On general topological spaces and $C(X, \mathbb R)$ , where for closed sets $A,B$ in $X$ , $I_A=I_B \implies A=B$

Question

Let $X$ be a metric space and $C(X, \mathbb R)$ be the ring of all real valued continuous functions from $X$ . For $A \subseteq X$ , let us define $I_A :=\{f \in C(X, \mathbb R) : f(x)=0 , \forall x \in A\}$ . I can prove that if $A,B$ are closed sets in $X$ , then $I_A=I_B \implies A=B$ . My question is , if we want $X$ to be a general topological space rather than just a metric space , then for which kind of topological spaces do we still have , for closed sets $A,B$ in $X$ , $I_A=I_B \implies A=B$ ? At the least , I am looking for an example of non-metrizable topological space for which it holds and a topological space where it doesn't hold . Please help . Thanks in advance .

Answer 1

A simple counterexample in general spaces is $\mathbb{R}$ with the cofinite topology, where all continuous real-valued functions $f$ are constant since $f$ can only be outside any $(a,b)$ finitely often and points are not open.

The simplest way to get a non-metrizable space for which this property nonetheless holds is to stick many metrizable spaces together. For instance, $[0,1]^\mathbb{R}$ with the product topology is not first countable, so not metrizable, but is compact Hausdorff and so normal, so that closed sets may be separated as you say. Indeed, normality is the property you want, by Urysohn's lemma.

EDIT: Actually, complete regularity is enough: this is the property that for every closed set $K$ and point $x$ there's a function vanishing on $K$ which is $1$ on $x$.