Prove the following.
Theorem. If \(J\) is uncountable, then \(\mathbb{R}^J\) is not normal.
Proof. (This proof is due to A. H. Stone, as adapted in [S-S].) Let \(X = (\mathbb{Z}_+)^J\); it will suffice to show that \(X\) is not normal, since \(X\) is a closed subspace of \(\mathbb{R}^J\). We use functional notation for the elements of \(X\), so that the typical element of \(X\) is a function \(\mathbf{x}: J \to \mathbb{Z}_+\).
If \(\mathbf{x} \in X\) and if \(B\) is a finite subset of \(J\), let \(U(\mathbf{x}, B)\) denote the set consisting of all those elements \(\mathbf{y}\) of \(X\) such that \(\mathbf{y}(\alpha) = \mathbf{x}(\alpha)\) for \(\alpha \in B\). Show the sets \(U(\mathbf{x}, B)\) are a basis for \(X\).
Define \(P_n\) to be the subset of \(X\) consisting of those \(\mathbf{x}\) such that on the set \(J - \mathbf{x}^{-1}(n)\), the map \(\mathbf{x}\) is injective. Show that \(P_1\) and \(P_2\) are closed and disjoint.
Suppose \(U\) and \(V\) are open sets containing \(P_1\) and \(P_2\), respectively. Given a sequence \(\alpha_1, \alpha_2, \ldots\) of distinct elements of \(J\), and a sequence
\[ 0 = n_0 < n_1 < n_2 < \dots \]
of integers, for each \(i \geq 1\) let us set
\[ B_i = \ts{\alpha_1, \dots, \alpha_{n_i}} \]
and define \(\mathbf{x}_i \in X\) by the equations
\[ \begin{array}{ll} \mathbf{x}_i(\alpha_j) = j & \text{for } 1 \leq j \leq n_{i-1}, \\ \mathbf{x}_i(\alpha) = 1 & \text{for all other values of } \alpha. \end{array} \]
Show that one can choose the sequences \(\alpha_j\) and \(n_j\) so that for each \(i\), one has the inclusion
\[ U(\mathbf{x}_i, B_i) \subset U. \]
[Hint: To begin, note that \(\mathbf{x}_1(\alpha) = 1\) for all \(\alpha\); now choose \(B_1\) so that \(U(\mathbf{x}_1, B_1) \subset U\).]
- Let \(A\) be the set \(\ts{\alpha_1, \alpha_2, \ldots}\) constructed in (c). Define \(\mathbf{y}: J \to \mathbb{Z}_+\) by the equations
\[ \mathbf{y}(\alpha_j) = j \quad \text{for } \alpha_j \in A, \]
\[ \mathbf{y}(\alpha) = 2 \quad \text{for all other values of } \alpha. \]
Choose \(B\) so that \(U(\mathbf{y}, B) \subset V\). Then choose \(i\) so that \(B \cap A\) is contained in the set \(B_i\). Show that
\[ U(\mathbf{x}_{i+1}, B_{i+1}) \cap U(\mathbf{y}, B) \]
is not empty.
Solution
Let \(X=(\mathbb Z_+)^J\), with each factor discrete.
In the product topology, a basic open set specifies open conditions in only finitely many coordinates. Since the factors are discrete, we may refine each specified coordinate condition to a singleton. Thus the sets \[ U(\mathbf x,B)=\{\mathbf y:\mathbf y(\alpha)=\mathbf x(\alpha)\text{ for all }\alpha\in B\}, \] with \(B\subset J\) finite, form a basis.
The complement of \(P_n\) is open. Indeed, \(\mathbf x\notin P_n\) exactly when there are distinct \(\alpha,\beta\in J\) and an integer \(m\ne n\) such that \[ \mathbf x(\alpha)=\mathbf x(\beta)=m. \] The cylinder imposing these two equalities is an open neighborhood of \(\mathbf x\) contained in \(X\setminus P_n\). Hence \(P_n\) is closed.
The sets \(P_1\) and \(P_2\) are disjoint. If \(\mathbf x\) belonged to both, then no two distinct coordinates could have the same value: a repetition at value \(1\) is forbidden by membership in \(P_2\), a repetition at value \(2\) is forbidden by membership in \(P_1\), and a repetition at any other value is forbidden by both. Thus \(\mathbf x:J\to\mathbb Z_+\) would be injective, impossible because \(J\) is uncountable.
- We construct the sequences recursively. For \(i=1\), \(\mathbf x_1\) is the constant function \(1\), hence lies in \(P_1\subset U\). By (a), choose a finite set \(C_1\subset J\) such that \[ U(\mathbf x_1,C_1)\subset U. \] Enumerate the elements of \(C_1\) among distinct points \(\alpha_1,\dots,\alpha_{n_1}\), enlarging if necessary so \(n_1>0\). Then \[ U(\mathbf x_1,B_1)\subset U(\mathbf x_1,C_1)\subset U. \]
Suppose \(\alpha_1,\dots,\alpha_{n_{i-1}}\) and \(n_{i-1}\) have been chosen. Define \(\mathbf x_i\) as in the problem. Outside the value \(1\), its values are among \(2,\dots,n_{i-1}\), each occurring at most once, so \(\mathbf x_i\in P_1\subset U\). Choose a finite \(C_i\subset J\) with \[ U(\mathbf x_i,C_i)\subset U. \] Since \(J\) is uncountable and only finitely many \(\alpha_j\) have been used, extend the sequence by finitely many new distinct \(\alpha_j\)’s so that \(C_i\subset B_i=\{\alpha_1,\dots,\alpha_{n_i}\}\) for some \(n_i>n_{i-1}\). Then \[ U(\mathbf x_i,B_i)\subset U. \] This completes the recursion.
- The function \(\mathbf y\) lies in \(P_2\): outside the value \(2\), its values on \(A=\{\alpha_j\}\) are distinct, and all points outside \(A\) have value \(2\). Hence \(\mathbf y\in V\). Choose a finite \(B\subset J\) with \[ U(\mathbf y,B)\subset V. \] Because \(B\cap A\) is finite, choose \(i\) such that \[ B\cap A\subset B_i. \]
The coordinate prescriptions defining \(U(\mathbf x_{i+1},B_{i+1})\) and \(U(\mathbf y,B)\) are compatible. On \(B\cap B_{i+1}\), every coordinate lies in \(B\cap A\subset B_i\), and for \(\alpha_j\in B_i\) both functions take the value \(j\). On coordinates of \(B\setminus A\), only \(\mathbf y\) prescribes a value, and on coordinates of \(B_{i+1}\setminus B\), only \(\mathbf x_{i+1}\) does. Hence one can define a function \(\mathbf z:J\to\mathbb Z_+\) agreeing with both prescriptions on \(B\cup B_{i+1}\), arbitrarily elsewhere. Then \[ \mathbf z\in U(\mathbf x_{i+1},B_{i+1})\cap U(\mathbf y,B). \] Thus every open \(U\supset P_1\) and \(V\supset P_2\) intersect, so the disjoint closed sets \(P_1,P_2\) cannot be separated by disjoint open neighborhoods. Hence \(X\) is not normal.
Finally, \(X=(\mathbb Z_+)^J\) is closed in \(\mathbb R^J\), since \[ X=\bigcap_{\alpha\in J}\pi_\alpha^{-1}(\mathbb Z_+) \] and \(\mathbb Z_+\) is closed in \(\mathbb R\). If \(\mathbb R^J\) were normal, its closed subspace \(X\) would be normal, contradiction. Therefore \(\mathbb R^J\) is not normal for uncountable \(J\).