How to prove that a topological space is nonhausdorff?by AdrianZ Tags: nonhausdorff, prove, space, topological 

#1
Feb2712, 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?




#2
Feb2712, 01:01 PM

Mentor
P: 16,590

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?? 



#3
Feb2712, 04:24 PM

P: 234

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 nonHausdorffness would be to have a counterexample to one of the properties that Hausdorff spaces have (e.g. that compact subsets are closed).




#4
Feb2812, 12:39 AM

Sci Advisor
P: 1,168

How to prove that a topological space is nonhausdorff?
You can check the above conditions and throwin metrizability as a sufficienttho 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. 



#5
Mar212, 03:49 AM

P: 320

for example Zariski topology, How do we show that it is nonHausdorff? I'm just interested to know how we could see if a space is Hausdorff or not.




#6
Mar212, 06:13 AM

Mentor
P: 16,590

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 nonempty 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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