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Lemma Let $X$ be a simplicial set and let $S$ be a subset of the geometric realization $|X|$. Suppose that, for every nondegenerate $n$-simplex $\sigma $ of $X$, the inverse image of $S$ under the composite map $| \Delta ^{n} | \rightarrow |X|$ contains only finitely many points of the interior $\mathring {| \Delta ^{n} |} \subseteq | \Delta ^{n} |$. Then $S$ is closed.

Proof. The geometric realization $|X|$ can be described as the colimit $\varinjlim _{\sigma : \Delta ^{n} \rightarrow X} | \Delta ^{n} |$, indexed by the category of simplices of $X$ (see Construction Consequently, to show that the subset $S \subseteq | X |$ is closed, it will suffice to show that the inverse image $| \sigma |^{-1}(S) \subseteq | \Delta ^{n} |$ is closed, for every $n$-simplex $\sigma : \Delta ^{n} \rightarrow X$. We proceed by induction on $n$. Using Proposition, we can reduce to the case where $\sigma $ is nondegenerate. In this case, our inductive hypothesis guarantees that $| \sigma |^{-1}(S)$ has closed intersection with the boundary $| \operatorname{\partial \Delta }^{n} | \subseteq | \Delta ^{n} |$. Since $| \sigma |^{-1}(S)$ contains only finitely many points in the interior of $| \Delta ^ n |$, it is closed. $\square$