# For a closed $G_\delta$ set $F\subseteq X,$ does there exist a continuous function $f:X\to [0,1]$ such that $f=0$ on $F$ and $f\neq 0$ outside $F?$

(All spaces are Hausdorff.)

This question is a variant of my previous question.

Let $$X$$ be a completely regular space, that is, for every closed set $$F\subseteq X$$ and $$x\not\in F,$$ there exists a continuous function $$g:X\to [0,1]$$ such that $$g(F) = \{0\}$$ and $$g(x) =1.$$

Question: For every closed $$G_\delta$$ set $$F\subseteq X,$$ does there exist a continuous function $$f:X\to [0,1]$$ such that $$f=0$$ on $$F$$ and $$f\neq 0$$ outside $$F?$$

A subset $$U\subseteq X$$ is a zero set if there exists a continuous function $$g:X\to [0,1]$$ such that $$g^{-1}(\{0\}) = U.$$

It is well-known that if $$X$$ is normal then all closed $$G_\delta$$ sets are zero sets.

However, I am not sure whether the same holds for completely regular space.

• I know that $X$ normal implies that all closed $G_\delta$'s are zero-sets, but I believe the reverse does not hold. If it did your question would be useless. Do you have a reference ? "well-known" it is not, it's not mentioned in Engelking, a standard reference for such matters. – Henno Brandsma Jan 5 '19 at 14:26
• Note that this can only be true when $F$ is closed and $G_\delta$. You cannot ask this for all closed sets. – Henno Brandsma Jan 5 '19 at 14:27
• @HennoBrandsma yes, you are right. I have edited my post. – Idonknow Jan 5 '19 at 14:59
• @HennoBrandsma by the way, if the converse holds, why would by question be useless? – Idonknow Jan 5 '19 at 15:00
• @PaulFrost that is the direction I agreed with, yes. The OP originally claimed that if all closed $G_\delta$ sets are zero-sets then $X$ is normal. This I doubt the truth of, as said. – Henno Brandsma Jan 5 '19 at 18:46

There is a well-known counterexample. Let $$S$$ be the Sorgenfrey line, that is the real line endowed with the Sorgenfrey topology (generated by the base consisting of half-intervals $$[a,b)$$, $$a). It is well-known that a product $$X=S\times S$$ is a Tychonoff but non-normal space (see, for instance, Examples 1.4.4 and 2.3.12 in [Eng]). Let $$D=\{(x,-x)\in S\times S: x\in S\}$$. Then any subset $$Y$$ of $$D$$ is closed in $$X$$, and we have $$2^{|D|}=2^{\frak c}$$ such subsets. Moreover, since $$Y=\bigcap_{n\in\Bbb N} Y+[0,1/n)\times [0,1/n)$$, $$Y$$ is a $$G_\delta$$-subset of $$X$$. On the other hand, let $$C$$ be a countable dense subset of $$X$$. By Theorem 2.1.9 in [Eng], each continuous real-valued function $$f$$ on $$X$$ is uniquely determined by its restriction on $$C$$, so there are at most $$\frak c^\omega=\frak c$$ such functions. Thus most subsets of $$D$$ are not zero-sets.