A space $X$ is called anti-compact if any compact subset of $X$ is finite. In this answer, there are three anti compact Haudorff which is not discrete and in this answer, Brian M. Scott has gave anti compact but not Haudorff. Since in Brian M. Scott's examples, the spaces are not first countable, so I am wondering whether is it true that every first countable anti compact space is always Hausdorff? Could some one prove it or give a counter example whenever it is not true.
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1You should always try some examples, first. – Noah Schweber Nov 25 '15 at 02:56
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Consider the indiscrete topology on two points . . .
Less stupidly: the topology on $\mathbb{N}$ where the open sets are exactly those of the form $\{x: x\le n\}$ for some $n$.
Noah Schweber
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