Difference between revisions of "Hausdorff space"

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==Definition==
 
==Definition==
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Given a [[Topological space]] {{M|(X,\mathcal{J})}} we say it is '''Hausdorff'''{{rITTBM}} or '''satisfies the Hausdorff axiom''' if:
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Given a [[topological space]] {{M|(X,\mathcal{J})}} we say it is ''Hausdorff''{{rITTBM}} or ''satisfies the Hausdorff axiom'' if:
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* For all {{M|a,b\in X}} that are distinct there exists [[Open set#Neighbourhood 2|neighbourhoods]] to {{M|a}} and {{M|b}}, {{M|N_a}} and {{M|N_b}} such that:
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* {{M|\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]}}<ref group="Note">Note that if {{M|X}} is the empty set, then there are no {{M|x_1,x_2\in X}}, so the statement is [[vacuously true]].</ref>
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** {{M|1=N_a\cap N_b=\emptyset}}
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** In words: for any two points in {{M|X}}, if the points are distinct then there exist [[neighbourhoods]] to each point such that the neighbourhoods are [[disjoint]]
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===Alternate definition===
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* We may also write it: {{M|\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]}}<ref group="Note">Again note that if {{M|X}} is the empty set, then there are no {{M|x_1,x_2\in X}}, so the statement is [[vacuously true]]. In the event {{M|X}} has one or more points notice that then {{M|X}} 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 {{M|x_1\eq x_2}} then we can pick this neighbourhood ({{M|X}} itself) and be done, as by the nature of [[logical implication]] we do not care about the truth or falsity of the {{M|N_1\cap N_2\eq\emptyset}} part.</ref><ref group="Note">These are easily seen to be equivalent, try it! You need to do the {{M|X}} is one point case, {{M|X}} is empty and then {{M|X}} contains 2 or more points - this is the easiest</ref>
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* {{M|1=\forall a,b\in X\exists A,B\in\mathcal{J}[a\ne b\implies A\cap B=\emptyset]}}{{rITTMJML}}
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{{Requires proof|Are these statements the same? Clearly {{M|\text{neighbourhood }\implies\text{open-set} }} as a neighbourhood to a point requires the existence of an open set containing that point (contained in the neighbourhood) and clearly {{M|\text{open-set}\implies\text{neighbourhood} }} as an open set ''is'' a neighbourhood - write this up.}}
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A topological space, say {{M|X}} as before, satisfying this property is said to be a ''Hausdorff space''{{rITTMJML}}
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A Hausdorff space is sometimes called a ''T2'' space
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===Equivalent definitions===
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* {{M|\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]}}{{rITTMJML}} - see '''''Claim 1'''''
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** In words: for all points in {{M|X}} if the points are distinct then there exists [[open sets]] acting as [[open neighbourhoods]] to each point such that these open neighbourhoods are [[disjoint]]
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** This, along the same thinking as for the definition, may be (and is commonly seen as) written: {{M|\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]}}
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==See next==
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* [[Topological separation axioms]]
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* [[A subspace of a Hausdorff space is Hausdorff]]
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==Proof of claims==
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{{Requires proof|grade=D|easy=true|msg=Easy to prove the first and only claim on this page. {{M|\impliedby}} is easily seen as open sets are neighbourhoods (see "''[[an open set is a neighbourhood to all of its points]]''", the other way requires:
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* If {{M|C\subseteq A}} and {{M|D\subseteq B}} then if {{M|A\cap B\eq\emptyset}} we have {{M|C\cap D\eq\emptyset}} - this could be worth factoring out
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but is otherwise really easy}}
 
==Further work for this page==
 
==Further work for this page==
 
* Link to a theorem about all metric spaces being Hausdorff.  
 
* Link to a theorem about all metric spaces being Hausdorff.  
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* [[A subspace of a Hausdorff space is Hausdorff]]
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* 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 [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 22:49, 22 February 2017 (UTC)
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==Notes==
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<references group="Note"/>
 
==References==
 
==References==
 
<references/>
 
<references/>
 
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{{Definition|Topology}}
 
{{Definition|Topology}}

Revision as of 22:49, 22 February 2017

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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]


A topological space, say [ilmath]X[/ilmath] as before, 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 Introduction to Topological Manifolds - John M. Lee

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Topology
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