Chapter 4.32, Problem 9E

Prove the following:

Theorem: If $J$ is uncountable, then ${ℝ}^J$ is not normal.

Proof: (This proof is due to A.H. Stone, as adapted in [S-S].) Let $X={\left({\mathbb{Z}}_{+}\right)}^J$; it will suffice to show that X is not normal, since X is a closed subspace of ${ℝ}^J$. We use functional notation for the elements of X, so that the typical element of X is a function $x:J\to {\mathbb{Z}}_{+}$.

(a) If $x\in X$ and if B is a finite subset of J, let $U\left(x,B\right)$ denote the set consisting of all those elements $y$ of X such that $y\left(\alpha \right)=x\left(\alpha \right)$ for $\alpha \in B$. Show the sets $U\left(x,B\right)$ are a basis for X.

(b) Define $P_n$ to be the subset of X consisting of those $x$ such that on the set $J-x^{-1}\left(n\right)$, the map $x$ is injective. Show that $P_1$ and $P_2$ are closed and disjoint.

(c) Suppose U and V are open sets containing $P_1$ and $P_2$, respectively. Given a sequence ${\alpha }_1,{\alpha }_2,\dots$ of distinct elements of $J$, and a sequence

$0=n_0<n_1<n_2<\dots$

of integers, for each $i\ge 1$let us set

$B_i=\left\{{\alpha }_1,\dots ,{\alpha }_{ni}\right\}$

And define $x_i\in X$ by the equations

$x_i\left({\alpha }_j\right)=j$ for $1\le j\le n_{i-1}$

$x_i\left(\alpha \right)=1$ for all other values of $\alpha$

Show that one can choose the sequences ${\alpha }_j$ and $n_j$ so that for each $i$, one has the inclusion

$U\left(x_i,B_i\right)\subset U$

[Hint: To begin, note that $x_1\left(\alpha \right)=1$ for all $\alpha$, now choose $B_1$ so that $U\left(x_1,B_1\right)\subset U$.]

(d) Let A be the set $\left\{{\alpha }_1,{\alpha }_2,\dots \right\}$ constructed in (c). Define $y:J\to {\mathbb{Z}}_{+}$ by the equations

$y\left({\alpha }_j\right)=j$ for ${\alpha }_j\in A$,

$y\left(\alpha \right)=2$ for other values of $\alpha$.

Choose B so that $U\left(y,B\right)\subset V$. Then choose $i$ so that $B\cap A$ is contained in the set $B_i$. Show that

$U\left(x_{i+1},B_{i+1}\right)\cap U\left(y,B\right)$

is not empty.