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May 20, 2015 at 13:02 vote accept Dominic van der Zypen
May 20, 2015 at 11:50 answer added Simon Henry timeline score: 4
May 20, 2015 at 11:39 comment added Simon Henry By separated I mean Hausdorff.I wouldn't say it is a duplicate, the linked question ask for conditions, this one for counterexamples.
May 20, 2015 at 10:57 comment added Johannes Hahn possible duplicate of Is every continuous function measurable?
May 20, 2015 at 7:35 review Close votes
May 20, 2015 at 12:25
May 20, 2015 at 7:05 comment added Dominic van der Zypen In the non-$T_1$ context, there may be compact sets that are not closed.
May 20, 2015 at 6:56 comment added Simon Henry Unless I'm missing something, If your topological spaces are separated the answer is clear: adding closed subset to $\tau$ will not change the $\sigma$-algebra it generates. So I guess the topological spaces are not assumed to be separated, and in this case I need to ask: does compact mean compact & separated or just compact ?
May 20, 2015 at 6:49 history asked Dominic van der Zypen CC BY-SA 3.0