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

  1. Feb 27, 2012 #1
    Is there a method or an algorithm or a theorem or whatever that tells us a topological space is not a Hausdorff space?
  2. jcsd
  3. Feb 27, 2012 #2
    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??
  4. Feb 27, 2012 #3
    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).
  5. Feb 28, 2012 #4


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    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.
  6. Mar 2, 2012 #5
    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.
  7. Mar 2, 2012 #6
    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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