Contents

# 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 $F \setminus U_x$ and $F \setminus U_y$ were distinct proper closed subsets of $F$ with

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

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.

## Strictness of implication

There are many examples of sober spaces which are not Hausdorff. For example, the spectrum of a ring which is not zero-dimensional is sober but not Hausdorff.

Any Hausdorff space is not only sober, but also $T_1$. However, even the converse to this fails. For example, let $X$ be the real line with a new point $p$ added, topologized such that any open set in the real line is open, and any cofinite set containing $p$ is open. Then $X$ is sober and $T_1$ but not Hausdorff.