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

1. Feb 27, 2012

Is there a method or an algorithm or a theorem or whatever that tells us a topological space is not a Hausdorff space?

2. Feb 27, 2012

### micromass

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. Feb 27, 2012

### Tinyboss

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).

4. Feb 28, 2012

### Bacle2

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.

5. Mar 2, 2012

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.

6. Mar 2, 2012

### micromass

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

$$\mathcal{T}=\{U\subseteq X~\vert~X\setminus U~\text{is finite}\}\cup \{\emptyset\}$$

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