Lemma 31.1 and Exercise 2, Section 31 of Munkres’ Topology

Question

Show that if $X$ is normal, every pair of disjoint closed sets have neighborhoods whose closures are disjoint.

Proof: Prob. 2, Sec. 31, in Munkres' TOPOLOGY, 2nd ed: Any pair of disjoint closed sets in a normal space have neighborhoods whose closures are disjoint.

In above proof, we didn’t use $T_1$ separation axiom, by Munkres Lemma 31.1, no need for $T_1$. So we only need $X$ to be $T_4$. Am I right?

Show that if $X$ is $T_4$, every pair of disjoint closed sets have neighborhoods whose closures are disjoint.

Rephrasing lemma 31.1 to my taste:

Let $(X, \mathcal{T}_X)$ be a topological space.

(a) $X$ is $T_3$ $\iff$ If $x\in X$ and $U\in \mathcal{N}_x$, then $\exists V\in \mathcal{N}_x$ such that $\overline{V} \subseteq U$.

(b) $X$ is $T_4$ $\iff$ If $A$ is closed in $X$ and $U\in \mathcal{N}_A$, then $\exists V\in \mathcal{N}_A$ such that $\overline{V} \subseteq U$.

Answer 1

According to that definition ($X$ normal=$X$ is $T_4$ and $T_1$ ), then $X$ just need to be $T_4,$

so doesn't need normality.

However, in Munkres (1975) book, normality is defined as:

"The space $X$ is said to be normal if for each pair $A,B$ of disjoint closed sets of $X,$ there exist disjoint open sets containing $A$ and $B,$ respectively."

rewording:

$∀F,G⊆X$ closed and $F∩G=∅,∃U,V∈τ:U∩V=∅,F⊆U$ and $G⊆V$