Hausdorff space

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Definition

Given a topological space [ilmath](X,\mathcal{J})[/ilmath] we say it is Hausdorff[1] 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][Note 1]
    • 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][Note 2][Note 3]
  • It may also be said that in a Hausdorff space that "points may be separated by open sets"[2]


A topological space satisfying this property is said to be a Hausdorff space[2]

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][2] - 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

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)

Notes

  1. Note that if [ilmath]X[/ilmath] is the empty set, then there are no [ilmath]x_1,x_2\in X[/ilmath], so the statement is vacuously true.
  2. Again note that if [ilmath]X[/ilmath] is the empty set, then there are no [ilmath]x_1,x_2\in X[/ilmath], so the statement is vacuously true. In the event [ilmath]X[/ilmath] has one or more points notice that then [ilmath]X[/ilmath] itself is an open set and an open set is a neighbourhood to all of its points, so there exists neighbourhoods, if we have points. Note lastly that if [ilmath]x_1\eq x_2[/ilmath] then we can pick this neighbourhood ([ilmath]X[/ilmath] itself) and be done, as by the nature of logical implication we do not care about the truth or falsity of the [ilmath]N_1\cap N_2\eq\emptyset[/ilmath] part.
  3. These are easily seen to be equivalent, try it! You need to do the [ilmath]X[/ilmath] is one point case, [ilmath]X[/ilmath] is empty and then [ilmath]X[/ilmath] contains 2 or more points - this is the easiest

References

  1. Introduction to Topology - Bert Mendelson
  2. 2.0 2.1 2.2 Introduction to Topological Manifolds - John M. Lee