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<n<\omega}, a sequence of functions \lbrace{f_n}\rbrace_{0<n<\omega}  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_0F_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}).

  2. Pingback: Lindelof Exercise 2 | Dan Ma's Topology Blog

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