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Dec 23, 2012 at 13:39 comment added Joel David Hamkins Eric, that is very nice!
Dec 23, 2012 at 13:10 comment added Marcus @Joel Great (and kind of unexpected). Thanks.
Dec 23, 2012 at 13:10 vote accept Marcus
Dec 23, 2012 at 12:21 comment added Eric Wofsey Essentially the same argument works when $X$ is any compact Hausdorff space. Just take a minimal sequence of closed neighborhoods of $U$ that generates the filter of all neighborhoods of $U$. By compactness, the intersection any initial segment of this must contain some point other than $U$.
Dec 23, 2012 at 12:15 history edited Joel David Hamkins CC BY-SA 3.0
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Dec 23, 2012 at 12:05 history edited Joel David Hamkins CC BY-SA 3.0
added 134 characters in body; added 23 characters in body
Dec 23, 2012 at 11:57 history answered Joel David Hamkins CC BY-SA 3.0