# A Theorem About Hereditarily Normality

Let $Y$ be either the closed unit interval $\mathbb{I}=[0,1]$ or $[0,\omega]=\omega+1$. Let $X$ be any one of the following spaces:

$X=[0,\omega_1]=\omega_1+1$,
$X=[0,\omega_1)=\omega_1$,
$X=$ Michael Line,
$X=$ Sorgenfrey Line.

The cross product $X \times Y$ is normal in all four cases. The $X$ factor, in the order listed, is compact, countably compact, paracompact and Lindelof. Thus they are all countably paracompact. According to the Dowker’s theorem, the product of a normal space $X$ and a compact metric space $Y$ is normal if and only if $X$ is countably paracompact. My goal here is to show that the four cases of $X \times Y$ here cannot be hereditarily normal. This is from a theorem due to Katetov. We prove the following theorem.

Theorem. If $X \times Y$ is hereditarily normal, then either $X$ is perfectly normal or every countably infinite subspace of $Y$ is closed and discrete.

Proof. Suppose that $X$ is not perfectly normal and $Y$ has a countably infinite subset that is not closed and discrete. Let $H \subset X$ be a closed set that is not a $G_\delta-$set. Let $C=\lbrace{y_n:n \in \omega}\rbrace \subset Y$ be an infinite set with an accumulation point $y$. We assume that $y \notin C$.

We show that the open subspace $U=(X \times Y)-(H \times \lbrace{y}\rbrace)$ is not normal. To this end, let $A=H \times (Y-\lbrace{y}\rbrace)$ and $B=(X-H) \times \lbrace{y}\rbrace$. The sets $A$ and $B$ are disjoint closed subspaces of the open subspace $U$. Suppose we have disjoint open sets $S$ and $T$ such that $A \subset S$ and $B \subset T$.

For each $n \in \omega$, let $O_n=\lbrace{x \in X:(x,y_n) \in S}\rbrace$. Each $O_n$ is open and $H \subset O_n$. Thus $H \subset \bigcap_n O_n$. Let $x \in \bigcap_n O_n$. Then $(x,y) \in \overline{S}$. This means $x \in H$. If $x \notin H$, then $(x,y) \in B \subset T$ (which is impossible). So we have $H=\bigcap_n O_n$, indicating that $H$ is a $G_\delta-$set, and leading to a contradiction. So the subspace $U=(X \times Y)-(H \times \lbrace{y}\rbrace)$ is not normal.

Corollary. For countably compact spaces (in particular compact spaces) $X$ and $Y$, if the Cartesian product $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are prefectly normal.

Proof. Note that both factors, being countably compact, cannot have closed and discrete infinite subsets.

Comment. Note that the converse of the corollary is not true. Let both factors be the double arrow space, which is perfectly normal. But the square of the double arrow space contains a copy of the Sorgenfrey Plane, which is not normal.

Reference
[Katetov] Katetov, M., [1948] Complete normality of Cartesian products, Fund. Math., 36, 271-274.