Condition in Strong Form of Urysohn Lemma superfluous? I am doing exercise 33.5 of Munkres:


Exercise 33.6: Theorem (Strong form of Urysohn Lemma). Let $X$ be a normal space. There is a continuous function $f : X \to [0,1]$ such that $f(x) = 0$ for all  $x\in A$ , and $f(x) = 1$ for all $x \in B$, and $0 < f(x) < 1$ otherwise iff $A$ and $B$ are disjoint closed $G_\delta$ sets in $X$.


Now the existence of such a function immediately gives that $A$ and $B$ are closed $G_\delta$ sets since $$A = f^{-1}\left( \bigcap_{n\geq 1} B_{\frac{1}{n}}(0)\right)$$
and similarly for $B$. Now for the other direction, I have also proved it but  only using the fact that one of $A$ or $B$ is $G_\delta$. How did I do this? Well by exercise 4 of the same section there is $f : X \to [0,1]$ such that $f(x) = 0$ for all $x \in A$ and $f(x) > 0$ for all $x \notin A$ since $A$ is $G_\delta$. By the Urysohn Lemma, there is $g(x) : X\to [0,1]$ such that $g|_B = 0$ and $g|_A = 1$. Then 
$$h(x) = \frac{g(x)}{f(x) + g(x)}$$
is the required function.


My question is: For this direction of the problem, is it a superfluous condition that we need both $A$ and $B$ to be $G_\delta$?


Thanks.
 A: No, though for many common spaces it is a superfluous condition.
A subset $A$ of a topological space is called a zero-set if there is a continuous function $f : X \to \mathbb{R}$ such that $A = f^{-1} [ \{ 0 \} ]$.  From Urysohn's Lemma one can almost immediately conclude that in a normal space the zero-sets are exactly the closed G$_\delta$ sets.

Added: Note that in the statement of the Strong Form of Urysohn's Lemma we require $f : X \to [ 0 , 1 ]$ to be continuous, $A = f^{-1} [ \{ 0 \} ]$ and $B = f^{-1} [ \{ 1 \} ]$.  This means that both $A$ and $B$ must be zero-sets, and so by the statement above they must be closed G$_\delta$.

There is a class of spaces strictly smaller than the normal spaces called perfectly normal, defined by being normal and having all closed sets G$_\delta$.  In these spaces the G$_\delta$ condition is clearly superfluous, and many of the most common normal spaces are actually perfectly normal: the real line (in fact all metric spaces), and the Sorgenfrey line, for example.
As an example of a normal not perfectly normal space, consider the one-point compactification of an uncountable discrete space: $\alpha X = X \cup \{ \infty \}$ where $X$ is uncountable, and topologised so that all points of $X$ are isolated, and the open neighbourhoods of $\infty$ are of the form $\{ \infty \} \cup A$ where $A$ is a co-finite subset of $X$.
A basic fact about this space is that given any continuous function $f : \alpha X \to \mathbb{R}$, the set $$\{ x \in X : f(x) \neq f ( \infty ) \}$$ is countable.  It follows that the closed set $\{ \infty \}$ is not a zero-set.

Added: In particular, given any finite (nonempty) $F \subseteq X$ there is no continuous function $f : \alpha X \to [ 0 , 1 ]$ such that $f^{-1} [ \{ 0 \} ] = \{ \infty \}$ and $f^{-1} [ \{ 1 \} ] = F$.

