Show that if $X$ is normal, every pair of disjoint closed sets 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 $\overline{V} \subseteq U$.
Best Answer
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 $G⊆V$