closed Hausdorff neighbourhoods, a theorem on
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|
|Date of creation||2013-03-22 18:30:46|
|Last modified on||2013-03-22 18:30:46|
|Last modified by||yark (2760)|