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.
Best Answer
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 $e$.