most recent 30 from http://mathoverflow.net 2013-05-20T01:36:28Z http://mathoverflow.net/feeds/question/39670 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39670/is-a-compactly-generated-hausdorff-space-functionally-hausdorff David Roberts 2010-09-23T00:29:21Z 2013-02-12T08:28:46Z

<p>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.</p>

http://mathoverflow.net/questions/39670/is-a-compactly-generated-hausdorff-space-functionally-hausdorff/39672#39672 Qiaochu Yuan 2010-09-23T00:45:06Z 2010-09-23T00:45:06Z

<p>There is an example at <a href="http://planetmath.org/?op=getobj&amp;from=objects&amp;id=5718" rel="nofollow">PlanetMath</a> 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.</p>

http://mathoverflow.net/questions/39670/is-a-compactly-generated-hausdorff-space-functionally-hausdorff/121569#121569 jbc 2013-02-12T08:28:46Z 2013-02-12T08:28:46Z

<p>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.</p>