Normality is not hereditary

Question

For $(S,\tau_{disc})$ denote $A(S) = S \cup \{\infty_S\}$ as its onepoint compactification. Let $S,T$ be discrete spaces. Then clearly $X := A(S) \times A(T)$ is compact Hausdorff since each onepoint compactification is compact Hausdorff. $X$ is also $T_1$ since $A(S)$ and $A(T)$ are $T_1$. Thus is $X$ normal. Now, let $Y := X \setminus \{(\infty_S,\infty_T)\}$. Since $X$ is $T_1$ we see that $Y$ is open, and thus as an open subspace of a compact Hausdorff space, also locally weakly compact. Now consider the sets $C := S \times \{\infty_T\}$ and $D := \{\infty_S\}\times T$. Clearly $C \cap D = \emptyset$ and both are closed in $Y$, since $$ C = S \times \{\infty_T\} = (A(S) \times \{\infty_T\}) \cap Y $$ where $A(S) \times \{\infty_T\}$ is closed in $X$ since this set equals $p_2^{-1}(\{\infty_T\})$, where $p_2$ is the projection from $X$ to $A(T)$. Similar $D$ is closed.

Now I want to show that $C$ and $D$ can't be separated by open sets in $Y$, when $S$ is countably infinite and $T$ is uncountable. What is the contradiction, if we assume that $C$ and $D$ can be separated in $Y$ ?

Answer 1

An open set containing $C$ or $D$ is the union of (open) neighbourhoods of the points of $C$ resp. $D$.

So how does a neighbourhood of $(s,\infty_T)$ resp. $(\infty_S,t)$ look like?. A basis of the neighbourhoods of $(x,y)$ is given by $U\times V$, where $U$ is a neighbourhood of $x$, and $V$ a neighbourhood of $y$.

We try to find disjoint open sets containing $C$ resp. $D$, so we will have greater chance of success if we choose the neighbourhoods small. Since $S$ and $T$ are discrete, we can use singleton sets for the neighbourhoods of that component, it doesn't get any smaller ;)

But a neighbourhood of $\infty_X,\; X \in \{S,T\}$, contains the complement of a compact subset of $S$ resp. $T$, and since the to spaces are discrete, that means a cofinite set.

So a neighbourhood of $C$ contains a set of the form

$$U = \bigcup_{s\in S} \{s\} \times (A(T)\setminus F_s),$$

where for each $s\in S$, $F_s$ is a finite subset of $T$. Since $S$ is countable,

$$R = \bigcup_{s\in S} F_s$$

is a countable subset of $T$, and hence

$$U \supset S \times (T\setminus R).$$

$T$ is uncountable, so $T\setminus R \neq \varnothing$, and for every $t \in T\setminus R$, we have $S\times{t} \subset U$. But an open set $V$ containing $D$ contains a neighbourhood of $(\infty_S, t)$, and thus it contains $(S\setminus G_t)\times \{t\}$ for some finite subset $G_t$ of $S$. Thus

$$\varnothing \neq (S\setminus G_t)\times\{t\} \subset U\cap V.$$