Hausdorff implies sober




topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory


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topological homotopy theory




Assuming excluded middle, then:

Every Hausdorff topological space is a sober topological space.

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


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

F=(FU x)(FU y) F = (F \setminus U_x) \cup (F \setminus U_y)

in contradiction to the assumption that FF 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 1T_1. However, even the converse to this fails. For example, let XX be the real line with a new point pp added, topologized such that any open set in the real line is open, and any cofinite set containing pp is open. Then XX is sober and T 1T_1 but not Hausdorff.


Last revised on September 20, 2018 at 15:51:10. See the history of this page for a list of all contributions to it.