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
Canonical name	TheUnionOfALocallyFiniteCollectionOfClosedSetsIsClosed
Date of creation	2013-03-22 16:14:45
Last modified on	2013-03-22 16:14:45
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	10
Author	yark (2760)
Entry type	Theorem
Classification	msc 54A99