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 intersection 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 sets
in Ac. Let’s call this union of closed sets C.
Each set that makes up C is contained by Ac, so C⊂Ac. If we could show Ac⊂C, 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 space X, any subset A of X is the intersection of all open sets containing A. Given x≠y, we’re looking for an open set containing x but not y, to show that X is T1.
{x}=⋂U openU∋xU |
by hypothesis. 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 |
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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 |