Proposition 4

The union of a locally finite family of closed subsets of a topological space $X$ is closed in $X$.

Let $(F_i)_{i \in I}$ be a locally finite family of closed subsets of $X$, and suppose that $x \in X$ does not belong to $F = \bigcup_{i \in I} F_i$; then $x$ has a neighbourhood $V$ which meets only those $F_i$ whose indices belong to a finite subset $J$ of $I$. For each $i \in J$ let $U_i$ be the complement of $F_i$; then $\complement F$ contains the set $V \cap \bigcap_{i \in J} U_i$, which is a neighbourhood of $x$ since each $U_i$ is open and contains $x$. Hence, by Proposition 1 of no. 2, $\complement F$ is open and therefore $F$ is closed in $X$.

We note that the union of an arbitrary family of closed subsets of $X$ is not necessarily closed; for example, on the rational line $\mathbf{Q}$, the set $]2, 1[$ is the union of the closed sets

but is not closed.