## A.H. Stone’s Theorem III

It is well-known that the product of arbitrary normal spaces need not be normal; the finite product even need not be normal. Robert Sorgenfrey demonstrated such an example with the Sorgenfrey plane. What is is true is that the finite product of metric spaces is normal. In fact, we can say more: the countable product of metric (metrizable) spaces is itself a metric (metrizable) space, hence normal.

As the uncountable product of metic (metrizable) spaces need not be a metric (metrizable) space, perhaps we should not be surprised that the uncountable product of normal spaces need not be normal. This fact motivates the consideration of fully normal/paracompact factor spaces. The concluding result of Stone’s paper is that the product of nonempty metric spaces is normal if and only its fully normal, which is true precisely when at most countably many of the factor spaces are not compact.

Before we get to that result, we’ll first spend a bit of time showing that the product of uncountably many metric spaces need not be normal by demonstrating a counterexample. This counterexample was presented by A.H. Stone in his original paper, and I have followed his exposition fairly closely.

Let ${N}$ denote the topological space consisting of the positive integers equipped with the discrete topology (i.e. the discrete metric). ${N}$ is trivially normal. Let ${\Lambda}$ be an uncountable set, say the reals. Define a topological space ${T}$ by equipping the Cartesian product ${\prod_{\lambda\in\Lambda}N_{\lambda}}$ (i.e. the set of all ${N}$-valued functions on ${\Lambda}$) with the product topology. We restrict our attention to basic open sets ${U}$ where a finite set ${R(U)}$ of coordinates ${\lambda}$ are restricted to fixed values ${\xi_{\lambda}}$.

For each ${k\in N}$, define ${A^{k}}$ to be the set of all ${x=(\xi_{\lambda})\in T}$ such that if ${n\neq k}$, then there is at most one index ${\lambda}$ with ${\xi_{\lambda}=n}$.

$\displaystyle A^{k}:=\left\{x=(\xi_{\lambda})\in T:\text{card}(x^{-1}(n))\leq 1\text{ if }n\neq k\right\}$

I claim that the sets ${\left\{A^{k}\right\}_{k=1}^{\infty}}$ are pairwise disjoint, closed sets in ${T}$. Consider ${A^{k}}$, for ${k}$ fixed. Since ${\Lambda}$ is uncountable (this is the only place where I can see that we use this hypothesis) and ${N}$ is countable, there exist at least two indices ${\lambda_{1},\lambda_{2}}$ such that ${\xi_{\lambda_{1}}=\xi_{\lambda_{2}}=n}$. By definition of ${A^{k}}$, we have ${n=k}$, which establishes disjointness. To see that ${A^{k}}$ is closed, let ${x=(\xi_{\lambda})\in T\setminus A^{k}}$. Then there are at least two indices ${\lambda_{1},\lambda_{2}}$ such that ${\xi_{\lambda_{1}}=\xi_{\lambda_{2}}=n\neq k}$. Hence,

$\displaystyle \left\{n\right\}_{\lambda_{1},\lambda_{2}}\times\prod_{\lambda\neq\lambda_{1},\lambda_{1}}N_{\lambda}$

is an open neighborhood of ${x}$ contained in ${T\setminus A^{k}}$. To prove that ${T}$ is not normal, we will show that there do not exist disjoint, open sets ${U}$ and ${V}$ containing ${A^{1}}$ and ${A^{2}}$, respectively, by means of a contradiction.

We inductively define sequences of points ${x_{n}\in A^{1}}$, of increasing integers ${0, and of indices ${\lambda_{n}\in\Lambda}$. Let ${x_{1}}$ be the point ${(\xi_{\lambda})}$ such that ${\xi_{\lambda}=1}$ for all ${\lambda\in\Lambda}$. It is evident that ${x_{1}\in A_{1}\subset U}$, so there exists a basic open neighborhood ${U_{1}\subset U}$ containing ${x_{1}}$ where ${R(U_{1})}$ denotes the set of ${m_{1}}$ coordinates ${\lambda_{1},\ldots,\lambda_{m_{1}}}$ such that ${y=(\eta_{\lambda})\in U_{1}\Leftrightarrow \eta_{\lambda}=1}$ for all ${\lambda\in R(U_{1})}$. Suppose we have defined points ${x_{1},\ldots,x_{n}\in A^{1}}$, integers ${m_{1}<\cdots, and indices ${\lambda_{1},\ldots,\lambda_{m_{n}}}$ that are restricted coordinates in a basic open neighborhood ${U_{n}\subset U}$ containing ${x_{n}}$. Define ${x_{n+1}}$ coordinate-wise by

$\displaystyle\xi_{\lambda}=\begin{cases}{k} & {\lambda=\lambda_{k}\text{ with }1\leq k\leq m_{n}}\\{1} & {\text{otherwise}}\end{cases}$

It is clear that ${x_{n+1}\in A^{1}\subset U}$, so there exists a basic open neighborhood ${U_{n+1}\subset U}$ containing ${x_{n+1}}$. By enlarging ${R(U_{n+1})}$ by finitely many elements if necessary, we may assume that ${R(U_{n})\subset R(U_{n+1})}$. Define ${m_{n+1}}$ to be the cardinality of ${R(U_{n+1})}$ and enumerate the ${m_{n+1}-m_{n}}$ elements of ${R(U_{n+1})\setminus R(U_{n})}$ by ${\lambda_{m_{n}+1},\ldots,\lambda_{m_{n+1}}}$. This completes the induction step.

Define a point ${y=(\eta_{\lambda})}$ coordinate-wise by

$\displaystyle \eta_{\lambda}=\begin{cases}{k} & {\lambda=\lambda_{k}\text{ for some }k\in N}\\{2} & {\text{otherwise}}\end{cases}$

It is evident that ${y\in A^{2}}$, so there is a basic open neighborhood ${V_{0}\subset V}$ containing ${y}$. ${R(V_{0})}$ is finite, so there exists an integer ${n}$ such that ${k>m_{n}\Rightarrow\lambda_{k}\in\Lambda\setminus R(V_{0})}$. Define a point ${z=(\zeta_{\lambda})}$ by

$\displaystyle\zeta_{\lambda}=\begin{cases}{k} & {\lambda=\lambda_{k}\text{ with }k\leq m_{n}}\\{1} & {\lambda=\lambda_{k}\text{ with }m_{n}

We conclude that ${z\in\left(U_{n+1}\cap V_{0}\right)\subset\left(U\cap V\right)}$, which yields the desired contradiction.

The preceding example gives us a necessary condition for a product of nonempty ${T_{1}}$ spaces to be normal. Indeed, the product is compact only if at most a countably many factor spaces are not limit point compact (i.e. every infinite subset has a limit point in the factor space). Indeed, a topological space ${X}$ which is not limit point compact necessarily has infinitely many points. Let ${S=\left\{x_{i}\right\}_{i=1}^{\infty}}$ be an enumeration of a countably infinite subset which does not have a limit in ${X}$. ${S}$ is closed, otherwise ${X\setminus S}$ contains a point for which every open neighborhood intersects ${S}$, implying ${S}$ has a limit point in ${X}$. In fact, every subset ${S'\subset S}$ does not have a limit point in ${X}$, implying that the subspace topology on ${S}$ is just the discrete topology. Taking ${\text{card}(\Lambda)}$ many such factor spaces yields a space that is homeomorphic to ${N}$.

Using the results collected over these three blog posts on Stone’s theorem and, perhaps, a reference to Munkres’s Topology, we can completely characterize the arbitrary product of metric spaces.

Theorem. The following statements about a product of nonempty metric spaces are equivalent:

1. The product is normal.
2. The product is fully normal (equivalently, paracompact).
3. At most countably many of the factor spaces are noncompact.

Following in a long tradition of math authors, I leave the proof of the theorem as an exercise to the reader.