closed Hausdorff neighbourhoods, a theorem on
Theorem.
If is a topological space in which
every point has a closed Hausdorff
neighbourhood,
then is Hausdorff.
Note. In this theorem (and the proof that follows) neighbourhoods are not assumed to be open. That is, a neighbourhood of a point is a set such that lies in the interior of .
Proof of theorem.
Let be a topological space in which
every point has a closed Hausdorff neighbourhood.
Suppose are distinct.
It suffices to show that and have disjoint neighbourhoods.
By assumption, there is a closed Hausdorff neighbourhood of .
If , then and
are disjoint neighbourhoods of and (as is closed).
So suppose . As is Hausdorff, there are disjoint sets that are open in , such that and . There are open sets and of such that and . Note that is a neighbourhood of , and is a neighbourhood of . As is a neighbourhood of , it follows that (that is, ) is a neighbourhood of . We have . So and are disjoint neighbourhoods of and . QED.
Title | closed Hausdorff neighbourhoods, a theorem on |
---|---|
Canonical name | ClosedHausdorffNeighbourhoodsATheoremOn |
Date of creation | 2013-03-22 18:30:46 |
Last modified on | 2013-03-22 18:30:46 |
Owner | yark (2760) |
Last modified by | yark (2760) |
Numerical id | 7 |
Author | yark (2760) |
Entry type | Theorem |
Classification | msc 54D10 |