I've read that

A topological space $X$ is completely regular iff it carries the initial (weak) topology w.r.t. $C(X,\mathbb{R})$ where $C(X, \mathbb{R})$ is the set of all bounded real-valued continuous functions $f: X \rightarrow \mathbb{R}$.

(Don't we need a topology on $X$ before we can talk about continuous functions on it?)

I'm having trouble understanding this statement because, as I understand things:

The initial topology induced by any family of functions $f_i: X \rightarrow \mathbb{R}$ is the smallest topology which makes each $f_i$ continuous. So the initial topology induced by $C(X, \mathbb{R})$ must just be the one we started with. Then any space must be completely regular, so what is there to define?

  • $\begingroup$ Here's a reference: Infinite Dimensional Analysis: A Hitchhiker's Guide, pg. 50 $\endgroup$ – Moya Apr 8 '16 at 2:00
  • $\begingroup$ Thanks! It's a bit more subtle than I thought. $\endgroup$ – Stanley Apr 8 '16 at 2:06
  • $\begingroup$ I'd still really appreciate an example of a space that's not completely regular (i.e. carries a larger topology than $\sigma(X,C)$) $\endgroup$ – Stanley Apr 8 '16 at 2:09

We do need a topology on $X$ and indeed, $X$ is by the assumption a topological space.

What the statement means is that $X$ is completely regular if and only if the original topology on $X$ coincides with the initial topology with respect to $C_b(X,{\bf R})$ (which is, in general, coarser).

For examples of Hausdorff spaces which are not completely regular, you may want to consult the $\pi$-base. There is also a related question here on math.se.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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