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

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?

• My guess is that you will need Urysohn's lemma, but I do not have a proof at the moment either. – Carsten S Mar 4 at 9:33
• I haven't reach there. It is in the next section. :( – Mathematician reloaded Mar 4 at 10:18
• I recomend you to have a look at Engelking's book General Topology. If I'm not wrong, there is a corollary stating extacly that. I'll check it when I come back. – Dog_69 Mar 4 at 20:18
• But, It is an exercise question in 'Foundations of Topology' by C.W Patty. This exercise gave in 5.3 and Urysohn lemma is in 5.4. @Dog_69 – Mathematician reloaded Mar 5 at 2:11
• How does the author define normal? – Henno Brandsma Mar 5 at 14:54

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 $$h^{-1}[\{1\}]=B$$.

• But, It is an exercise question in 'Foundations of Topology' by C.W Patty. This exercise gave in 5.3 and Urysohn lemma is in 5.4. – Mathematician reloaded Mar 5 at 1:48
• @Mathematicianreloaded Urysohn’s lemma should be part of the text, not an exercise IMO. – Henno Brandsma Mar 5 at 5:07
• I mean this statement given as Theorem.5.30. But Urysohn's lemma given as Theorem 5.36. How can I use a theorem before studying the theorem? – Mathematician reloaded Mar 5 at 14:36
• @Mathematicianreloaded If it's a theorem in the text you already know its proof. Or do you have to prove all the statements in the book yourself, Texas style? What other facts about functions and normality are known in the book? Tietze? – Henno Brandsma Mar 5 at 14:41
• Nope. proof is not given. They given as exercise 5.3.11. – Mathematician reloaded Mar 5 at 14:46