1
$\begingroup$

The topological question:

Are there Hausdorff topological spaces $X$ which are compactly generated (=Kelly spaces = $k$-spaces, that is, a subset is closed if its intersection with every compact set is closed) which fail to satisfy Tietze's extension theorem for compact subsets (that is, there is a real-valued continuous function defined on some compact subset without continuous extension to the whole space)?

A possibility to answer this question would be to exhibit a compactly generated space which is not completely Hausdorff (continuous functions do not separate points). Unfortunately, neither the space book nor the literature I have consulted (in particular Counterxamples in Topology) contained an answer.

I have chosen the Functional Analysis tag because such a space would yield a very natural example of a projective (or inverse) limit of Banach spaces $C(K)$ ($K$ compact) with surjective linking maps $C(L)\to C(K)$, $f\to f|_K$ for $K\subseteq L$ (recall that compact spaces are normal so that Tietze's theorem applies) such that the canonical map from the projective limit (which is $C(X)$ because of compact generation) to the "steps" is not surjective.


Edit. Recalling the implications completely Hausdorff $\Rightarrow$ $T_{2\, 1/2}$ and second countable $\Rightarrow$ compactly generated one gets a list of 6 examples from spacebook namely

Double Origin Topology

Irrational Slope Topology

Minimal Hausdorff Topology

Prime Integer Topology

Relatively Prime Integer Topology

Simplified Arens Square (this one is particularly simple).

$\endgroup$
  • $\begingroup$ I am pretty sure that there is a counterexample in Buchwalter's work on compactologies but at the moment I don't have access to his articles and so can't verify this. By the way, on the positive side he shows the (not very deep) result that this cannot happen for countable spectra. $\endgroup$ – alpha Feb 26 '14 at 18:04
  • $\begingroup$ @alpha: of course, I meant surjective. The countable case is clear. In particular, the space can't be $\sigma$-compact. $\endgroup$ – Jochen Wengenroth Feb 26 '14 at 20:45
4
$\begingroup$

The answer to this question give an example of a topological space which is compactly generated, but is not functionally Hausdorff.

Is a compactly generated Hausdorff space functionally Hausdorff?

$\endgroup$
  • 1
    $\begingroup$ For instance, a space which is countable, second countable, hausdorff and connected. A real valued map is then constant (because a connected countable subset of the reals is a point). There are some known examples with these properties (one is in Counterexamples in Topology if I remember well). $\endgroup$ – Mathieu Baillif Feb 27 '14 at 12:48

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.