the union of a locally finite collection of closed sets is closed

The union of a collection of closed subsets of a topological space need not,of course, be closed. However, we do have the following result:

Theorem.

The union of a locally finite collection of closed subsetsof a topological space is itself closed.

Proof.

Let $𝒮$ be a locally finite collection of closed subsetsof a topological space $X$, and put $Y=\cup 𝒮$.Let $x\in X\setminus Y$.By local finiteness there is an open neighbourhood $U$ of $x$that meets only finitely many members of $𝒮$,say $A_1,\mathrm{\dots },A_n$.So $U\setminus Y=U\setminus {\bigcup }_{i=1}^n{A}_i$, which is open.Thus $U\setminus Y$ is an open neighbourhood of $x$ that does not meet $Y$.It follows that $Y$ is closed.∎

One use for this result can be found in the entry on gluing together continuous functions.