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Let $(E, \mathcal{T})$ a compact Hausdorff space. It is well known that every topology $\mathcal{U}$ coarser than $\mathcal{T}$ such that $(E, \mathcal{U})$ is Hausdorff is equal to $\mathcal{T}$.

Is the converse true ?

(that is: if $\mathcal{T}$ is a coarsest topology amongst Hausdorff topology on $E$, then $(E, \mathcal{T})$ is compact)

Thanks in advance.

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No. See for example in this paper from A. Smythe and C. A. Wilkins. – martini Mar 27 '12 at 19:41
you can find the paper by searching "A. Smythe and C. A. Wilkins" on google. That link didn't work for some reason. – Dustan Levenstein Mar 27 '12 at 19:46
Perhaps this google-Link will do. It does for me, at least. – martini Mar 27 '12 at 19:48
Thanks, it works. – francis-jamet Mar 27 '12 at 19:51
Your question is probably sufficiently answered by the first comment. But a key word you might look for is "minimal Hausdorff space." Also, note that some spaces don't have "minimal Hausdorff" coarsenings -- the rational numbers for example. – thatguy Nov 15 '13 at 14:34

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