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In Hatcher's Algebraic topology p.520 he gives the following proposition and proof about CW-complexes which I'll copy partially for clarity sake.

Propostion A.1: A compact subspace of a CW complex is contained in a finite subcomplex.

Proof: First we show that a compact set $C$ in a CW complex $X$ can meet only finitely many cells of $X$. Suppose on the contrary that there is an infinite sequence of points $x_i\in C$ all lying in distinct cells. Then the set $S:=\{x_1,x_2,x_3,\dots\}$ is closed in $X$. Namely, assuming $S\cap X^{n-1}$ is closed in $X^{n-1}$ by induction on $n$, then for each cell $e_{\alpha}^n$ of $X$, $\varphi_{\alpha}^{-1}(S)$ is closed in $\partial D_{\alpha}^n$, and $\Phi_{\alpha}^{-1}(S)$ consists of at most one more point in $D_{\alpha}^n$, so $\Phi_{\alpha}^{-1}(S)$ is closed in $D_{\alpha}^n$. Therefore $S\cap X^n$ is closed in $X^n$ for each $n$, hence $S$ is closed in $X$...

I'm having trouble with the part where he starts the induction, I think that $\varphi^{-1}(S)$ is just short for $\varphi^{-1}_{\alpha}(S\cap X^{n-1})$ since the other intersection gives you the whole space $\partial D_{\alpha}^n$ and because $\varphi_{\alpha}$ is continuous it must be the case that the preimage of a closed set is closed.

I do not follow why $\Phi_{\alpha}^{-1}(S)$ must have at most one more point in $D_{\alpha}^n$; using this part as a fact how does the closedness of $\Phi_{\alpha}^{-1}(S)$ follows?

EDIT: As suggested ,$\varphi_{\alpha}:\partial D_\alpha^n\to X^{n-1}$ is the attaching map, $\Phi_\alpha:D_\alpha^n\to X$ is the characteristic map of the $n$-cell $e_\alpha^n$.

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    $\begingroup$ You must clarify notation. What are $\varphi_\alpha$ and $\phi_\alpha$? Characteristic maps of cells, attaching maps? $\endgroup$ Commented Apr 7, 2020 at 11:46
  • $\begingroup$ They are the attaching maps and characteristic maps respectively. $\endgroup$ Commented Apr 7, 2020 at 13:51
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    $\begingroup$ You should edit your question to include that information. People do not want to read comments to understand details. Specify precisely domain and range of these maps. $\endgroup$ Commented Apr 7, 2020 at 14:14

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This is a bit late, but since no answer was chosen by the OP, I post this one.

We want to show that this is also true for any $n$-cell $e^n_\alpha$, the preimage of $S$ under its characteristic map $\Phi_\beta(S) $ is closed in the closed disk $D^n_\beta$.

There is, associated with this cell, an attaching map $\varphi_\alpha:S^{n-1}=\partial D^n_\alpha\to X^{n-1}$, which is continuous. By hypothesis $S\cap X^{n-1}$ is closed in $X^{n-1}$, then $\varphi^{-1}_\alpha(S)$ is closed in $\partial D^n_\alpha$ (As Adam remarked $\varphi_\alpha^{-1}(S)=\varphi^{-1}_\alpha(S\cap X^{n-1})$).

The boundary $\partial D^n_\alpha$ is closed in $D^n_\alpha$. Hence, so is $\varphi_\alpha^{-1}(S)$. Since there is at most one point of $S$ in $e^n_\alpha$ which is homeomorphic to $D^n_\alpha$.

Then $\Phi^{-1}_\alpha(S)=\varphi^{-1}_\alpha(S)\cup A$, where $A$ is either empty or a singleton. In any case, A is closed and so is $\Phi^{-1}(S)$ being the union of two closed sets. This implies that $S\cap X^n$ is closed in $X^n$. This is the characterization of the topology of $X$ given by Hatcher earlier in the appendix.

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  • $\begingroup$ so question, why is $S \cap X^n$ closed for each $n$? lol $\endgroup$ Commented Apr 10, 2023 at 21:21
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Note that $\varphi_\alpha^{-1}(S)$ and $\varphi_\alpha^{-1}(S\cap X^{n-1})$ coincide since $\varphi_\alpha: S^{n-1}\to X^{n-1}$. $\Phi^{-1}_\alpha(S)$ can contain at most one more point since all the $x_i$ lie in distinct cells and $\Phi_\alpha|_{(D_\alpha^n)^\circ }$ is a homeomorphism with its image.

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  • $\begingroup$ I was under the impression that $\varphi_{\alpha}^{-1}(S)$ is not well defined since $S\subset X\neq X^{n-1}$ where the latter is the function codomain. I think I understand what you say, but is kind of weird. You identify $\varphi^{-1}_\alpha(S)$ with its image under the inclusion map to $D_\alpha^n$; then because the point may be in the border the which could be a gluing point for the cells there is at least one more point in the rest of the cell, is that it? $\endgroup$ Commented Apr 7, 2020 at 0:11
  • $\begingroup$ It is somewhat debatable if we should require $A$ to be in the codomain of $f$ in order to define $f^{-1}(A)$. I do not quite understand the rest of you comment $\endgroup$ Commented Apr 7, 2020 at 0:18
  • $\begingroup$ You say that $\Phi_\alpha^{-1}(S)$ can contain at most on more point since the image of $\Phi_\alpha$ is a cell, how does it relate to $\varphi^{-1}_\alpha(S)$ ? $\endgroup$ Commented Apr 7, 2020 at 0:28

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