$X$ is completely regular iff it carries the initial topology w.r.t. $C(X,\mathbb{R})$

Question

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?

Here's a reference: Infinite Dimensional Analysis: A Hitchhiker's Guide, pg. 50 — Moya, Apr 08 '16 at 02:00
I'd still really appreciate an example of a space that's not completely regular (i.e. carries a larger topology than $\sigma(X,C)$) — Stanley, Apr 08 '16 at 02:09

score 3 · Accepted Answer · edited Apr 01 '23 at 20:33

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.

$X$ is completely regular iff it carries the initial topology w.r.t. $C(X,\mathbb{R})$

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