Difference between revisions of "Hausdorff space"
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==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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==References== | ==References== | ||
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{{Topology navbox|plain}} | {{Topology navbox|plain}} | ||
{{Definition|Topology}} | {{Definition|Topology}} | ||
Revision as of 12:34, 10 October 2016
Grade: A
This page is currently being refactored (along with many others)
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Page was 1 year and 1 day since modification, basically a stub, seriously needs an update.
Definition
Given a Topological space [ilmath](X,\mathcal{J})[/ilmath] we say it is Hausdorff[1] or satisfies the Hausdorff axiom if:
- For all [ilmath]a,b\in X[/ilmath] that are distinct there exists neighbourhoods to [ilmath]a[/ilmath] and [ilmath]b[/ilmath], [ilmath]N_a[/ilmath] and [ilmath]N_b[/ilmath] such that:
- [ilmath]N_a\cap N_b=\emptyset[/ilmath]
Alternate definition
- [ilmath]\forall a,b\in X\exists A,B\in\mathcal{J}[a\ne b\implies A\cap B=\emptyset][/ilmath][2]
(Unknown grade)
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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Are these statements the same? Clearly [ilmath]\text{neighbourhood }\implies\text{open-set} [/ilmath] as a neighbourhood to a point requires the existence of an open set containing that point (contained in the neighbourhood) and clearly [ilmath]\text{open-set}\implies\text{neighbourhood} [/ilmath] as an open set is a neighbourhood - write this up.
Further work for this page
- Link to a theorem about all metric spaces being Hausdorff.
- A subspace of a Hausdorff space is Hausdorff
References
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