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.



Is a compactly generated Hausdorff space functionally Hausdorff?
There is an example at [PlanetMath](http://planetmath.org/?op=getobj&from=objects&id=5718) 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.


