Why is a perfectly normal space completely normal?

Question

Suppose that $X$ is perfectly normal space.

To show that $X$ is completely normal, I must show that every subspace of $X$ is normal. To that effect, let $Y$ be a subspace of $X$.

Let $A, B$ be disjoint closed sets in Y. So there exist closed sets $A_1, B_1$ in X (these need not be disjoint) such that $A=A_1\cap Y, B=B_1\cap Y$. Since X is perfectly normal, every closed set in it is $G_\delta$. It follows that there exist open sets $U_i$'s and $V_i$'s in X such that $A_1=\cap U_i, B_1= \cap V_i$. This gives: $A=\cap (U_i\cap Y), B=\cap (V_i\cap Y)$.

But I'm not sure how to proceed further. The problem is that $A_1, B_1$ need not be disjoint. If they were disjoint then by normality, they could be separated by disjoint open sets.

My question is how to complete the solution from here.

Note: I found the same question being asked here. But it makes no sense as it talks about closure $A$ and closure $B$ and I don't understand how those help here.

Answer 1

Perfect normality is a hereditary property (see here). And perfectly normal spaces are normal. So every subspace of a perfectly normal space is normal, which is the definition of completely normal.

Answer 2

Convention: It's assumed throughout that normal is $T_4$ , i.e., it is understood that singleton sets are closed.

Definition: $X$ is completely normal.$\iff$ Every suspace $Y\subset X$ is normal.

Lemma: $X$ completely normal. $\iff$ For every separated sets $A$ and $B$ in $X$, there exist disjoint open sets $U\supset A, V\supset B$ in $X$.

Proof: $(\Rightarrow):$ Let $A$ and $B$ be separated sets in $X$. $\implies A\cap \bar B=\emptyset, B\cap \bar A=\emptyset$.

$\implies$ $A, B$ are subsets of $(X-\bar B\cap \bar A)=:Y.$ Let's denote closure of $C$ in $Y$ as $\bar C|_Y$.

Note that $\bar A|_Y\cap \bar B|_Y=\bar A\cap \bar B\cap Y=\emptyset\implies$ there exist disjoint open sets $U$ and $V$ in $Y$ (hence in $X$ since $Y$ is open) such that $U\supset \bar A|_Y, V\supset \bar B|_Y$. $A=A\cap Y\subset \bar A\cap Y\subset U\implies A\subset U$. Similarly, $B\subset V$.

$(\Leftarrow):$ Let $Y\subset X$. Suppose that $A, B$ are disjoint closed sets in $Y$.

$\bar B|_Y\overbrace{=}^{\text{because B is closed in Y}}B=\bar B\cap Y$

$A\cap \bar B|_Y=\emptyset=A\cap \bar B$. Similarly, $B\cap \bar A=\emptyset.$

$\implies A$ and $B$ are separated. $\implies $ There exist disjoint open sets $U\supset A, V\supset B$. Since $U$ and $V$ are also open in $Y$, we are done.

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Now to prove the statement in OP, suppose that $A$ and $B$ are separated sets in $X$. Since $X$ is perfectly normal, every closed set in it is a $G_\delta$ set so by Urysohn's lemma, there exist continuous maps $f,g: X\to [0,1]$ such that $f$ vanishes only on $\bar A$ and $g$ vanishes only on $\bar B$.

Consider the continuous map $h:=f-g$. Note that

$h^{-1}((-\infty, 0))=\{t\in X: h(t)<0\}\supset A\implies A$ is contained in open set $h^{-1}((-\infty, 0))$. Similarly, $B$ is contained in open set $h^{-1}((0,\infty))\supset B$. The open sets $h^{-1}((0,\infty))$ and $h^{-1}((-\infty,0))$ are disjoint. By the lemma, we are done. QED.