Question is the title. I suspect the answer is no, without some further conditions (clearly, normal is sufficient). Pointers to counterexamples would be appreciated, but not necessary.
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asked Sep 23, 2010 at 0:29
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2 Answers
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There is an example at PlanetMath of a Hausdorff space which is not completely Hausdorff / functionally Hausdorff. On the other hand it is second-countable, hence first-countable and hence compactly generated.
answered Sep 23, 2010 at 0:45
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$\begingroup$ Thanks! I'd seen this example, but hadn't clicked to the fact it was compactly generated. $\endgroup$– David Roberts ♦Commented Sep 23, 2010 at 0:51
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Since you implicitly ask for sufficient conditions for a compactly generated space to be completely Hausdorff, it might interest you that one such is that the family of compacta be countably generated. As an appendix to the negative example given by Qiaochu Yuan, any Hausdorff space which is not completely Hausdorff provides a counterexample, since a space is completely Hausdorff if and only if the associated compactly generated space is completely Hausdorff. Both of these facts are in the 1969 paper "Topologies et compactologies" by Henri Buchwalter.
