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The following is from Theorem 5.14 of John Lee's Introduction to Topological Manifolds.

Let $X$ be a CW complex. A subset of $X$ is compact if and only if it is closed in $X$ and contained in a finite subcomplex.

The proof for the only if direction is given as: suppose $K \subset X$ is compact. If $K$ intersects infinitely many cells, by choosing one point of $K$ in each such cell we obtain an infinite closed and discrete subset of $K$, which is impossible.

My question is why is this subset, say $A$, a closed subset of $K$? I think this is because $A'=\emptyset$, but I can't give a rigorous proof of this. I would greatly appreciate any help.

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Let us prove a slightly more general theorem:

Let $A$ is a subset of $X$ such that $A \cap e$ is finite [this allows being empty] for each cell $e$ of $X$. Then $A$ is closed in $X$.

The set $A$ is closed iff $A \cap \bar e$ is closed in $\bar e$ for all cells $e$ of $X$. We can prove this by induction over the dimension of cells.

The base case $\dim e = 0$ is trivial.

Assume that $A \cap \bar e$ is closed in $\bar e$ for all cells $e$ of $X$ with $\dim e \le n$. This means that $A \cap X^n$ is closed in the $n$-skeleton $X^n$ and hence closed in $X$.

We have to show that $A \cap \bar e$ is closed in $\bar e$ for all cells $e$ of $X$ with $\dim e = n+1$.

Let $e$ be such a cell. We have $A \cap \bar e = (A \cap X^n) \cap \bar e \cup (A \cap e)$. Since $A \cap X^n$ is closed in $X$, $(A \cap X^n) \cap \bar e$ is closed in $\bar e$. But $A \cap e$ is finite, thus it is closed in $\bar e$. Since unions of closed sets are closed, we see that $A \cap \bar e$ is closed in $\bar e$.

By the way, the above argument also shows that $A$ is a discrete subspace of $X$. In fact, all subsets of $A$ are closed in $X$ and hence also closed in $A$. This means $A$ carries the discrete topology.

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Remember how the weak topology works! It suffices for $A$ to be closed in each cell. Do you see why this is true? This will prove that $A$ is closed in $X$ and hence in $K$.

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  • $\begingroup$ Yes the definition I have is $A$ is closed iff $A\cap \bar{e}$ is closed in $\bar{e}$ for each (open) cell $e$. but the difficulty I have is well if $A \cap e$ is a singleton for each chosen cell $e$, how do I guarantee that $A \cap \bar{e}$ will still be singleton? $\endgroup$ Commented Jan 15, 2024 at 6:33
  • $\begingroup$ Aren't you done from showing that $A \cap e$ is a singleton? You only need to be closed in each cell. $\endgroup$ Commented Jan 15, 2024 at 6:35
  • $\begingroup$ I need $A \cap \bar{e}$ to be closed in $\bar{e}$. $\endgroup$ Commented Jan 15, 2024 at 6:46
  • $\begingroup$ Ok I think I need to use (C) condition as well. So for each cell $e$, $A \cap e$ is at most singleton. Since $\bar{e}$ is contained in a union of finitely many cells, so is $\bar{e} - e$. Therefore, $A \cap \bar{e}= (A \cap e) \cup (A \cap \bar{e}-e)$ is at most a union of finite points, hence closed in $\bar{e}$. $\endgroup$ Commented Jan 15, 2024 at 6:51
  • $\begingroup$ Oh, my bad! Yes, you need compactness to see that each closed cell is contained a union of finitely many (open) cells. Now the weak topology will do. $\endgroup$ Commented Jan 15, 2024 at 7:04

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