# 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 subsets of a topological space is itself closed.

###### Proof.

Let $\cal S$ be a locally finite collection of closed subsets of a topological space $X$, and put $Y=\cup\cal S$. Let $x\in X\setminus Y$. By local finiteness there is an open neighbourhood $U$ of $x$ that meets only finitely many members of $\cal S$, say $A_{1},\dots,A_{n}$. 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.

Title the union of a locally finite collection of closed sets is closed TheUnionOfALocallyFiniteCollectionOfClosedSetsIsClosed 2013-03-22 16:14:45 2013-03-22 16:14:45 yark (2760) yark (2760) 10 yark (2760) Theorem msc 54A99