closed Hausdorff neighbourhoods, a theorem on


Theorem. If X is a topological spaceMathworldPlanetmath in which every point has a closed HausdorffPlanetmathPlanetmath neighbourhood, then X 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 x is a set A such that x lies in the interior of A.

Proof of theorem. Let X be a topological space in which every point has a closed Hausdorff neighbourhood. Suppose a,bX are distinct. It suffices to show that a and b have disjoint neighbourhoods. By assumptionPlanetmathPlanetmath, there is a closed Hausdorff neighbourhood N of b. If aN, then XN and N are disjoint neighbourhoods of a and b (as N is closed).

So suppose aN. As N is Hausdorff, there are disjoint sets U0,V0N that are open in N, such that aU0 and bV0. There are open sets U and V of X such that U0=UN and V0=VN. Note that U is a neighbourhood of a, and V is a neighbourhood of b. As N is a neighbourhood of b, it follows that VN (that is, V0) is a neighbourhood of b. We have UV0=U0V0=. So U and V0 are disjoint neighbourhoods of a and b. 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