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How to prove that a topological space is non-hausdorff?

by AdrianZ
Tags: nonhausdorff, prove, space, topological
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AdrianZ
#1
Feb27-12, 11:25 AM
P: 320
Is there a method or an algorithm or a theorem or whatever that tells us a topological space is not a Hausdorff space?
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micromass
#2
Feb27-12, 01:01 PM
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This depends on the space in question.
I think finding a counterexample to the definition shouldn't be so hard. Another thing you can do is to find a sequence that converges to more than one point (but if such a sequence does not exist, then the space can still be Hausdorff).

You have any specific space in mind??
Tinyboss
#3
Feb27-12, 04:24 PM
P: 233
If you don't have enough information to actually exhibit a pair of points that can't be separated by disjoint open neighborhoods, then the only other way I can think of to show non-Hausdorff-ness would be to have a counterexample to one of the properties that Hausdorff spaces have (e.g. that compact subsets are closed).

Bacle2
#4
Feb28-12, 12:39 AM
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How to prove that a topological space is non-hausdorff?

You can check the above conditions and throw-in metrizability as a sufficient--tho not necessary --condition for Hausdorffness.

Still, I think the question is too broad* ; and you may find a better answer if you know how the space is presented/described to you.


*tho, don't get me wrong, I like broads.
AdrianZ
#5
Mar2-12, 03:49 AM
P: 320
for example Zariski topology, How do we show that it is non-Hausdorff? I'm just interested to know how we could see if a space is Hausdorff or not.
micromass
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Mar2-12, 06:13 AM
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Quote Quote by AdrianZ View Post
for example Zariski topology, How do we show that it is non-Hausdorff? I'm just interested to know how we could see if a space is Hausdorff or not.
What do you mean with the Zariski topology here?? Do you mean the topology consisting of all cofinite sets, or do you mean the topology associated with a commutative ring??

Let's say you mean the former, then we have an infinite set X and a topology

[tex]\mathcal{T}=\{U\subseteq X~\vert~X\setminus U~\text{is finite}\}\cup \{\emptyset\}[/tex]

Take two arbitrary non-empty open sets U and V. Then [itex]U\cap V[/itex] is nonempty (check this). So the space is Hausdorff because there don't exist disjoint open sets!


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