Welcome to MathOverflow

Question

9

3

I've looked on the web and haven't found a simple example.

flag	asked Feb 14 2012 at 10:33 Bob Solovay 93●3

4

From Wikipedia: en.wikipedia.org/wiki/Weak_Hausdorff_space A space $X$ is weak Hausdorff if whenever $Y$ is compact Hausdorff and $f:Y\rightarrow X$ is continuous, then $f(Y)$ is closed in $X$. – Matthew Daws Feb 14 2012 at 11:16

Welcome to MathOverflow, Prof. Solovay. Gerhard "And Happy Valentines Day, Too" Paseman, 2012.02.14 – Gerhard Paseman Feb 14 2012 at 20:15

Stephen S · Accepted Answer · 2012-02-15 09:29:33Z UTC

16

The one-point compactification of $\mathbb{Q}$ has the property that every compact subset is closed. So it is certainly a weak Hausdorff space. But it isn't Hausdorff, as $\mathbb{Q}$ isn't locally compact.

Addendum

Another example is the cocountable topology on an uncountable set. No two points have disjoint neighbourhoods, and the only compact subsets are the finite subsets.

edited Feb 15 2012 at 9:29

answered Feb 14 2012 at 11:38

Stephen S
784●1●7●9

1

I don't know what you mean by "the one point compactification of Q". Of course, I am familiar with the one point compactification of a locally compact space. What is the topology of this space. What are two points that don't have disjoint neighborhoods? – Bob Solovay Feb 15 2012 at 2:21

1

@BobSolovay: I was not sure of this myself, but consulting Kelley's book tells me that for any top. space X the 1-point compactification X* has as its open sets all the open subsets of X, together with all subsets of X* whose complements are closed compact subsets of X. (TBC) – Yemon Choi Feb 15 2012 at 3:46

Thus, if I have understood things correctly: when we give Q its subspace topology from R (thus not locally compact) we find that the open neighbourhoods of the point at infinity correspond to cofinite subsets of Q, and hence they meet every non-empty open subset of Q. (Open subsets of Q in this topology are either empty or infinite.) – Yemon Choi Feb 15 2012 at 3:48

3

@Bob Solovay: Yemon Choi's definition of the one-point compactification is correct. @Yemon Choi: But your second comment is wrong, since $\mathbb{Q}$ has infinite compact subsets (e.g., any convergent sequence together with its limit forms a compact set). But every compact subset of $\mathbb{Q}$ has empty interior, so it's true that every neighbourhood of the point at infinity meets every non-empty open set. – Stephen S Feb 15 2012 at 9:05

Nathan · Answer 2 · 2012-02-14 19:08:55Z UTC

3

Steen & Seebach's counterexample #99: Maximal Compact Topology is another example. This is also a KC space (every compact set is closed) but not Hausdorff.

answered Feb 14 2012 at 19:08

Nathan
139●1●4

Example of a weak Hausdorff space that is not Hausdorff?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.