Subset of a CW complex containing one point from each cell is a closed discrete subset.

Question

Suppose that $K$ is a compact subset of a CW complex space $X$. If $K$ intersects infinitely many cells, by choosing one point of $K$ in each such cell we obtain an infinite subset of $K$, say $A$. In this case, how can we show that $A$ is a closed discrete subset of $K$?

$A$ being closed means that for every cell $e$ in $X$, $A \cap \bar{e}$ is closed in $\bar{e}$. Since $\bar{e}$ is contained in finitely many cells, we would always have $A \cap \bar{e}$ be some finite set. Since finite sets are closed in $\bar{e}$, this shows that $A$ is closed.

Discrete means that every singleton of $A$ is open in $K$. The problem I have is that each cell is not guaranteed to be open so I can't think of a neighborhood for each point.
I would greatly appreciate any help with this part.

Answer 1

To say that $A$ is discrete means that it subspace topology is the discrete topology, meaning that every singleton is open in $A$, not in $K$. Actually, it is easier to instead show that every subset of $A$ is closed in $A$. It suffices to show that an arbitrary subset $B\subseteq A$ is closed in $X$. But you can prove this in exactly the same way that you proved $A$ is closed in $X$, since $B\cap\overline{e}$ is still finite for each cell $e$.

Answer 2

You cannot show that $A$ is closed and discrete, because no such sets are infinite. Any CW complex is metrizable (it's Hausdorff, paracompact, and locally metrizable; now apply Smirnov metrization), so $K$ is metrizable too. But an infinite subset of a compact metrizable space has a limit point, and thus is not discrete.

Answer 3

Its not hard to prove the more general result that if a subset $S$ (compact or not) intersects each cell in at most finitely many points, then it is discrete. You only need basic facts about $CW$ complexes and some topological definitions.

Here's the spoiler (proof), if you get really stuck:

Use the fact that every cell $e$ is contained in a finite subcomplex, for then $S\cap \overline e$ is also finite and therefore closed in $\overline e.$ But then, since we're working in a $CW$ complex $X$, this means that $S$ is closed in $X,$ too. Repeating this argument for an arbitrary $\textit{subset}\ S'$ of $S$ shows that every subset of $S$ is closed in $X$ and now by definition, this means that $S$ is discrete.