# Let $\{A_i\}_{i \in I}$ be a locally finite collection of closed subsets of $X$. Show that the union $\bigcup_{i \in I} A_i$ is closed in $X$.

general-topology

Let $$\{A_i\}_{i \in I}$$ be a locally finite collection of closed subsets of a topological space $$X$$. Show that the union $$\bigcup_{i \in I} A_i$$ is closed in $$X$$.

We are to show that $$\left(\bigcup_{i \in I} A_i \right)^c = \bigcap_{i \in I}A_i^c$$ is open. So let $$x \in \bigcap_{i \in I}A_i^c$$. Since $$\{A_i\}_{i \in I}$$ is locally finite there exists neighbourhood of $$x$$ say $$O$$ such that $$O \cap A_n \ne \emptyset$$ for $$n \in \Bbb N$$.

What results do I have to use now? I'm trying to show that we can find an open set containing $$x$$ that's a contained in $$\bigcap_{i \in I}A_i^c$$.

Suppose $$x \notin \bigcup_{i \in I} A_i$$, so $$x$$ is in none of the $$A_i$$.
The family being locally finite means there is some open neighbourhood $$U_x$$ of $$x$$ so that $$F:=\{i \in I\mid A_i \cap U_x\neq \emptyset\}$$ is a finite set of indices.
For each $$i \in F$$ we have $$x \notin A_i$$ so some $$O_i$$ open around $$x$$ exists such that $$O_i \cap A_i =\emptyset$$.
Now note that $$O:=U_x \cap \bigcap_i O_i$$ is an open subset containing $$x$$ (as all sets are open and $$F$$ is finite!) so that $$O \cap \left(\bigcup_{i \in I} A_i\right) = \emptyset$$ showing that the union is indeed closed.
($$y \in O$$ means $$y \in U_x$$ so that $$y \notin A_i$$ for $$i \notin F$$ already. For $$i \in F$$ we use $$y \in O_i$$ so $$y \notin A_i$$ too).