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Let $(E, \mathcal{T})$ be 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.

  • 5
    No. See for example in [this](http://journals.cambridge.org/download.php?file=%2FJAZ%2FJAZ3_02%2FS1446788700027907a.pdf&code=884c1d5b377d31121d84a553e6b92ef1) paper from A. Smythe and C. A. Wilkins.2012-03-27
  • 2
    you can find the paper by searching "A. Smythe and C. A. Wilkins" on google. That link didn't work for some reason.2012-03-27
  • 2
    Perhaps [this google-Link](http://www.google.de/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CCcQFjAA&url=http%3A%2F%2Fjournals.cambridge.org%2Fproduction%2Faction%2FcjoGetFulltext%3Ffulltextid%3D4922028&ei=XRlyT5OOH8vltQbKkp35DQ&usg=AFQjCNFPXyAHYGnpU8-IkgzSSZ1Cq3Aqnw&sig2=nFOlnn_oFOqb3H5BoDZ0vg) will do. It does for me, at least.2012-03-27
  • 1
    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.2013-11-15

1 Answers 1