Hausdorff space
From Maths
This page is currently being refactored (along with many others)
Please note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.
The message provided is:
This page is waiting for a final review, then this notice will be removed.
The message provided is:
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)
This page is waiting for a final review, then this notice will be removed.
Contents
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 todo list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
This proof has been marked as an page requiring an easy proof
The message provided is:
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
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 handwaved away as trivial" but it certainly isn't worth my time now Alec (talk) 22:49, 22 February 2017 (UTC)
Notes
 ↑ 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.
 ↑ 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.
 ↑ 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
 ↑ Introduction to Topology  Bert Mendelson
 ↑ ^{2.0} ^{2.1} ^{2.2} Introduction to Topological Manifolds  John M. Lee
