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I just read an article about anti-compact. The article states that one of an example anti-compact, i.e.,if all compact sets are finite, Hausdorff but not discrete is the real line generated by usual open sets and cocountable sets.

My question is how to prove that the real line with that topology is Hausdorff?

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Given distinct $x,y\in \mathbb R$, use the usual open sets that would separate them in the usual topology, and these will also separate them in the new topology since it is strictly finer. In other words, do the same thing you would do for the standard topology.

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    $\begingroup$ Yes, any refinement of a Hausdorff space is Hausdorff. $\endgroup$ – Thomas Andrews Sep 9 '15 at 0:22

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