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

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 |