Hausdorff space

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Page was 1 year and 1 day since modification, basically a stub, seriously needs an update.

• Add Example:The real line with the finite complement topology is not Hausdorff as an example of a familiar set with an unfamiliar topology
• Huge overhaul - removed utter nonsense Alec (talk) 22:49, 22 February 2017 (UTC)

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Definition

Given a topological space [ilmath](X,\mathcal{J})[/ilmath] we say it is Hausdorff or satisfies the Hausdorff axiom if:

• [ilmath]\forall x_1,x_2\in X\big[x_1\neq x_2\implies \big(\exists N_1\in\mathcal{N}(x_1,X)\exists N_2\in\mathcal{N}(x_2,X)[N_1\cap N_2\eq\emptyset]\big)\big][/ilmath]
  • In words: for any two points in [ilmath]X[/ilmath], if the points are distinct then there exist neighbourhoods to each point such that the neighbourhoods are disjoint
• We may also write it: [ilmath]\forall x_1,x_2\in X\exists N_1\in\mathcal{N}(x_1,X)\exists N_2\in\mathcal{N}(x_2,X)[x_1\neq x_2\implies N_1\cap N_2\eq\emptyset][/ilmath]
• It may also be said that in a Hausdorff space that "points may be separated by open sets"

A topological space satisfying this property is said to be a Hausdorff space

A Hausdorff space is sometimes called a T2 space

Equivalent definitions

• [ilmath]\forall x_1,x_2\in X\big[x_1\neq x_2\implies \big(\exists U_1\in\mathcal{O}(x_1,X)\exists U_2\in\mathcal{O}(x_2,X)[U_1\cap U_2\eq\emptyset]\big)\big][/ilmath] - see Claim 1
  • In words: for all points in [ilmath]X[/ilmath] if the points are distinct then there exists open sets acting as open neighbourhoods to each point such that these open neighbourhoods are disjoint
  • This, along the same thinking as for the definition, may be (and is commonly seen as) written: [ilmath]\forall x_1,x_2\in X\exists U_1\in\mathcal{O}(x_1,X)\exists U_2\in\mathcal{O}(x_2,X)[x_1\neq x_2\implies U_1\cap U_2\eq\emptyset][/ilmath]

See next

• Topological separation axioms
• A subspace of a Hausdorff space is Hausdorff

Proof of claims

Grade: D

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

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Easy to prove the first and only claim on this page. [ilmath]\impliedby[/ilmath] is easily seen as open sets are neighbourhoods (see "an open set is a neighbourhood to all of its points", the other way requires:

• If [ilmath]C\subseteq A[/ilmath] and [ilmath]D\subseteq B[/ilmath] then if [ilmath]A\cap B\eq\emptyset[/ilmath] we have [ilmath]C\cap D\eq\emptyset[/ilmath] - this could be worth factoring out

but is otherwise really easy

This proof has been marked as an page requiring an easy proof

Further work for this page

• Link to a theorem about all metric spaces being Hausdorff.
• Proof of the equivalent form claims - I did say "no proof will be hand-waved away as trivial" but it certainly isn't worth my time now Alec (talk) 22:49, 22 February 2017 (UTC)