a space is T1 if and only if every subset A is the intersection of all open sets containing A


Say we have X, a T1-space, and A, a subset of X. We aim to show that the intersectionMathworldPlanetmath of all open sets containing A equals A. By de Morgan’s laws, that would be true if the complement of A, Ac, equalled the union of all closed setsPlanetmathPlanetmath in Ac. Let’s call this union of closed sets C.

Each set that makes up C is contained by Ac, so CAc. If we could show AcC, we’d be done.

Since X is T1, each singleton in Ac is closed (http://planetmath.org/ASpaceIsT1IfAndOnlyIfEverySingletonIsClosed). Their union, a subset of C, contains Ac, so we’re through.

Now suppose we know that in some topological spaceMathworldPlanetmath X, any subset A of X is the intersection of all open sets containing A. Given xy, we’re looking for an open set containing x but not y, to show that X is T1.

{x}=U openUxU

by hypothesisMathworldPlanetmathPlanetmath. If all open sets containing x contained y, y would be in the intersection; since y isn’t in the intersection, X must be T1.

Title a space is T1 if and only if every subset A is the intersection of all open sets containing A
Canonical name ASpaceIsT1IfAndOnlyIfEverySubsetAIsTheIntersectionOfAllOpenSetsContainingA
Date of creation 2013-03-22 14:20:18
Last modified on 2013-03-22 14:20:18
Owner waj (4416)
Last modified by waj (4416)
Numerical id 4
Author waj (4416)
Entry type Proof
Classification msc 54D10
Related topic T1Space
Related topic ASpaceIsT1IfAndOnlyIfEverySingletonIsClosed