Is a compact simplicial complex necessarily finite?

Question

I'm aware that a finite simplicial complex is compact, and I am wondering whether the converse is true.

If we have the topological realisation of a simplicial complex (not necessarily finite), $|K|$, can we say that this is contained in a union of all it's insides (which are open) and therefore we can find a finite sub cover of this, and so it has a finite number of faces? However I'm slightly concerned that not all the insides are open, because 0-simplices only contain a single point?

Thanks

Answer 1

There is a general fact

Lemma: Let $X$ be the union of its subspaces $X_\alpha$ with the coherent topology, i.e. a subset $K$ is closed if $K\cap X_α$ is closed in $X_α$ for each $α$. Let's say $β<α$ iff $X_β\subset X_α$ Further assume:
(a) $X_α\cap X_\beta=X_\gamma$ for some $γ$ depending on $α,β$
(b) $\{β\mid β<α\}$ is finite for each $α$
(c) Each $X_α$ is $T_1$
Define $e_α=X_α-\bigcup_{β<α} X_β$. Then each subset $B$ contains a closed discrete subset $S$ with exactly one point from each $e_α$ met by $B$.
Proof: Choose one $x_α$ from $e_α\cap B$ whenever possible and let $S$ be the set of all these $x_α$. Note that $X_α\cap e_β\ne\emptyset$ implies $X_β\subseteq X_α$. From this and from (b) it follows that the $e_α$ partition $X$ and also that each $X_α$ contains only a finite number of $e_β$. We conclude that $T\cap X_α$ is closed in $X_α$ for each $T\subseteq S$ and each $α$, thus closed in $X$. Hence $S$ is closed discrete in $X$.

Corollary: Under the hypotheses of the lemma, each limit point compact $K⊆X$ is contained in some $X_{α_1}\cup...\cup X_{α_n}$

Now if $X$ is a simplicial complex, we can take the closed simplices as the $X_α$ and see that they satisfy the conditions of the lemma. This implies that a compact subset is contained in finitely many simplices. In particular, a compact simplicial complex is finite.