Let $A$ be a compact subset of a CW complex $X$. I want to show that $A$ intersects only finitely many open cells of $X$. I know you can prove this by taking the set containing one point from the intersection of $A$ with each open cell, and showing that this set is compact and discrete, and thus must be finite. (See for example this question.)

I'm curious where I have made a mistake in my attempt at a different proof of the same result:

Let $\{e_{\alpha}\}_{\alpha \in \mathcal{B}}$ be the collection of open cells of $X$ which intersect $A$ non-trivially. Since $X$ is a disjoint union of open cells, it must be that $A \subset \bigcup_{\alpha \in \mathcal{B}} e_{\alpha}$, and thus $\{e_{\alpha}\}_{\alpha \in \mathcal{B}}$ forms an open cover of $A$. Since $A$ is compact, there exists a finite subset $\{e_1, \cdots, e_k\} \subset \{e_{\alpha}\}_{\alpha \in \mathcal{B}}$ such that $A \subset \bigcup_{i=1}^k e_{i}$. Since for all $\alpha \in \mathcal{B}, e_{\alpha} \cap A \neq \emptyset$, we must have that $e_{\alpha} \cap \bigcup_{i=1}^k e_{i} \neq \emptyset$ for all $\alpha \in \mathcal{B}$. As all the open cells in $X$ are disjoint, if the indexing set $\mathcal{B}$ is infinite then there exists $1\leq j\leq k$ such that infinitely many $\alpha \in \mathcal{B}$ satisfy that $e_{\alpha} \cap e_{j} \neq \emptyset$. Again by disjointness, for infinitely many $\alpha \in \mathcal{B}, e_{\alpha} = e_j$. This is impossible since all the cells in $\{e_{\alpha}\}_{\alpha \in \mathcal{B}}$ are distinct.

I'm really worried that this does not use the closure-finiteness of CW complexes which is so crucial in the first proof.

  • $\begingroup$ Closure finiteness essentially comes free from compactness by considering the size of covers. You are working with the cover definition of compactness directly; your idea is not a bad one at all as far I can tell (though I don't feel qualified to judge your proof given I'm studying the same material) $\endgroup$ – Brevan Ellefsen Aug 20 at 3:47

Your proof is not correct because the open cells of $X$ are in general no open subspaces of $X$.

The only open $n$-cells which are open as subspaces are those which do not meet the image of any characteristic map of a cell of dimension $> n$. For example, a $0$-cell is open as a subspace iff it is an isolated point of $X$.

The name "open $n$-cell" is perhaps misleading. The reason for this notation is that they are homeomorphic copies of open $n$-balls.

PS. If all open cells are open subspaces, then also all complements of open cells are open, i.e. all open cells are connected components of $X$. This is true only for $0$-dimensional CW-complexes. For any open $n$-cell $e$ of dimension $n > 0$ we have $\overline{e} \cap X^{(n-1)} \ne \emptyset$, i.e. $e \subsetneqq \overline{e}$ which is impossible for a connected component $e$.


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