Is every $G_\delta $, zero-set?

Question

Zero-set means a set of the form:

$Z(f) = \{ x \in X | f(x) = 0 \}\quad\text{for some } f \in C(X)$

$C(X)$ is the ring of continuous function on $X$.

I know that every zero-set is $G_\delta $, i.e, a countably intersection of open sets.

Is every $G_\delta $, zero-set? if not, can you give me an simple example.

Answer 1

Trivially not, in general: there is no continuous function on $\mathbb{R}$ whose zero set is $(0,1)$, since if $f$ were such a function, then it would satisfy $$f(1) = \lim_{x\to1}f(x) = \lim_{x\to1^-}0 = 0,$$ so $1 \in Z(f)$, so $Z(f) \neq (0,1)$, but as an open set, $(0,1)$ is clearly $G_\delta$: it is the intersection of countably many copies of itself.

Answer 2

$Z(f)$ has to be closed, so any $G_\delta$ set which is not closed will be a counterexample.

As a very specific example, in $\mathbb{R}$, the set $(0,1)$ is $G_\delta$ (indeed it's open) but it cannot be the zero set of any continuous function.

It's also not true in general that every closed $G_\delta$ set is a zero set, not even if $X$ is Hausdorff. See Is a closed $G_\delta$ set in a Hausdorff space always a zero set?. (It is true if $X$ is $T_6$, and maybe some weaker condition would also suffice.)

Answer 3

A zero-set of $X$ is always a closed $G_\delta$ by construction, in any $X$. A closed $G_\delta$ set is a zero-set when $X$ is metric or more generally $T_4$.

This need not hold in all spaces, though. You'll meet counterexamples naturally later in the study of rings of continuous functions.