# The Uncountable Product Of The Countable Discrete Space Is Not Normal

It is well known that normality is very easily destroyed by taking Cartesian product. In a previous post, we showed that the Cartesian product of the Sorgenfrey Line with itself is not normal. Here, we present an example showing that the uncountable product of metric spaces can fail to be normal. Specifically we show that the product $\mathbb{N}^{\omega_1}$ is not normal where $\mathbb{N}$ is the set of all natural numbers and $\omega_1$ is the first uncountable ordinal. The set $\omega$ is the first infinite ordinal and is identical to $\mathbb{N}$.

For $j \in \lbrace{0,1}\rbrace$, let $H_j=\lbrace{f \in \mathbb{N}^{\omega_1}:}$ for each $i<\omega$ where $i \neq j$, $f(\alpha)=i$ for at most one $\alpha<\omega_1 \rbrace$. It is clear that $H_0$ and $H_1$ are disjoint and closed. We show that $H_0$ and $H_1$ cannot be separated by disjoint open sets.

Let $U_0$ and $U_1$ be open sets in $\mathbb{N}^{\omega_1}$ such that $H_0 \subset U_0$ and $H_1 \subset U_1$. We will produce a function $h$ such that $h \in U_0 \cap U_1$.

For any finite set $F \subset \omega_1$ and for any $f \in \mathbb{N}^{\omega_1}$, let $\mathcal{B}(F,f)$ be defined as the following:

$\displaystyle . \ \ \ \ \mathcal{B}(F,f)=\lbrace{t \in \mathbb{N}^{\omega_1}:t = f \text{ on the set }F}\rbrace$.

Because each factor in the product space is the discrete space $\mathbb{N}$, $\mathcal{B}(F,f)$ is a basic open set at the point $f$.

Define an inductive process producing a sequence of finite sets $\lbrace{F_n}\rbrace_{0, a sequence of functions $\lbrace{f_n}\rbrace_{0  such that for each $n$, $F_n \subset \omega_1$, $f_n \in H_0$, $\mathcal{B}(F_n,f_n) \subset U_0$.

To start off, let $f_0$ be the zero function on $\omega_1$. Since $f_o \in H_0$, there is some finite $F_1$ such that $\mathcal{B}(F_1,f_0) \subset U_0$$F_1=\lbrace{\alpha_0,\alpha_1,...,\alpha_{s(1)}}\rbrace$. Define $f_1$ by $f_1(\alpha_i)=i$ for all $\alpha_i \in F_1$ and $f_1(\alpha)=0$ for all $\alpha \in \omega_1-F_1$.

Now consider the $n^{th}$ step in the induction process where $n>1$. Since $f_{n-1} \in H_0$, choose some finite $F_n$ such that $\mathcal{B}(F_n,f_{n-1}) \subset U_0$. $F_n$ can be chosen in such a way that it extends $F_{n-1}$. That is , $F_n=\lbrace{\alpha_0,\alpha_1,...,\alpha_{s(n-1)},\alpha_{s(n-1)+1}, ..., \alpha_{s(n)}}\rbrace$. Now, define $f_n$ by $f_n(\alpha_i)=i$ for all $\alpha_i \in F_n$ and $f_n(\alpha)=0$ for all $\alpha \in \omega_1-F_n$. To help with a later step, note that:

$\displaystyle (*) \ \ \ \ f_{n-1}(\alpha_i)=i$ for all $\alpha_i \in F_{n-1}$ and $f_{n-1}(\alpha_i)=0$ for all $\alpha_i \in F_n-F_{n-1}$.

Let $F=\cup_{n}F_n$. Define $g$ by $g(\alpha_i)=i$ for all $\alpha_i \in F$ and $g(\alpha)=1$ for all $\alpha \in \omega_1-F$. Since $g \in H_1$, there is some finite $G \subset \omega_1$ such that $\mathcal{B}(G,g) \subset U_1$. Since $G$ is finite, choose some integer $n$ such that $G \cap F \subset F_{n-1}$. Now define $h$ by $h(\alpha_i)=i$ for all $\alpha_i \in F_{n-1}$, $h(\alpha_i)=0$ for all $\alpha_i \in F_n-F_{n-1}$, and $h(\alpha)=1$ for all $\alpha \in \omega_1-F_n$.

Note that the function $h$ agrees with $g$ on the finite set $G$. Thus $h \in \mathcal{B}(G,g) \subset U_1$. Note that the function $h$ agrees with $f_{n-1}$ on the finite set $F_n$ (see $(*)$). Thus $h \in \mathcal{B}(F_n,f_{n-1}) \subset U_0$. The proof that $\mathbb{N}^{\omega_1}$ is not normal is now complete.

## 2 thoughts on “The Uncountable Product Of The Countable Discrete Space Is Not Normal”

1. Hi,
great post as always but I found a small mistake: in the end of the paragraph “Define an inductive process…” I think you meant B(F_n,f_{n-1}).