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Dec 11, 2018 at 23:17 comment added Joel David Hamkins Yes, that is right, this counterexample seems totally fine.
Dec 11, 2018 at 23:10 comment added aposyndetic but $S$ contains $0$
Dec 11, 2018 at 22:39 comment added aposyndetic But isn't $(-\infty,0)\cup S$ open, with open complement?
Dec 11, 2018 at 18:35 comment added aposyndetic I think it is false because what if $X=\mathbb R$ in the usual topology and $S=\mathbb Q \cap (-\infty,0]$.
Dec 11, 2018 at 18:15 review Close votes
Dec 12, 2018 at 20:36
Dec 11, 2018 at 17:45 review First posts
Dec 11, 2018 at 17:56
Dec 11, 2018 at 17:42 history asked aposyndetic CC BY-SA 4.0