Difference between revisions of "Hausdorff space"

From Maths
Jump to: navigation, search
m (Added reminder to include something)
(Adding phrasing, removed pointless use of X, happy now I think!)
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
−
{{Refactor notice|grade=A|msg=Page was 1 year and 1 day since modification, basically a stub, seriously needs an update.
+
{{Refactor notice|review=true|msg=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}}
+
* 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 [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 22:49, 22 February 2017 (UTC)}}
 +
__TOC__
 
==Definition==
 
==Definition==
−
Given a [[Topological space]] {{M|(X,\mathcal{J})}} we say it is '''Hausdorff'''{{rITTBM}} or '''satisfies the Hausdorff axiom''' if:
+
Given a [[topological space]] {{M|(X,\mathcal{J})}} we say it is ''Hausdorff''{{rITTBM}} or ''satisfies the Hausdorff axiom'' if:
−
* 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:
+
* {{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>
−
** {{M|1=N_a\cap N_b=\emptyset}}
+
** 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]]
−
===Alternate definition===
+
* 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>
−
* {{M|1=\forall a,b\in X\exists A,B\in\mathcal{J}[a\ne b\implies A\cap B=\emptyset]}}{{rITTMJML}}
+
* It may also be said that in a Hausdorff space that "''points may be separated by open sets''"{{rITTMJML}}
−
{{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.}}
+
  
 +
 +
A topological space satisfying this property is said to be a ''Hausdorff space''{{rITTMJML}}
 +
 +
A Hausdorff space is sometimes called a ''T2'' space
 +
===Equivalent definitions===
 +
* {{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'''''
 +
** 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]]
 +
** 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]}}
 +
==See next==
 +
* [[Topological separation axioms]]
 +
* [[A subspace of a Hausdorff space is Hausdorff]]
 +
==Proof of claims==
 +
{{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:
 +
* 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
 +
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.  
−
* [[A subspace of a Hausdorff space is 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 [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 22:49, 22 February 2017 (UTC)
 +
==Notes==
 +
<references group="Note"/>
 
==References==
 
==References==
 
<references/>
 
<references/>
 
{{Topology navbox|plain}}
 
{{Topology navbox|plain}}
 
{{Definition|Topology}}
 
{{Definition|Topology}}

Latest revision as of 22:52, 22 February 2017

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

This page is waiting for a final review, then this notice will be removed.

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.
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:
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

v • d • e
 
Topology
Overview of the structures studied in topology
Primitives
Global properties of topological spaces
Local properties of topological spaces
Constructs
Subtypes of topological spaces
inner product spaces [ilmath]\subset[/ilmath] normed spaces [ilmath]\subset[/ilmath] metric spaces [ilmath]\subset[/ilmath] topological spaces
Maps between topological spaces
Retrieved from "https://www.maths.kisogo.com/index.php?title=Hausdorff_space&oldid=4218"