# Contents

## Statement

###### Proposition

Assuming excluded middle, then:

More specifically, in a Hausdorff topological space the irreducible closed subspaces are precisely the singleton subspaces.

###### Proof

The second statement clearly implies the first. To see the second statement, suppose that $F$ is an irreducible closed subspace which contained two distinct points $x \neq y$. Then by the Hausdorff property there are disjoint neighbourhoods $U_x, U_y$, and hence it would follow that the relative complements $F \backslash U_x$ and $F \backslash U_y$ were distinct proper closed subsets of $F$ with

$F = (F \backslash U_x) \cup (F \backslash U_y)$

in contradiction to the assumption that $F$ is irreducible.

This proves by contradiction that every irreducible closed subset is a singleton. Conversely, generally the topological closure of every singleton is irreducible closed.

## References

Revised on April 14, 2017 13:27:55 by Urs Schreiber (46.183.103.8)