# Example of a weak Hausdorff space that is not Hausdorff?

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

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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 '12 at 11:16
Welcome to MathOverflow, Prof. Solovay. Gerhard "And Happy Valentines Day, Too" Paseman, 2012.02.14 –  Gerhard Paseman Feb 14 '12 at 20:15

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.

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.

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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 '12 at 2:21
@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 '12 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 '12 at 3:48
@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 '12 at 9:05