# Kerodon

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Proposition 3.5.1.10. Let $X$ be a simplicial set. Then a subset $K \subseteq |X|$ is compact if and only if it is closed and contained in $|X'| \subseteq |X|$, for some finite simplicial subset $X' \subseteq X$.

Proof of Proposition 3.5.1.10. Let $X$ be a simplicial set. If $X' \subseteq X$ is a finite simplicial subset, then the geometric realization $| X' |$ is a continuous image of a finite disjoint union $\coprod _{i \in I} | \Delta ^{n_ i} |$ (Proposition 3.5.1.7), and is therefore compact. It follows that any closed subset $K \subseteq |X'|$ is also compact. Conversely suppose that $K \subseteq |X|$ is compact. Since $|X|$ is Hausdorff, the set $K$ is closed. We wish to show that $K$ is contained in $|X'|$ for some finite simplicial subset $X' \subseteq X$. Suppose otherwise. Then we can choose an infinite collection of nondegenerate simplices $\{ \sigma _{j}: \Delta ^{n_ j} \rightarrow X \} _{j \in J}$ for which each of the corresponding cells $\mathring {| \Delta ^{n_ j} |} \hookrightarrow |X|$ contains some point $x_ j \in K$. Applying Lemma 3.5.1.12, we deduce that for every subset $J' \subseteq J$, the set $\{ x_ j \} _{j \in J'}$ is closed in $|X|$. In particular, $\{ x_ j \} _{j \in J}$ is an infinite closed subset of $K$ endowed with the discrete topology, contradicting our assumption that $K \subseteq |X|$ is compact. $\square$