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Nov 12, 2023 at 15:52 comment added Pietro Majer (one may also say that A is even sequentially compact, because it is the product of two seq. comp. topological spaces)
Jul 8, 2010 at 9:42 comment added KP Hart You may want to replace `non-normal' by normal in the first line.
Jul 6, 2010 at 14:30 comment added Henno Brandsma Adde: we cannot have a countably compact $A$ with non-countably compact closure for a non-normal space, like here, because cl(A) would then be pseudocompact (having a dense countably compact subset) and hence (by normality) countably compact. But, can we still have a relatively countably compact $A$ in a $T_4$ space $X$ with non countably compact closure?
Jul 6, 2010 at 4:29 comment added Henno Brandsma Ah, the Tychonoff plank. I tried to define a similar example on $\omega \times \omega$, but failed... Is it essential that such a space be non-normal, as this is?
Jul 6, 2010 at 4:26 vote accept Henno Brandsma
Jul 6, 2010 at 1:38 history edited Joel David Hamkins CC BY-SA 2.5
Improved presentation. Added crude diagram.
Jul 6, 2010 at 0:29 history answered Joel David Hamkins CC BY-SA 2.5