\Phi }}_i$, we have the following data:

• A map $f_{n-1}\circ {\mathrm{\Phi }}_i:S^{n-1}\to I$
• A subset $A^{\prime }$ of $D^n$, which is $A\cap e_i$, along with all points on $S^{n-1}$ whose image via ${\mathrm{\Phi }}_i$ lies inside $A$
• A corresponding subset $B^{\prime }$ of $D^n$

Consider $S^{n-1}\cup A^{\prime }\cup B^{\prime }=P$ as a subset of $D^n$. This is a closed subset. Define $g:P\to I$ such that $g(A^{\prime })=0$, $g(B^{\prime })=1$ and $g{|}_{S^{n-1}}=f_{n-1}\circ {\mathrm{\Phi }}_i$. The well-definedness and continuity of $g$ follow from the properties of $f_{n-1}$, the fact that all three subsets are closed, and the gluing lemma.

Since $D^n$ is normal, the function $g:P\to I$ extends to a continuous map $g^{\prime }:D^n\to I$. The restriction of $g^{\prime }$ to the interior of $D^n$, is the extension we require.

Related results

• CW implies perfectly normal
• CW implies paracompact: This is a stronger result because any paracompact Hausdorff space is normal