# Compact subset in a $\Delta$-complex structure

Suppose that $X$ is a topological space equipped with a delta-complex structure. Suppose that $K\subset X$ is compact. Prove that $K$ meets only finitely many open simplices.

I managed to find a similar proposition (Proposition A.1 in the Appendix) in Hatcher but the statement is for CW complexes.

I have an argument but I'm not sure if it's sufficient or correct. Would appreciate if somebody can comment on it.

Assume to the contrary that $K$ meets infinitely many open simplices, $\Delta_{\alpha_i},\alpha_i\in J$. Form an open cover $\{A_i\}_{i\in\mathbb N}$ s.t. $(K\cap\Delta_{\alpha_i})\subset A_i$. Then this cover has no finite subcover and hence $\bigcup (K\cap\Delta_{\alpha_i})$ is not compact. But each $\Delta_{\alpha_i}$ is compact and hence $K$ is not compact, which is a contradiction.

How do you know that the cover $\{A_i\}_\mathbb N$ has no finite subcover?
Hatcher provides a direct proof of this fact for $\Delta$-complexes on page 130. Let $C$ be a compact set intersecting infinitely many open simplices, that is, simplices with their proper faces deleted. Then it would contain an infinite sequence of points $x_i$, each lying in a different open simplex. Then the sets $U_i=X-\bigcup_{j\neq i}\{x_j\}$ are open. To see this, we have to show that $\sigma_\alpha^{-1}(U_i)$ is open in each $\Delta^{n}_\alpha$. But this preimage is just the complement of finitely many points, since a simplex has only a finite number of faces. Thus each $U_i$ is open. These $U_i$ form an open cover with no finite subcover.