Lemma 9.2.18. Let \[ Y_0 \xhookrightarrow {} Y_1 \xhookrightarrow {} Y_2 \xhookrightarrow {} \dots \] be a sequence of closed embeddings of \(T_1\) topological spaces, and let \(Y := \colim _n Y_n\) be the resulting colimit in \(\Top \). Then every compact subset \(K \subseteq Y\) is contained in some \(Y_n\). In particular, every continuous map \(f\colon X \to Y\) from a compact space factors through some \(Y_n\).
Proof. Assume for contradiction that \(K\) is not contained in any \(Y_n\), and choose \(x_n \in K \setminus Y_n\). Set \(A:=\{x_n\mid n\geq 0\}\). This set is infinite: if it were finite, all its points would lie in some common stage \(Y_N\), contradicting \(x_n\notin Y_n\) for \(n\geq N\). For every \(m\), the intersection \(A\cap Y_m\) is contained in the finite set \(\{x_0,\dots ,x_{m-1}\}\), hence is closed in the \(T_1\)-space \(Y_m\). The same holds for every subset of \(A\). By the definition of the colimit topology, every subset of \(A\) is therefore closed in \(Y\). Thus \(A\) is an infinite closed discrete subspace of \(K\), contradicting compactness of \(K\).
The last claim follows by applying the first to the compact subset \(f(X) \subseteq Y\). □
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