$(X,\mathscr T)$ is perfectly normal iff it is normal and each closed subsets of $X$ is a $G_{\delta}$ set.(Without using Urysohn lemma)

Question

A topological space $(X,\mathscr T)$ is perfectly normal iff it is normal and each closed subsets of $X$ is a $G_{\delta}$ set.

Proof. Suppose A topological space $(X,\mathscr T)$ is perfectly normal. If we consider two disjoint non-empty closed subsets $C$ and $D$. There exists a continuous function $f:X\to I=[0,1]: C=f^{-1}(0)$ and $D=f^{-1}(1).$ $[0,.3)$ and $(.6,1]$ are open in relative topology on $I.$ So $C\subseteq f^{-1}([0,.3))$ and $D\subseteq f^{-1}((.6,.1]).$ By the well-definition of $f,$ we have $ f^{-1}((.6,.1])\cap f^{-1}([0,.3))=\emptyset.$ So, $(X,\mathscr T)$ is normal. Consider the closed subset $C$ of $X$. Our next aim is to prove $C$ as the intersection of countable number of open sets. ...................................................................................................................................................................................................................................................................................................................................... ....................................................................................

Conversaly, A topological space $(X,\mathscr T)$ is normal and each closed subsets of $X$ is a $G_{\delta}$ set. If we consider two disjoint non-empty closed subsets $C$ and $D$. Our aim is to prove there exists a continuous function $f:X\to I=[0,1]:C=f^{-1}(0)$ and $D=f^{-1}(1).$ If we consider two disjoint non-empty closed subsets $C$ and $D$. There exists countable open subsets $\{U_i\}_{i\in \mathbb N}$ and $\{U_i\}_{i\in \mathbb N}: C=\bigcap_{i\in \mathbb N}U_i$ and $D=\bigcap_{i\in \mathbb N}V_i.$ Since $(X,\mathscr T)$ is normal, there exists disjoint non-empty open disjoint subsets of $X$, $U$ and $V$ such that $C\subseteq U$ and $D\subseteq V.$ With these data how could I construct a continuous function from $X$ to $I$? How do I complete the proof without Urysohn lemma?

Answer 1

You do need Urysohn's lemma for this.

Suppose $X$ is perfectly normal in the functional exact separation sense.

Then to see $X$ is normal take two disjoint closed subsets $C$ and $D$ of $X$, and find $f: X \to [0,1]$ with $f^{-1}[\{0\}] = C$ and $f^{-1}[\{1\}] = D$. Then $C \subseteq U=f^{-1}[[0,\frac12)]$ and the latter set is open and also $D \subseteq V =f^{-1}[[(\frac12,1]]$ and the latter set is also open and clearly $U \cap V=\emptyset$.

To see that all closed sets are $G_\delta$, let $A$ be closed. WLOG $A \neq X$ so we have a closed $B=\{p\}$ for some $p \notin A$ (assuming normal implies $T_1$ here). Anyway, we have a continuous function $f: X \to [0,1]$ with $f^{-1}[\{0\}]= A$ and then $A = \bigcap_{n \in \mathbb{N}^+} f^{-1}\left[[0, \frac{1}{n})]\right]$ shows that $A$ is a $G_\delta$ set.

Now suppose that $X$ is normal and all closed sets are $G_\delta$ sets. Let $A$ and $B$ be disjoint closed sets of $X$.

First we handle $A$: $A=\bigcap_n U_n$, where all $U_n$ are open. Then apply Urysohn's lemma (I see no way to avoid it) to find an $f_n: X \to [0,1]$ such that $f[A]=\{0\}$ and $f[X\setminus U_n]=\{1\}$. Then define $f(x) = \sum_n \frac{1}{2^n} f_n(x)$ defines a continuous function $X \to [0,1]$ with $f^{-1}[\{0\}]=A$.

Similarly we can find a continuous $g: X \to [0,1]$ with $g^{-1}[\{0\}] = B$. Now use $h=\frac{f}{f+g}$ as a continuous function $X \to [0,1]$ that obeys $h^{-1}[\{0\}]=A$ and $h^{-1}[\{1\}]=B$.