Any compact subset $K$ of a CW-complex $X$ meets finitely many cells.
The proof of the above result in the lecture note I am following argues along the following lines $:$
Proof $:$ For each cell $e_{\alpha}$ meeting $K,$ pick a point $k_{\alpha} \in K \cap e_{\alpha}.$ Let $S = \{k_{\alpha}\}_{\alpha \in I},$ where $I$ is an indexing set consisting of all those $\alpha$ for which $K \cap e_{\alpha} \neq \varnothing.$ Let $X^{(n)}$ denote the $n$-skeleton of $X.$ Then clearly $S \cap X^{(0)}$ is closed in $X^{(0)}$ since $X^{(0)}$ is a discrete set. Assuming that $S \cap X^{n-1}$ is closed in $X^{(n-1)}.$ Now $S \cap X^{(n)}$ is the union of $S \cap X^{(n-1)}$ and the points $k_{\alpha}$ lying in the interior of the $n$-cells. So $S \cap X^{(n)}$ is also closed in $X^{(n)}$ as both these sets are closed in $X^{(n)}.$ Hence by induction $S$ is closed in $X^{(n)}.$ So $S$ is closed in $X,$ since $X$ is endowed with the weak topology.
By similar argument we can in fact show that any subset of $S$ is also closed in $X.$ Thus $S$ is discrete and $S$ being a closed subset of $K$ it follows that $S$ is compact. So $S$ has to be finite as it is both compact and discrete.
This completes the proof.
In the above proof I don't understand the bold faced sentence. Would anybody please help me understanding it?
Thanks in advance.
Best Answer
$X^{n-1} \cap S$ is closed in $X^{n-1}$ by inductive hypothesis, hence it is closed in $X^{n}$, since for any characteristic map of an $n$-cell, $\Phi$, $\Phi^{-1}(X^{n-1} \cap S) = \phi^{-1}(X^{n-1} \cap S)$ where $\phi = \Phi \restriction \partial D^n$ is the attaching map $\phi : \partial D^n \rightarrow X^{n-1}$. Since $\phi$ is continuous $\phi^{-1}(X^{n-1} \cap S)$ is a closed subset of $\partial D^n$ and thus a closed subset of $D^n$. Now the other set, namely the set of points $k_{\alpha}$ contained in $e^n_{\alpha} \cap S$ (i.e. exactly one $k_{\alpha}$ for every $e^n_{\alpha} \cap S \neq \emptyset$) has empty preimage by any characteristic map of a cell of dimension $< n$ or cell of dimension $n$ with index not equal to $\alpha$, and singleton preimage by any characteristic map of a cell of dimension $n$ with index $\alpha$, and is thus closed in $X^n$. The union of both these sets is thus closed in $X^n$.