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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.

GEdgar
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  • What topological properties does $X$ have? – Sungjin Kim Jan 13 '19 at 17:36
  • For example, an open set is a $G_\delta$, and there are likely to be open sets that are not closed, and therefore not zero-sets. – GEdgar Jan 13 '19 at 17:41
  • even a closed $G_\delta$ need not be a zero set https://math.stackexchange.com/q/292948 As a side,you never so far accepted any answer: Are you aware that you could accept answers (which is good for whoever posted the answer,as well as for others that read the question later)? You could also vote answers up or down,see A quick tour https://math.stackexchange.com/tour Helpful tips to write a good question https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question/9960#9960 MathJax https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Mirko May 23 '19 at 11:14

3 Answers3

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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.

user3482749
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  • what you mean about "$f(1) = \lim_{x\to1}f(x) = \lim_{x\to1^-}0 = 0,$? –  Jan 13 '19 at 18:41
  • For a continuous function $f$, the value of $f$ at any point is equal to its two-sided limit at that point, which is equal to each of the one-sided limits there. Applying this with the left-handed limit at $1$ for any continuous function such that $Z(f) \supseteq (0,1)$ gives that $f(1) = 0$. – user3482749 Jan 13 '19 at 18:47
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$Z(f)$ 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.)

Nate Eldredge
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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.

Henno Brandsma
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