# When all powers of a space are normal

It follows from the Tychonoff Theorem that if a topological space $X$ is a compact Hausdorff space, then the product space $X^\tau$ is normal for any cardinal number $\tau$. The converse is also true. If $X^\tau$ is normal for any cardinal number $\tau$, then $X$ must be compact. This is a theorem that is due to Noble (see [5]). The proof in [5] is a corollary resulted from a long chain of previous results of Noble and others. Many authors had produced simpler and more direct proofs of Noble’s theorem (e.g. [2], [3] and [4]). All these more direct proofs make use of the fact that the product space $\omega^{\omega_1}$ is not normal (due to A. H. Stone). All of them except [2] make use of other strong topological results in order to derive Noble’s theorem. In [2], Engelking established Noble’s theorem by an elementary proof. In this post, we present the proof in [2] in full details. Noble’s theorem is also given in Engelking’s textbook as an exercise (see 3.12.15 on p. 233 in [1]).

Before proceeding to the main theorem, let’s set up some notation for working with the product space $\prod_{\alpha \in S} X_\alpha$. For $x \in \prod_{\alpha \in S} X_\alpha$, the $\alpha^{th}$ coordinate of $x$ is denoted by $x_\alpha$ or $(x)_\alpha$. For $M \subset S$, the map $P_M: \prod_{\alpha \in S} X_\alpha \rightarrow \prod_{\alpha \in M} X_\alpha$ is the natural projection map. In the product space $\prod_{\alpha \in S} X_\alpha$, standard basic open sets are of the form $\prod_{\alpha \in S} O_\alpha$ where $O_\alpha= X_\alpha$ for all but finitely many $\alpha$. We use $supp(\prod_{\alpha \in S} O_\alpha)$ to denote the set of the finite number of $\alpha \in S$ where $O_\alpha \ne X_\alpha$.

Noble’s Theorem

If each power of a space $X$ is normal, then $X$ is compact.

Proof
Suppose that $X^\tau$ is normal for all cardinal numbers $\tau$. Suppose that $X$ is not compact. Then there exists a collection $\mathcal{H}=\left\{H_\alpha: \alpha \in S \right\}$ of closed subsets of $X$ such that $\mathcal{H}$ has the finite intersection property but has empty intersection. Let $H=\prod_{\alpha \in S} H_\alpha$, which is a subspace of the product space $\prod_{\alpha \in S} X_\alpha$ where each $X_\alpha=X$. We can also denote the product space $\prod_{\alpha \in S} X_\alpha$ by $X^\tau$ where $\tau=\lvert S \lvert$.

Let $K=\left\{k \in X^{\lvert S \lvert}: \forall \ \beta, \gamma \in S, \ k_\beta=k_\gamma \right\}$. Note that $K$ is commonly referred to as the diagonal of the product space in question. Both $H$ and $K$ are closed sets in the product space $X^\tau$. Because the collection $\mathcal{H}$ has empty intersection, $H$ and $K$ are disjoint closed sets. Since $X^\tau$ is normal, there exist disjoint open subsets $U$ and $V$ of $X^\tau$ such that $H \subset U$ and $K \subset V$.

Let $x_1 \in H$. Let $O_1$ be a basic standard open set with $x \in O_1 \subset U$. Let $S_1=supp(O_1)$. Then we have $P^{-1}_{S_1}(P_{S_1}(x_1)) \subset U$. Since $\mathcal{H}$ has the finite intersection property, choose $a_1 \in \bigcap_{\alpha \in S_1} H_\alpha$. Then define $x_2 \in H$ such that $(x_2)_\alpha=a_1$ for all $\alpha \in S_1$ and $(x_2)_\alpha=(x_1)_\alpha$ for all $\alpha \in S-S_1$.

Let $O_2$ be a basic standard open set with $x_2 \in O_2 \subset U$. Let $S_2=supp(O_2)$. By making $O_2$ a smaller open set if necessary, we can have $S_1 \subset S_2$. Then we have $P^{-1}_{S_2}(P_{S_2}(x_2)) \subset U$. Choose $a_2 \in \bigcap_{\alpha \in S_2} H_\alpha$. Then define $x_3 \in H$ such that $(x_3)_\alpha=a_2$ for all $\alpha \in S_2$ and $(x_3)_\alpha=(x_2)_\alpha$ for all $\alpha \in S-S_2$.

After this inductive process is completed, we can obtain:

• a sequence $x_1,x_2,x_3,\cdots$ of points of $H=\prod_{\alpha \in S} H_\alpha$,
• a sequence $S_1 \subset S_2 \subset S_3, \subset \cdots$ of finite subsets of the index set $S$,
• a sequence $a_1,a_2,a_3,\cdots$ of points of $X$

such that for each $n \ge 2$, $(x_n)_\alpha=a_{n-1}$ for all $\alpha \in S_{n-1}$ and $P^{-1}_{S_n}(P_{S_n}(x_n)) \subset U$.

By A. H. Stone’s theorem (Theorem 5 in [6]), $X$ cannot contain a closed copy of $\mathbb{N}$ (the space of the positive integers with the discrete topology). A proof that $\mathbb{N}^{\omega_1}$ is also found in this post. Let $A=\left\{a_1,a_2,a_3,\cdots \right\}$. Either $A$ is infinite or finite.

Case 1
Assume that $A$ is an infinite set. Then $A$ has a limit point $a$, meaning that every open subset of $X$ containing $a$ contains some $a_n$ different from $a$. For each $n \ge 2$, define $y_n \in \prod_{\alpha \in S} X_\alpha$ such that

• $(y_n)_\alpha=(x_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$
• $(y_n)_\alpha=a$ for all $\alpha \in S-S_n$

It is the case that $y_n \in P^{-1}_{S_n}(P_{S_n}(x_n)) \subset U$ for all $n$, since $y_n$ agrees with $x_n$ on the finite set $S_n$. Let $t \in K$ such that $t_\alpha=a$ for all $\alpha \in S$. It follows that $t$ is a limit point of $\left\{y_2,y_3,y_4,\cdots \right\}$. Thus $t \in \overline{U}$. Since $t \in K \subset V$, the open set $V$ would have to contain points of $U$. But $U$ and $V$ are supposed to be disjoint open subsets of the product space $\prod_{\alpha \in S} X_\alpha$. Thus we have a contradiction.

Case 2
Assume that $A$ is a finite set. Then for some $m$, $a_j=a_m$ for all $j \ge m$. For each $n \ge 2$, define $y_n \in \prod_{\alpha \in S} X_\alpha$ such that

• $(y_n)_\alpha=(x_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$
• $(y_n)_\alpha=a_m$ for all $\alpha \in S-S_n$

Let $t \in K$ such that $t_\alpha=a_m$ for all $\alpha \in S$. Then $y_n=t$ for all $n \ge m+1$. As in Case 1, $y_n \in P^{-1}_{S_n}(P_{S_n}(x_n)) \subset U$ for all $n$, implying that $t \in K \cap U$. This is a contradiction, since $K \subset V$ and $U$ and $V$ are supposed to be disjoint.

Both cases lead to a contradiction. Thus if all powers of $X$ is normal, $X$ must be compact. This completes the proof of the theorem.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Engelking, R., An elementary proof of Noble’s theorem on normality of powers, Comment. Math. Univ. Carolinae, 29.4, 677-678, 1988.
3. Franklin, S. P., Walker, R. C., Normalit of powers implies compactness, Proc. Amer. Math. Soc., 36, 295-296, 1972.
4. Keesling, J., Normality and infinite product spaces, Adv. in. Math., 9, 90-92, 1972.
5. Noble, N., Products with closed projections, II, Trans. Amer. Math. Soc., 160, 169-183, 1971.
6. Ross, K. A., Stone, A. H. Products of separable spaces, Amer. Math. Monthly, 71, 398-403, 1964.

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$\copyright \ 2014 \text{ by Dan Ma}$