a topological space $X$ is called $T_B$ if every compact subset is closed.
Is $X \times X$, $T_B$ when $X$ is $T_B$ ?
If it is not true, which conditions do need to $X \times X$ be $T_B$?
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A counterexample is any compact non-Hausdorff KC-space, i.e. a space where the compact sets and the closed sets coincide. Then the diagonal $\triangle$ is not closed in $X\times X$ as that would imply Hausdorffness of $X$, but $\triangle\approx X$ is compact. Hence $X\times X$ is not KC. – Stefan Hamcke Sep 21 '13 at 17:37
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Do you mean if X is Hausdorff and KC ,$X \times X$ will be KC ($T_B)$? – Alireza Sep 21 '13 at 19:49
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@Alireza: Of course: every Hausdorff space is automatically $KC$ (or $T_B$, if you prefer), and the product of two Hausdorff spaces in Hausdorff. Here I gave a the specific example of the one-point compactification of $\Bbb Q$ as a compact $KC$ space whose square is not $KC$; of course $\Bbb Q^*$ is not Hausdorff. – Brian M. Scott Sep 22 '13 at 05:00