# Compact subspace of a CW complex: find the mistake in my proof

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.

• 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) – 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 $$e$$.