Sigma-products of separable metric spaces are collectionwise normal

Let \prod_{\alpha \in \omega_1} \mathbb{R}=\mathbb{R}^{\omega_1} be the Cartesian product of \omega_1 many copies of the real line. This product product space is not normal since it contains \prod_{\alpha \in \omega_1} \omega=\omega^{\omega_1} as a closed subspace. The subspace of \mathbb{R}^{\omega_1} consisting of points which have at most countably many non-zero coordinates is collectionwise normal. Such spaces are called \Sigma-products. In this post, we show that the \Sigma-product of separable and metrizable spaces is always collectionwise normal. To place the result proved in this post in a historical context, see the comments at the end of the post.

Consider the product space X=\prod_{\alpha \in A} X_\alpha. Let a \in X. The \Sigma-product of the spaces \left\{X_\alpha \right\}_{\alpha \in A} about the base point a is the following subspace of X:

    \Sigma_{\alpha \in A} X_\alpha(a)=\left\{x \in X: x_\alpha \ne a_\alpha \text{ for at most countably many } \alpha \in A \right\}

When the base point a is understood, we denote the space by \Sigma_{\alpha \in A} X_\alpha. First we want to eliminate cases that are not interesting. If the index set A is countable, then the \Sigma-product is simply the Cartesian product. We assume that the index set A is uncountable. If all but countably many of the factors consist of only one point, then the \Sigma-product is also the Cartesian product. So we assume that each X_\alpha has at least 2 points. When these two assumptions are made, the resulting \Sigma-products are called proper.

The collectionwise normality of \Sigma_{\alpha \in A} X_\alpha is accomplished in two steps. First, \Sigma_{\alpha \in A} X_\alpha is shown to be normal if each factor X_\alpha is a separable metric space (Theorem 1). Secondly, observe that normality in \Sigma-product is countably productive, i.e., if Y=\Sigma_{\alpha \in A} X_\alpha is normal, then Y^\omega is also normal (Theorem 2). Then the collectionwise normality of \Sigma_{\alpha \in A} X_\alpha follows from a theorem attributed to Corson. We have the following theorems.

Theorem 1

    For each \alpha \in A, let X_\alpha be a separable and metrizable space. Then \Sigma_{\alpha \in A} X_\alpha is a normal space.

Theorem 2

    For each \alpha \in A, let X_\alpha be a separable and metrizable space. Let Y=\Sigma_{\alpha \in A} X_\alpha. Then Y^\omega is a normal space.

Theorem 3 (Corson’s Theorem)

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. If Y \times Y is normal, then Y is collectionwise normal.

    For a proof of Corson’s theorem, see this post.

The above three theorems lead to the following theorem.

Theorem 4

    For each \alpha \in A, let X_\alpha be a separable and metrizable space. Then \Sigma_{\alpha \in A} X_\alpha is a collectionwise normal space.

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Proofs

Before proving the theorems, let’s set some notations. For each B \subset A, \pi_B is the natural projection from \prod_{\alpha \in A} X_\alpha into \prod_{\alpha \in B} X_\alpha. The standard basic open sets in the product space X=\prod_{\alpha \in A} X_\alpha are of the form \prod_{\alpha \in A} O_\alpha where O_\alpha=X_\alpha for all but finitely many \alpha \in A. We use supp(\prod_{\alpha \in A} O_\alpha) to denote the set of finitely many \alpha \in A such that O_\alpha \ne X_\alpha.

The following lemma is used (for a proof, see Lemma 1 in this post).

Lemma 5

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. Then the following conditions are equivalent.

    1. Y is normal.
    2. For any pair of disjoint closed subsets H and K of Y, there exists a countable B \subset A such that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing.
    3. For any pair of disjoint closed subsets H and K of Y, there exists a countable B \subset A such that \pi_B(H) and \pi_B(K) are separated in \pi_B(Y), meaning that \overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing.

Proof of Theorem 1

Let X=\prod_{\alpha \in A} X_\alpha be the product space in question. Let Y=\Sigma_{\alpha \in A} X_\alpha be defined using the base point b \in X. Note that Y is dense in the product space X=\prod_{\alpha \in A} X_\alpha. In light of Lemma 5, to show Y is normal, it suffices to show that for each pair of disjoint closed subsets H and K of Y, there exists a countable B \subset A such that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing. Let H and K be disjoint closed subsets of Y.

Before building up to a countable set B, let’s set some notation that will be used along the way. For each y \in Y, let S(y) denote the set of all \alpha \in A such that y_\alpha \ne b_\alpha. For any set T \subset Y, let S(T)=\bigcup_{y \in T} S(y).

To start, choose \gamma \in A and let A_1=\left\{\gamma \right\}. Consider \pi_{A_1}(H) and \pi_{A_1}(K). They are subsets of \prod_{\alpha \in A_1} X_\alpha, which is a hereditarily separable space. Choose a countable D_1 \subset \pi_{A_1}(H) and a countable E_1 \subset \pi_{A_1}(K) such that \overline{D_1} = \pi_{A_1}(H) and \overline{E_1} = \pi_{A_1}(K). For each u \in D_1, choose f(u) \in H such that \pi_{A_1}(f(u))=u. For each v \in E_1, choose g(v) \in H such that \pi_{A_1}(g(v))=v. Let H_1 and K_1 be defined by:

    H_1=\left\{f(u): u \in D_1 \right\}
    K_1=\left\{g(v): v \in E_1 \right\}

Clearly \pi_{A_1}(H)=\overline{\pi_{A_1}(H_1)} and \pi_{A_1}(K)=\overline{\pi_{A_1}(K_1)}. Let A_2=A_1 \cup S(H_1) \cup S(K_1).

Now perform the next step inductive process. Consider \pi_{A_2}(H) and \pi_{A_2}(K). As before, we can find countable dense subsets of these 2 sets. Choose a countable D_2 \subset \pi_{A_2}(H) and a countable E_2 \subset \pi_{A_2}(K) such that \overline{D_2} = \pi_{A_2}(H) and \overline{E_2} = \pi_{A_2}(K). For each u \in D_2, choose f(u) \in H such that \pi_{A_2}(f(u))=u. For each v \in E_2, choose g(v) \in H such that \pi_{A_2}(g(v))=v. Let H_2 and K_2 be defined by:

    H_2=\left\{f(u): u \in D_2 \right\} \cup H_1
    K_2=\left\{g(v): v \in E_2 \right\} \cup K_1

Clearly \pi_{A_2}(H) \subset \overline{\pi_{A_2}(H_2)} and \pi_{A_2}(K) \subset \overline{\pi_{A_2}(K_2)}. To prepare for the next step, let A_3=A_1 \cup A_2 \cup S(H_2) \cup S(K_2).

Continue the inductive process and when completed, the following sequences are obtained:

  • a sequence of countable sets A_1 \subset A_2 \subset A_3 \subset \cdots \subset A
  • a sequence of countable sets H_1 \subset H_2 \subset H_3 \subset \cdots \subset H
  • a sequence of countable sets K_1 \subset K_2 \subset K_3 \subset \cdots \subset K

such that

  • \pi_{A_j}(H) \subset \overline{\pi_{A_j}(H_j)} and \pi_{A_j}(K) \subset \overline{\pi_{A_j}(K_j)} for each j
  • A_{j+1}=(\bigcup_{i \le j} A_i) \cup S(H_j) \cup S(K_j) for each j

Let B=\bigcup_{j=1}^\infty A_j, H^*=\bigcup_{j=1}^\infty H_j and K^*=\bigcup_{j=1}^\infty K_j. We have the following claims.

Claim 1
\pi_B(H) \subset \overline{\pi_B(H^*)} and \pi_B(K) \subset \overline{\pi_B(K^*)}.

Claim 2
\overline{\pi_B(H^*)} \cap \overline{\pi_B(K^*)}=\varnothing.

Proof of Claim 1
It suffices to show one of the set inclusions. We show \pi_B(H) \subset \overline{\pi_B(H^*)}. Let h \in H. We need to show that \pi_B(h) is a limit point of \pi_B(H^*). To this end, let V=\prod_{\alpha \in B} V_\alpha be a standard basic open set containing \pi_B(h). Then supp(V) \subset A_j for some j. Let V_j=\prod_{\alpha \in A_j} V_\alpha. Then \pi_{A_j}(h) \in V_j. Since \pi_{A_j}(H) \subset \overline{\pi_{A_j}(H_j)}, there is some t \in H_j such that \pi_{A_j}(t) \in V_j. It follows that \pi_B(t) \in V. Thus every open set containing \pi_B(h) contains a point of \pi_B(H^*).

Proof of Claim 2
Suppose that x \in \overline{\pi_B(H^*)} \cap \overline{\pi_B(K^*)}. Define y \in Y=\Sigma_{\alpha \in A} X_\alpha such that y_\alpha=x_\alpha for all \alpha \in B and y_\alpha=b_\alpha for all \alpha \in A-B. It follows that y \in \overline{H} \cap \overline{K}=H \cap K, which is a contradiction since H and K are disjoint closed sets. To see that y \in H \cap K, let W=\prod_{\alpha \in A} W_\alpha be a standard basic open set containing y. Let W_1=\prod_{\alpha \in B} W_\alpha. Since x \in W_1, there exist h \in H^* and k \in K^* such that \pi_B(h) \in W_1 and \pi_B(k) \in W_1. Note that the supports S(h) \subset B and S(k) \subset B. For the coordinates outside of B, both h and k agree with the base point b and hence with y. Thus h \in W and k \in W. We have just shown that every open set containing y contains a point of H and a point of K. This means that y \in H \cap K, a contradiction. This completes the proof of Claim 2.

Both Claim 1 and Claim 2 imply that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing. By Lemma 5, Y=\Sigma_{\alpha \in A} X_\alpha is normal. \blacksquare

Proof of Theorem 2

Let Y=\Sigma_{\alpha \in A} X_\alpha be the \Sigma-product about the base point b \in \prod_{\alpha \in A} X_\alpha. The following countable product

    \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a)

is a product of separable metric spaces. So any \Sigma-product that can be defined within the product space (a) is normal (by Theorem 1). In particular, consider the \Sigma-product defined about the base point c=(b, b, b, \cdots) (countably many coordinates). Denote this \Sigma-product by T. Observe that T is homeomorphic to the following countable product of Y=\Sigma_{\alpha \in A} X_\alpha:

    \Sigma_{\alpha \in A} X_\alpha \times \Sigma_{\alpha \in A} X_\alpha \times \Sigma_{\alpha \in A} X_\alpha \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (b)

Thus T can be identified with Y^\omega. We can conclude that Y^\omega is normal. \blacksquare

Proof of Theorem 4

Let Y=\Sigma_{\alpha \in A} X_\alpha be the \Sigma-product of the separable metric spaces X_\alpha. By Theorem 1, Y is normal. By Theorem 2, Y^\omega is normal. In particular, Y \times Y is normal. Clearly, Y is a dense subspace of the product space X=\prod_{\alpha \in A} X_\alpha. By Corson’s theorem (Theorem 3), Y is collectionwise normal. \blacksquare

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A Brief History

The notion of \Sigma-products was introduced by Corson in [1] where he proved that the \Sigma-product of complete metric spaces is normal. Corson then asked whether the \Sigma-product of copies of the rationals is normal. In 1973, Kombarov and Malyhin [4] showed that the \Sigma-product of separable metric spaces is normal. In 1977, Gulko [2] and Rudin [6] independently proved the \Sigma-product of metric spaces is normal. In 1978, Kombarov [3] generalized Gulko and Rudin’s result by showing that any \Sigma-product of paracompact p-spaces \left\{X_\alpha: \alpha \in A \right\} is collectionwise normal if and only if all spaces X_\alpha have countable tightness. A useful resource is Przymusinski’s chapter in the Handbook of Set-Theoretic Topology [5], which has a section on \Sigma-products.

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Reference

  1. Corson H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
  2. Gulko S. P., On the properties of sets lying in \Sigma-products, Dokl. Acad. Nauk. SSSR, 237, 505-508, 1977 (in Russian).
  3. Kombarov A. P., On the tightness and normality of \Sigma-products, Dokl. Acad. Nauk. SSSR, 239, 775-778, 1978 (in Russian).
  4. Kombarov A. P., Malyhin V. I., On \Sigma-products, Soviet Math. Dokl., 14 (6), 1980-1983, 1973.
  5. Przymusinski, T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
  6. Rudin M. E., Book Review, Bull. Amer. Math. Soc., 84, 271-272, 1978.

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\copyright \ 2014 \text{ by Dan Ma}

A theorem attributed to Corson

In this post, we prove a theorem that is attributed to Corson. It had been reported in the literature (see [1] and [2] for two instances) and on the Internet that this theorem can be deduced from a paper of Corson [3]. Instead of having an indirect proof, we give a full proof of this theorem. One application of this theorem is that we can use it to show the collectionwise normality of a \Sigma-product of separable metric spaces (see this blog post). We prove the following theorem.

Theorem 1 (Corson’s Theorem)

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. If Y \times Y is normal, then Y is collectionwise normal.

Another way to state this theorem is through the angle of finding normal spaces that are collectionwise normal. The above theorem can be re-stated: any dense normal subspace Y of a product of separable metric spaces must be collectionwise normal if one additional condition is satisfied: the square of Y is also normal. Thus we have the following theorem:

Theorem 1a

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a normal and dense subspace of X. If Y \times Y is normal, then Y is collectionwise normal.

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A Brief Background Discussion

A space S is said to be collectionwise normal if for any discrete collection \mathcal{A} of closed subsets of S, there exists a pairwise disjoint collection \mathcal{U} of open subsets of S such that for each A \in \mathcal{A}, there is exactly one U \in \mathcal{U} such that A \subset U. Here’s some previous posts on the definitions and a background discussion on collectionwise normality.

There is one circumstance where normality implies collectionwise normality. If all closed and discrete subsets of a normal space are countable, then it is collectionwise normal. We have the following theorem.

Theorem 2

    Let S be a normal space. If all closed and discrete subsets of S are countable, then S is collectionwise normal.

Proof of Theorem 2
We first establish the following lemma.

Lemma 2a
Let L be a normal space. Let \left\{C_1,C_2,C_3,\cdots \right\} be a countable discrete collection of closed subsets of L. Then there exists a pairwise disjoint collection \left\{O_1,O_2,O_3,\cdots \right\} of open subsets of L such that C_j \subset O_j for each j.

Proof of Lemma 2a
For each j, choose disjoint open subsets U_j and V_j such that C_j \subset U_j and \bigcup_{n \ne j} C_n \subset V_j. Let O_1=U_1. For each j>1, let O_j=U_j \cap \bigcap_{n \le j-1} V_n. It follows that \left\{O_1,O_2,O_3,\cdots \right\} is pairwise disjoint such that C_j \subset O_j for each j. This completes the proof for Lemma 2a.

Suppose that all closed and discrete subsets of the normal space S are countable. It follows that any discrete collection of closed subsets of S is countable. Then the collectionwise normality of S follows from Lemma 2a. \blacksquare

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Proving Corson’s Theorem

If Y \times Y is normal, then clearly Y is normal. In light of Theorem 2, to show Y is collectionwise normal, it suffices to show that every closed and discrete subspace of Y is countable. Thus Theorem 1 is established by combining Theorem 2 and the following theorem.

Theorem 3

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. If Y \times Y is normal, then every closed and discrete subspace of Y is countable.

Before proving Theorem 3, we state one more lemma that is needed. For C \subset A, \pi_C is the natural projection map from X=\prod_{\alpha \in A} X_\alpha into \prod_{\alpha \in C} X_\alpha. The map \pi_C \times \pi_C refers to the projection map from \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha into \prod_{\alpha \in C} X_\alpha \times \prod_{\alpha \in C} X_\alpha defined by (\pi_C \times \pi_C)(x,y)=(\pi_C(x),\pi_C(y)).

Lemma 4

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. Then the following conditions are equivalent.

    1. Y \times Y is normal.
    2. For any pair of disjoint closed subsets H and K of Y \times Y, there exists a countable C \subset A such that \overline{(\pi_C \times \pi_C)(H)} \cap \overline{(\pi_C \times \pi_C)(K)}=\varnothing.
    3. For any pair of disjoint closed subsets H and K of Y \times Y, there exists a countable C \subset A such that H_1=(\pi_C \times \pi_C)(H) and K_1=(\pi_C \times \pi_C)(K) are separated in \pi_C(Y) \times \pi_C(Y), meaning that H_1 \cap \overline{K_1}=\varnothing=\overline{H_1} \cap K_1.

Lemma 4 deals with the dense subspace Y of X=\prod_{\alpha \in A} X_\alpha and the dense subspace Y \times Y of \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha. So the map \pi_C should be restricted to Y and the map \pi_C \times \pi_C is restricted to Y \times Y. For a proof of Lemma 4, see the proof of Lemma 2 in this previous post.

Proof of Theorem 3
Let T=\left\{t_\alpha: \alpha<\omega_1 \right\} be an uncountable closed and discrete subset of Y. We define two disjoint closed subsets H and K of Y \times Y such that for each countable set C \subset A, (\pi_C \times \pi_C)(H) \cap \overline{(\pi_C \times \pi_C)(K)} \ne \varnothing. By Lemma 4, Y \times Y is not normal. Consider the following two subsets of Y \times Y:

    H=\left\{(t_\alpha,t_\alpha): \alpha<\omega_1 \right\}

    K=\left\{(t_\delta,t_\rho): \delta,\rho<\omega_1 \text{ and } \delta \ne \rho \right\}

Clearly H and K are disjoint. It is clear that H is a closed subset of Y \times Y. Because T is closed and discrete in Y, K is a closed subset of Y \times Y. Thus H and K are disjoint closed subsets of Y \times Y.

Let C \subset A be countable. Note that (\pi_C \times \pi_C)(H) and (\pi_C \times \pi_C)(K) are:

    (\pi_C \times \pi_C)(H)=\left\{(\pi_C(w_\alpha),\pi_C(w_\alpha)): \alpha<\omega_1 \right\}

    (\pi_C \times \pi_C)(K)=\left\{(\pi_C(w_\delta),\pi_C(w_\rho)): \delta,\rho<\omega_1, \delta \ne \rho \right\}

Consider P=\left\{\pi_C(w_\alpha): \alpha<\omega_1 \right\}. Clearly, P \times P=(\pi_C \times \pi_C)(H). We consider two cases: P is uncountable or P is countable.

Case 1: P is uncountable.
Note that \prod_{\alpha \in C} X_\alpha is the product of countably many separable metric spaces and is therefore a hereditarily Lindelof space. As a subspace of \prod_{\alpha \in C} X_\alpha, \pi_C(Y) is also hereditarily Lindelof. Since P is an uncountable subspace of \pi_C(Y), there must exist a point p \in P such that every open set (open in \pi_C(Y)) containing p must contain uncountably many points of P. Note that (p,p) \in (\pi_C \times \pi_C)(H).

Let O be an open subset of \pi_C(Y) with p \in O. Then there exist \gamma, \rho<\omega_1 with \gamma \ne \rho such that \pi_B(w_\gamma) \in O and \pi_B(w_\rho) \in O. Then (\pi_B(w_\gamma), \pi_B(w_\gamma)) is a point of (\pi_C \times \pi_C)(H) that is in O \times O. The point (\pi_B(w_\gamma), \pi_B(w_\delta)) is a point of (\pi_C \times \pi_C)(K) that is in O \times O. This means that (p,p) \in \overline{(\pi_C \times \pi_C)(K)}. Thus we have (\pi_C \times \pi_C)(H) \cap \overline{(\pi_C \times \pi_C)(K)} \ne \varnothing.

Case 2: P is countable.
Then there exists p \in P such that p=\pi_C(w_\alpha) for uncountably many \alpha. Choose \gamma, \rho<\omega_1 such that \gamma \ne \rho and p=\pi_C(w_\gamma) and p=\pi_C(w_\rho). Then (p,p) \in (\pi_C \times \pi_C)(H) and (p,p) \in (\pi_C \times \pi_C)(K).

In either case, we can say that (\pi_C \times \pi_C)(H) \cap \overline{(\pi_C \times \pi_C)(K)} \ne \varnothing. By Lemma 4, Y \times Y is not normal. \blacksquare

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Reference

  1. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
  2. Baturov, D. P., On perfectly normal dense subspaces of products, Topology Appl., 154, 374-383, 2007.
  3. Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
  4. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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\copyright \ 2014 \text{ by Dan Ma}

Revisit a lemma dealing with normality in products of separable metric spaces

In this post we continue to discuss a lemma that has been discussed previously in this post. The lemma characterizes the dense normal subspaces of a product of separable metric spaces. The lemma discussed here has been sharpened over the version in the previous post. Two versions of the lemma are given (Lemma 1 and Lemma 2). Any one of these two versions can be used to prove that the \Sigma-product of separable metric spaces is normal (see this blog post).

Lemma 1

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. Then the following conditions are equivalent.

    1. Y is normal.
    2. For any pair of disjoint closed subsets H and K of Y, there exists a countable B \subset A such that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing.
    3. For any pair of disjoint closed subsets H and K of Y, there exists a countable B \subset A such that \pi_B(H) and \pi_B(K) are separated in \pi_B(Y), meaning that \overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing.

The closure in condition 2 and condition 3 is taken in \pi_B(Y). The map \pi_B is the natural projection from the full product space X=\prod_{\alpha \in A} X_\alpha into the subproduct \prod_{\alpha \in B} X_\alpha.

Proof of Lemma 1
1 \Longrightarrow 2
Let H and K be disjoint closed subsets of Y. Since Y is normal, there exists a continuous function f: Y \rightarrow [0,1] such that f(H) \subset \left\{0 \right\} and f(H) \subset \left\{1 \right\}. By Theorem 1 in this previous post, the continuous function f depends on countably many coordinates. This means that there exists a countable B \subset A and there exists a continuous g:\pi_B(Y) \rightarrow [0,1] such that f= g \circ \pi_B. The continuity on the full product space is now reduced to the continuity on a countable subproduct. Now O_H=g^{-1}([0,0.2)) and O_K=g^{-1}((0.8,1]) are disjoint open sets in \pi_B(Y). Since f= g \circ \pi_B, it is the case that \pi_B(H) \subset O_H and \pi_B(K) \subset O_K. Since g is continuous, we have

    \overline{O_H}=\overline{g^{-1}([0,0.2))} \subset g^{-1}(\overline{[0,0.2)})=g^{-1}([0,0.2]) \ \ \ \ \ \ \ \ (a)

    \overline{O_K}=\overline{g^{-1}((0.8,1])} \subset g^{-1}(\overline{(0.8,1]})=g^{-1}([0.8,1]) \ \ \ \ \ \ \ \ (b)

Note that \overline{\pi_B(H)} \subset \overline{O_H} and \overline{\pi_B(K)} \subset \overline{O_K}. If \overline{\pi_B(H)} \cap \overline{\pi_B(K)} \ne \varnothing, then g^{-1}([0,0.2]) \cap g^{-1}([0.8,1]) \ne \varnothing. Thus \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing.

The direction 2 \Longrightarrow 3 is immediate.

The direction 3 \Longrightarrow 1 follows from Lemma 1 in this previous post (see the direction 2 \rightarrow 1 of Lemma 1 in the previous post). \blacksquare

The following lemma is another version of Lemma 1 which may be useful in some circumstances. For B \subset A, let \pi_B \times \pi_B be the projection map from \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha into \prod_{\alpha \in B} X_\alpha \times \prod_{\alpha \in B} X_\alpha defined by (\pi_B \times \pi_B)(x,y)=(\pi_B(x),\pi_B(y)).

Lemma 2

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. Then the following conditions are equivalent.

    1. Y \times Y is normal.
    2. For any pair of disjoint closed subsets H and K of Y \times Y, there exists a countable C \subset A such that \overline{(\pi_C \times \pi_C)(H)} \cap \overline{(\pi_C \times \pi_C)(K)}=\varnothing.
    3. For any pair of disjoint closed subsets H and K of Y \times Y, there exists a countable C \subset A such that (\pi_C \times \pi_C)(H) and (\pi_C \times \pi_C)(K) are separated in \pi_C(Y) \times \pi_C(Y).

Proof of Lemma 2
1 \Longrightarrow 2
Let H and K be disjoint closed subsets of Y \times Y. Since Y \times Y is normal, there exists a continuous function f: Y \times Y \rightarrow [0,1] such that f(H) \subset \left\{0 \right\} and f(H) \subset \left\{1 \right\}. By Theorem 2 in this previous post, the continuous function f depends on countably many coordinates. This means that there exists a countable C \subset A and there exists a continuous g:\pi_C(Y) \times \pi_C(Y) \rightarrow [0,1] such that f= g \circ (\pi_C \times \pi_C). Now O_H=g^{-1}([0,0.2)) and O_K=g^{-1}((0.8,1]) are disjoint open sets in \pi_C(Y) \times \pi_C(Y). Since f= g \circ (\pi_C \times \pi_C), it is the case that (\pi_C \times \pi_C)(H) \subset O_H and (\pi_C \times \pi_C)(K) \subset O_K.

Since g is continuous, conditions (a) and (b) in the proof of Lemma 1 also hold here. Note that \overline{(\pi_C \times \pi_C)(H)} \subset \overline{O_H} and \overline{(\pi_C \times \pi_C)(K)} \subset \overline{O_K}. It follows that \overline{(\pi_C \times \pi_C)(H)} \cap \overline{(\pi_C \times \pi_C)(K)}=\varnothing.

The direction 2 \Longrightarrow 3 is immediate.

3 \Longrightarrow 1
Let H and K be disjoint closed subsets of Y \times Y. By condition 3, there exists a countable C \subset A such that F_H=(\pi_C \times \pi_C)(H) and F_K=(\pi_C \times \pi_C)(K) are separated in M=\pi_C(Y) \times \pi_C(Y). Note that \overline{F_H} \cap F_K=\varnothing and F_H \cap \overline{F_K}=\varnothing. Consider the following subspace of M.

    W=M-\overline{F_H} \cap \overline{F_K}

The space W is an open subspace of M. The space M is a subspace of a product of countably many separable metric spaces. Thus both M and W are also second countable and hence normal.

For L \subset W, let Cl_W(L) denote the closure of L in the space W. Both Cl_W(F_H) and Cl_W(F_K) are disjoint closed subsets of W. Let G_H and G_K be disjoint open subsets of W with Cl_W(F_H) \subset G_H and Cl_W(F_K) \subset G_K. Then \pi_B^{-1}(G_H) \cap Y and \pi_B^{-1}(G_K) \cap Y are disjoint open subsets of Y separating H and K. \blacksquare

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Remark

The countable sets in both Lemma 1 and Lemma 2 can be expanded to larger countable sets. For example,

    for Lemma 1, for any disjoint closed subsets H and K of Y:

    1. If for some countable set B, \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing, then \overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing for any countable E \subset A with B \subset E.
    2. If for some countable set B, \overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing, then \overline{\pi_E(H)} \cap \pi_E(K)=\pi_E(H) \cap \overline{\pi_E(K)}=\varnothing for any countable E \subset A with B \subset E.

It is straightforward to verify these facts. For the sake of completeness, we verify condition 2. Suppose that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing for some countable B \subset A. Let E \subset A be countable with B \subset E. We show \overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing. Suppose x \in \overline{\pi_E(H)} \cap \overline{\pi_E(K)}. Then \pi_B(x) \notin \overline{\pi_B(H)} \cap \overline{\pi_B(K)}. Choose some standard basic open set O=\prod_{\alpha \in B} O_\alpha with \pi_B(x) \in O such that O \cap \overline{\pi_B(H)}=\varnothing and O \cap \overline{\pi_B(K)}=\varnothing. Consider O_1=\prod_{\alpha \in E} O_\alpha such that O_\alpha=X_\alpha for all \alpha \in C-B. Clearly x \in O_1. Then there exist h \in H and k \in K such that \pi_E(h) \in O_1 and \pi_E(h) \in O_1. It follows that \pi_B(h) \in O_1 and \pi_B(h) \in O, a contradiction. Thus \overline{\pi_E(H)} \cap \overline{\pi_E(K)}=\varnothing.

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\copyright \ 2014 \text{ by Dan Ma}
Revised 3/31/2014.

A factorization theorem for products of separable spaces

Let X=\prod_{\alpha \in A} X_\alpha be a product space. Let f:X \rightarrow T be a continuous function where T is a topological space. In this post, we discuss what it means for the continuous function f to depend on countably many coordinates and then discuss some conditions that we can impose on the product space and on the range space T to ensure that every continuous f defined on the product space will depend on countably many coordinates. This notion of a continuous function depending on countably many coordinates is equivalent to factoring the continuous function into the composition of a projection map and a continuous function defined on a countable subproduct (see Lemma 1 below).

Let’s set some notation about the product space we work with in this post. Let X=\prod_{\alpha \in A} X_\alpha be a product space. Let T be a topological space. Let f:X \rightarrow T be continuous. For any B \subset A, \pi_B is the natural projection from the full product space X=\prod_{\alpha \in A} X_\alpha into the subproduct \prod_{\alpha \in B} X_\alpha. Standard basic open sets of X=\prod_{\alpha \in A} X_\alpha are of the form \prod_{\alpha \in A} O_\alpha where each O_\alpha is open in X_\alpha and that O_\alpha=X_\alpha for all but finitely many \alpha \in A. We use supp(\prod_{\alpha \in A} O_\alpha) to denote the finite set of \alpha \in A where O_\alpha \ne X_\alpha.

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Factoring a Continuous Map

The function f is said to depend on countably many coordinates if there exists a countable set B \subset A such that for any x,y \in X, if x_\alpha=y_\alpha for all \alpha \in B, then f(x)=f(y). Suppose f is instead defined on a subspace Y of X. The function f is said to depend on countably many coordinates if there exists a countable B \subset A such that for any x,y \in Y, if x_\alpha=y_\alpha for all \alpha \in B, then f(x)=f(y).

We have the following lemmas.

Lemma 1

    Let X=\prod_{\alpha \in A} X_\alpha be a product space. Let T be a topological space. Let f:X \rightarrow T be continuous. Then the following are equivalent.

    1. There exists a countable B \subset A such that for any x,y \in X, if x_\alpha=y_\alpha for all \alpha \in B, then f(x)=f(y).
    2. There exists a countable B \subset A such that f=g \circ \pi_B where g: \prod_{\alpha \in B} X_\alpha \rightarrow T is continuous.

Lemma 1a

    Let X=\prod_{\alpha \in A} X_\alpha be a product space. Let T be a topological space. Let Y be a dense subspace of X. Let f:Y \rightarrow T be continuous. Then the following are equivalent.

    1. There exists a countable B \subset A such that for any x,y \in Y, if x_\alpha=y_\alpha for all \alpha \in B, then f(x)=f(y).
    2. There exists a countable B \subset A such that f=g \circ \pi_B where g: \pi_B(Y) \rightarrow T is continuous.

It is straightforward to verify Lemma 1 and Lemma 1a. We use condition 1 to define what it means for a function to be dependent on countably many coordinates. Both lemmas indicate that either condition is a valid definition. These two lemmas also indicate why the notion being discussed can be called a factorization notion.

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When a Continuous Map Can Be Factored

We discuss some conditions that we can place on the product space X=\prod_{\alpha \in A} X_\alpha and on the range space T so that any continuous map depends on countably many coordinates. We prove the following theorem.

Theorem 1

    Let X=\prod_{\alpha \in A} X_\alpha be a product space such that each factor X_\alpha is a separable space. Let T be a second countable space (i.e. having a countable base). Then for any dense subspace Y of X, any continuous function f:Y \rightarrow T depends on countably many coordinates, i.e., either one of the conditions in Lemma 1a holds.

Before stating the main theorem, we need one more lemma. Let W \subset X=\prod_{\alpha \in A} X_\alpha. The set W is said to depend on countably many coordinates if there exists a countable B \subset A such that for any x \in W and for any y \in X, if x_\alpha=y_\alpha for all \alpha \in B, then y \in W.

When we try to determine whether a function f:Y \rightarrow T, where Y \subset X, can be factored, we will need to decide whether a set W \subset Y depends on countably many coordinates. Let W \subset Y \subset X=\prod_{\alpha \in A} X_\alpha. The set W is said to depend on countably many coordinates if there exists a countable B \subset A such that for any x \in W and for any y \in Y, if x_\alpha=y_\alpha for all \alpha \in B, then y \in W. We have the following lemma.

Lemma 2

    Let X=\prod_{\alpha \in A} X_\alpha be a product space with the countable chain condition. Let Y be a dense subspace of X.

    1. Let U be an open subset of X. Then \overline{U} depends on countably many coordinates.
    2. Let W be an open subset of Y. Then \overline{W} depends on countably many coordinates (closure in Y).

Proof of Lemma 2
Proof of Part 1
Let U \subset X be open. Let \mathcal{B} be a collection of pairwise disjoint open subsets of the open set U \subset X such that \mathcal{B} is maximal with this property, i.e., if you throw one more open set into \mathcal{B}, it will be no longer pairwise disjoint. Let V=\bigcup \mathcal{B}. Since \mathcal{B} is maximal, \overline{V}=\overline{U}. Since X has the countable chain condition, \mathcal{B} is countable.

Let B=\bigcup \left\{supp(O): O \in \mathcal{B} \right\}. The set B is a countable subset of A since B is the union of countably many finite sets. We have the following claims.

Claim 1
The open set V depends on the coordinates in B.

Let x \in V and y \in X such that x_\alpha=y_\alpha for all \alpha \in B. We need to show that y \in V. Firstly, x \in O for some O \in \mathcal{B}. It follows that x_\alpha=y_\alpha for all \alpha \in supp(O). Thus y \in O \subset V. This completes the proof of Claim 1.

Claim 2
The set \overline{V} depends on the coordinates in B.

Let x \in \overline{V} and y \in X such that x_\alpha=y_\alpha for all \alpha \in B. We need to show y \in \overline{V}. To this end, let O=\prod_{\alpha \in A} O_\alpha be a standard basic open set with y \in O. The goal is to find some q \in O \cap V. Define G=\prod_{\alpha \in A} G_\alpha such that G_\alpha=O_\alpha for all \alpha \in B and G_\alpha=X_\alpha for all \alpha \in A-B. Then x \in G. Since x \in \overline{V}, there exists some p \in V \cap G. Define q such that q_\alpha=p_\alpha for all \alpha \in B and q_\alpha=y_\alpha for all \alpha \in A-B. Since supp(V) \subset B, q \in V. On the other hand, q \in O. This completes the proof of Claim 2.

As noted above, \overline{V}=\overline{U}. Thus \overline{U} depends on countably many coordinates, namely the coordinates in the set B. This completes the proof of Part 1.

Proof of Part 2
For any S \subset X, let \overline{S} denote the closure of S in Y. Let Cl_X(S) denote the closure of S in X. Let W \subset Y be open. Let W_1 be open in X such that W=W_1 \cap Y. By Part 1, Cl_X(W_1) depends on countably many coordinates, say the coordinates in the countable set B \subset A. This means that for any x \in Cl_X(W_1) and for any y \in X, if x_\alpha=y_\alpha for all \alpha \in B, then y \in Cl_X(W_1). Thus for any x \in \overline{W} and for any y \in Y, if x_\alpha=y_\alpha for all \alpha \in B, then y \in Cl_X(W_1). If we have y \in \overline{W}, then we are done. So we only need to show that if y \in Y and y \in Cl_X(W_1), then y \in \overline{W}. This is why we need to assume Y is dense in X.

Let y \in Y and y \in Cl_X(W_1). Let C be an open subset of Y with y \in C. There exists an open subset D of X such that C=D \cap Y. Then D \cap Cl_X(W_1) \ne \varnothing. Note that D \cap W_1 is open and D \cap W_1 \ne \varnothing. Since Y is dense in X, D \cap W_1 must contain points of Y. These points of Y are also points of W. Thus C contains points of W. It follows that y \in \overline{W}. This concludes the proof of Part 2. \blacksquare

Proof of Theorem 1
Let Y be a dense subspace of X=\prod_{\alpha \in A} X_\alpha. Let f:Y \rightarrow T be continuous. Let \mathcal{M} be a countable base for the separable metrizable space T. By Lemma 2 Part 2, for each M \in \mathcal{M}, \overline{f^{-1}(M)} depends on countably many coordinates, say the countable set B_M. Let B=\bigcup_{M \in \mathcal{M}} B_M.

We claim that B is a countable set of coordinates we need. Let x,y \in Y such that x_\alpha=y_\alpha for all \alpha \in B. We need to show that f(x)=f(y). Suppose f(x) \ne f(y). Choose \left\{M_1,M_2,M_3,\cdots \right\} \subset \mathcal{M} such that

  • \left\{f(x) \right\}=\bigcap_{j=1}^\infty M_j=\bigcap_{j=1}^\infty \overline{M_j}
  • \overline{M_{j+1}} \subset M_j for each j

This is possible since T is a second countable space. Then f(y) \notin \overline{M_{k}} for some k. Furthermore, y \notin f^{-1}(\overline{M_{k}}). Since f is continuous, \overline{f^{-1}(M_{k})} \subset f^{-1}(\overline{M_{k}}). Therefore, y \notin \overline{f^{-1}(M_{k})}. On the other hand, \overline{f^{-1}(M_{k})} depends on the countably many coordinates in B_{M_k}. We assume above that x_\alpha=y_\alpha for all \alpha \in B. Thus x_\alpha=y_\alpha for all \alpha \in B_{M_k}. This means that y \in \overline{f^{-1}(M_{k})}, a contradiction. It must be that case that f(x)=f(y). \blacksquare

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Another Version

We state another version of Theorem 1 that will be useful in some situations.

Theorem 2

    Let X=\prod_{\alpha \in A} X_\alpha be a product space such that each factor X_\alpha is a separable space. Let T be a second countable space. Let Y be a dense subspace of X. Let f:Y \times Y \rightarrow T be any continuous function. Then the function f depends on countably many coordinates, which means either one of the following two conditions:

    1. There exists a countable set C \subset A such that for any (x,y),(p,q) \in Y \times Y, if x_\alpha=p_\alpha and y_\alpha=q_\alpha for all \alpha \in C, then f(x,y)=f(p,q).
    2. There exists a countable set C \subset A and there exists a continuous g:\pi_C(Y) \times \pi_C(Y) \rightarrow T such that f=g \circ (\pi_C \times \pi_C).

The map \pi_C \times \pi_C is the projection map from \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha into the subproduct \prod_{\alpha \in C} X_\alpha \times \prod_{\alpha \in B} X_\alpha defined by (\pi_C \times \pi_C)(x,y)=(\pi_C(x),\pi_C(y)). In Theorem 2, we only need to consider \pi_C \times \pi_C being defined on the subspace Y \times Y.

Theorem 2 follows from Theorem 1. It is only a matter of fitting Theorem 2 in the framework of Theorem 1. Note that the product \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha is identical to the product \prod_{\alpha \in A \cup A^*} X_\alpha where A^* is a disjoint copy of the index set A. For (x,y) \in \prod_{\alpha \in A} X_\alpha \times \prod_{\alpha \in A} X_\alpha, let x \times y \in \prod_{\alpha \in A \cup A^*} X_\alpha be defined by (x \times y)_\alpha=x_\alpha for all \alpha \in A and (x \times y)_\alpha=y_\alpha for all \alpha \in A^*.

With the identification of (x,y) with x \times y, we have a setting that fits Theorem 1. The product \prod_{\alpha \in A \cup A^*} X_\alpha is also a product of separable spaces. The set Y \times Y is a dense subspace of the product \prod_{\alpha \in A \cup A^*} X_\alpha. In this new setting, we view a point in Y \times Y as x \times y. The map f:Y \times Y \rightarrow T is still a continuous map. We can now apply Theorem 1.

Let B \subset A \cup A^* be a countable set such that for all x \times y,p \times q \in Y \times Y, if (x \times y)_\alpha=(p \times q)_\alpha for all \alpha \in B, then f(x \times y)=f(p \times q). Specifically, if x_\alpha=p_\alpha for all \alpha \in B \cap A and y_\alpha=q_\alpha for all \alpha \in B \cap A^*, then f(x,y)=f(p,q).

Choose a countable set C \subset A such that B \cap A \subset C and B \cap A^* \subset C^*. Here, C^* is the copy of C in A^*. We claim that C is a countable set we need in condition 1 of Theorem 2. Let (x,y),(p,q) \in Y \times Y such that x_\alpha=p_\alpha and y_\alpha=q_\alpha for all \alpha \in C. This implies that x_\alpha=p_\alpha for all \alpha \in B \cap A and y_\alpha=q_\alpha for all \alpha \in B \cap A^*. Then f(x,y)=f(p,q). Thus condition 1 of Theorem 2 holds. It is also straightforward to verify that Condition 1 and Condition 2 are equivalent.

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Remarks

The notion of factorizing a continuous map defined on a product space is an old topic. Theorem 1 discussed in this post is based on Theorem 4 found in [6]. Theorem 4 found in [6] is to factor continuous maps defined on a product of separable spaces. Theorem 1 in this post is modified to consider continuous maps defined on a dense subspace of a product of separable spaces. This modification will make it more useful. The references listed below represent a small sample of papers or books that have involves theorems of factoring functions defined on products. The work in [3] and [5] have more systematic treatment.

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Reference

  1. Brandenburg H., Husek M., On mappings from products into developable spaces, Topology Appl., 26, 229-238, 1987.
  2. Engelking R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  3. Engelking R., On functions defined on Cartesian products, Fund. Math., 59, 221-231, 1966.
  4. Keesling J., Normality and infinite product spaces, Adv. in. Math., 9, 90-92, 1972.
  5. Noble N., Ulmer M., Factoring functions on Cartesian products, Trans. Amer. Math. Soc., 163, 329-339, 1972.
  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}

A lemma dealing with normality in products of separable metric spaces

In this post we prove a lemma that is a great tool for working with product spaces of separable metrizable spaces. As an application of the lemma, we give an alternative proof for showing the non-normality of the product space of uncountably many copies of the discrete space of the non-negative integers.

Consider the product space X=\prod_{\alpha \in A} X_\alpha where each X_\alpha is a separable and metrizable space. The lemma we discuss here is a tool that can shed some light on normality of dense subspaces of the product space X. The lemma is stated in two equivalent forms (Lemma 1 and Lemma 2).

Before stating the lemmas, let’s fix some notations. For any B \subset A, the map \pi_B is the natural projection from the full product X=\prod_{\alpha \in A} X_\alpha to the subproduct \prod_{\alpha \in B} X_\alpha. The standard basic open sets in the product space X=\prod_{\alpha \in A} X_\alpha are of the form \prod_{\alpha \in A} O_\alpha where O_\alpha=X_\alpha for all but finitely many \alpha \in A. We use supp(\prod_{\alpha \in A} O_\alpha) to denote the set of finitely many \alpha \in A such that O_\alpha \ne X_\alpha.

Given a space W, and given F,G \subset W, the sets F and G are separated if F \cap \overline{G}=\varnothing=\overline{F} \cap G.

Lemma 1

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. For any sets H,K \subset Y, the following two conditions are equivalent:

    1. There exist disjoint open subsets U and V of Y such that H \subset U and K \subset V.
    2. There exists a countable B \subset A such that the sets \pi_B(H) and \pi_B(K) are separated in the space \pi_B(Y).

Lemma 2

    Let X=\prod_{\alpha \in A} X_\alpha be a product of separable metrizable spaces. Let Y be a dense subspace of X. Then Y is normal if and only if for each pair of disjoint closed subsets H and K of Y, there exists a countable B \subset A such that \pi_B(H) and \pi_B(K) are separated in \pi_B(Y).

If Lemma 1 holds, it is clear that Lemma 2 holds. We prove Lemma 1. The lemmas indicate that to separate disjoint sets in the full product, it suffices to separate in a countable subproduct. In this sense normality in dense subspaces of a product of separable metrizable spaces only depends on countably many coordinates.

This lemma seems to have been around for a long time. We cannot find any reference of this lemma in Engelking’s topology textbook (see [4]). We found three references. One is Corson’s paper (see [3]), in which the lemma is mentioned in relation to the non-normality of \mathbb{N}^{\omega_1} and is attributed to a paper of M. Bockstein in 1948. Another is a paper of Baturov (see [2]), in which the lemma is used to prove a theorem about normality in dense subspace of M^{\omega_1} where M is a separable metric space. In [2] the lemma is attributed to Uspenskii. Another reference is Arkhangelskii’s book on function space (see Lemma I.6.1 on p. 43 in [1]) where the lemma is used in proving some facts about normality in function spaces C_p(X).

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Proof of Lemma 1

1 \Longrightarrow 2
Let U and V be disjoint open subsets of Y with H \subset U and K \subset V. Let U_1 and V_1 be open subsets of X such that U=U_1 \cap Y and V=V_1 \cap Y. Since Y is dense in X, U_1 \cap V_1=\varnothing.

Let \mathcal{U} be a maximal pairwise disjoint collection of standard basic open sets, each of which is a subset of U_1. Let \mathcal{V} be a maximal pairwise disjoint collection of standard basic open sets, each of which is a subset of V_1. These two collections can be obtained using a Zorn lemma argument. The product space X has the countable chain condition since it is a product of separable spaces. So both \mathcal{U} and \mathcal{V} are countable. Let B be the union of finite sets each one of which is a supp(O) where O \in \mathcal{U} \cup \mathcal{V}. The set B is countable too.

Let U^*=\cup \mathcal{U} and V^*=\cup \mathcal{V}. Note that U^* \cap V^*=\varnothing. We have the following observations:

    \pi^{-1}_B(\pi_B(U^*))=U^* \subset U_1 and \pi^{-1}_B(\pi_B(V^*))=V^* \subset V_1

The above observations lead to the following observations:

    \pi^{-1}_B(\pi_B(U^*)) \cap \pi^{-1}_B(\pi_B(V^*)) \subset U_1 \cap V_1=\varnothing

implying that \pi_B(U^*) \cap \pi_B(V^*)=\varnothing. Both \pi_B(U^*) and \pi_B(V^*) are open subsets of \pi_B(X) and are dense in \pi_B(X), respectively.

We claim that \pi_B(U_1) \cap \pi_B(V_1)=\varnothing. Suppose that y \in \pi_B(U_1) \cap \pi_B(V_1). Then \pi_B(V_1) contains a point of \pi_B(U^*), say t. With t \in \pi_B(U^*), t=\pi_B(q) for some q \in O where O \in \mathcal{U}. Note that supp(O) \subset B. Thus \pi^{-1}_B(\pi_B(q))=\pi^{-1}_B(t)=O \subset U_1. On the other hand, t \in \pi_B(V_1) implies that t=\pi_B(w) for some w \in V_1. It follows that w \in U_1 \cap V_1, a contradiction. Therefore \pi_B(U_1) \cap \pi_B(V_1)=\varnothing.

We have \pi_B(H) \subset \pi_B(U) \subset \pi_B(U_1) and \pi_B(K) \subset \pi_B(V) \subset \pi_B(V_1). This implies that \overline{\pi_B(H)} \cap \pi_B(K)=\varnothing and \pi_B(H) \cap \overline{\pi_B(K)}=\varnothing (closure in \pi_B(X)). Then \pi_B(H) and \pi_B(K) are separated in \pi_B(Y) as well. This concludes the proof for the 1 \Longrightarrow 2 direction.

2 \Longrightarrow 1
Suppose that B \subset A is countable such that \pi_B(H) and \pi_B(K) are separated in the space \pi_B(Y). Note that \pi_B(H) \subset \pi_B(Y) and \pi_B(K) \subset \pi_B(Y). Then we have the following:

    \pi_B(H) \cup \pi_B(K) \subset \pi_B(Y) - (\overline{\pi_B(H)} \cap \overline{\pi_B(K)}) \ \ \ \ \text{closures in } \pi_B(Y)

Consider W=\pi_B(Y) - (\overline{\pi_B(H)} \cap \overline{\pi_B(K)}). The space W is an open subspace of \pi_B(Y). Furthermore, \pi_B(Y) is a subspace of \prod_{\alpha \in B} X_\alpha, which is a separable and metrizable space. Thus the space W is metrizable and hence normal.

For L \subset W, let Cl_W(L) denote the closure of L in the space W. Note that Cl_W(\pi_B(H)) and Cl_W(\pi_B(K)) are disjoint and closed sets in W. Let G_H and G_K be disjoint open subsets of W such that Cl_W(\pi_B(H)) \subset G_H and Cl_W(\pi_B(K)) \subset G_K. Then \pi^{-1}_B(G_H) \cap Y and \pi^{-1}_B(G_K) \cap Y are disjoint open subsets of Y such that H \subset \pi^{-1}_B(G_H) \cap Y and K \subset \pi^{-1}_B(G_H) \cap Y. \blacksquare

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Remark

The proof of Lemma 1 does not need the full strength of separable metric in each factor of the product space. The above proof only makes two assumptions about the product space: the product space X=\prod_{\alpha \in A} X_\alpha has the countable chain condition (CCC) and that any countable subproduct is normal, i.e., \prod_{\alpha \in B} X_\alpha is normal for any countable B \subset A.

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Example

As an application of the above lemma, we give another proof of the non-normality of the product space of uncountably many copies of the discrete space of the non-negative integers. See this post for a version of A. H. Stone’s original proof.

Let \mathbb{N} be the set of all nonnegative integers and let \omega_1 be the first uncountable ordinal (i.e. the set of all countable ordinals). We provide an alternative proof that \mathbb{N}^{\omega_1} is not normal. In A. H. Stone’s proof, the following disjoint closed sets cannot be separated in \mathbb{N}^{\omega_1}:

    H=\left\{x \in \mathbb{N}^{\omega_1}: \forall \ n \ne 0, x_\alpha=n \text{ for at most one } \alpha<\omega_1  \right\}

    K=\left\{x \in \mathbb{N}^{\omega_1}: \forall \ n \ne 1, x_\alpha=n \text{ for at most one } \alpha<\omega_1  \right\}

We can also use Lemma 1 to show that H and K cannot be separated. Note that for each countable B \subset \omega_1, the sets \pi_B(H) and \pi_B(K) have non-empty intersection. Hence they cannot be separated in \pi_B(\mathbb{N}^{\omega_1}). By Lemma 1, H and K cannot be separated in the full product space \mathbb{N}^{\omega_1}.

To see that \pi_B(H) \cap \pi_B(K) \ne \varnothing, choose a function g:\omega_1 \rightarrow \mathbb{N} such that g^{-1}(0) \cap B=\varnothing. Let g_B:B \rightarrow \mathbb{N} be defined by g_B(\alpha)=g(\alpha) for all \alpha \in B. Then g_B \in \pi_B(H) \cap \pi_B(K).

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Reference

  1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  2. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
  3. Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
  4. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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\copyright \ 2014 \text{ by Dan Ma}

The normality of the product of the first uncountable ordinal with a compact factor

The product of a normal space with a compact space needs not be normal. For example, the product space \omega_1 \times (\omega_1+1) is not normal where \omega_1 is the first uncountable ordinal with the order topology and \omega_1+1 is the immediate successor of \omega_1 (see this post). However, \omega_1 \times I is normal where I=[0,1] is the unit interval with the usual topology. The topological story here is that I has countable tightness while the compact space \omega_1+1 does not. In this post, we prove the following theorem:

Theorem 1

    Let Y be an infinite compact space. Then the following conditions are equivalent:

    1. The product space \omega_1 \times Y is normal.
    2. Y has countable tightness, i.e., t(Y)=\omega.

Theorem 1 is a special case of the theorem found in [4]. The proof for the direction of countable tightness of Y implies \omega_1 \times Y is normal given in [4] relies on a theorem in another source. In this post we attempt to fill in some of the gaps. For the direction 2 \Longrightarrow 1, we give a complete proof. For the direction 1 \Longrightarrow 2, we essentially give the same proof as in [4], proving it by using a series of lemmas (stated below).

The authors in [2] studied the normality of X \times \omega_1 where X is not necessarily compact. The necessary definitions are given below. All spaces are at least Hausdorff.

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Definitions and Lemmas

Let X be a topological space. The tightness of X, denoted by t(X), is the least infinite cardinal number \kappa such that for any A \subset X and for any x \in \overline{A}, there exists a B \subset A such that x \in \overline{B} and \lvert B \lvert \le \kappa. When t(X)=\omega, we say X has countable tightness or is countably tight. When t(X)>\omega, we say X has uncountably tightness or is uncountably tight. An handy example of a space with uncountably tightness is \omega_1+1=\omega_1 \cup \left\{\omega_1 \right\}. This space has uncountable tightness at the point \omega_1. All first countable spaces and all Frechet spaces have countable tightness. The concept of countable tightness and tightness in general are discussed in more details here.

A sequence \left\{x_\alpha: \alpha<\tau \right\} of points of a space X is said to be a free sequence if for each \alpha<\tau, \overline{\left\{x_\beta: \beta<\alpha \right\}} \cap \overline{\left\{x_\beta: \beta \ge \alpha \right\}}=\varnothing. When a free sequence is indexed by the cardinal number \tau, the free sequence is said to have length \tau. The cardinal function F(X) is the least infinite cardinal \kappa such that if \left\{x_\alpha \in X: \alpha<\tau \right\} is a free sequence of length \tau, then \tau \le \kappa. The concept of tightness was introduced by Arkhangelskii and he proved that t(X)=F(X) (see p. 15 of [3]). This fact implies the following lemma.

Lemma 2

    Let X be compact. If t(X) \ge \tau, then there exists a free sequence \left\{x_\alpha \in X: \alpha<\tau \right\} of length \tau.

A proof of Lemma 2 can be found here.

The proof of the direction 1 \Longrightarrow 2 also uses the following lemmas.

Lemma 3

    For any compact space Y, \beta (\omega_1 \times Y)=(\omega_1+1) \times Y.

Lemma 4

    Let X be a normal space. For every pair H and K of disjoint closed subsets of X, H and K have disjoint closures in \beta X.

For Lemma 3, see 3.12.20(c) on p. 237 of [1]. For Lemma 4, see 3.6.4 on p. 173 of [1].

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Proof of Theorem 1

1 \Longrightarrow 2
Let X=\omega_1 \times Y. Suppose that X is normal. Suppose that Y has uncountable tightness, i.e., t(Y) \ge \omega_1. By Lemma 2, there exists a free sequence \left\{y_\alpha \in Y: \alpha<\omega_1 \right\}. For each \beta<\omega_1, let C_\beta=\left\{y_\alpha: \alpha>\beta \right\}. Then the collection \left\{\overline{C_\beta}: \beta<\omega_1 \right\} has the finite intersection property. Since Y is compact, \bigcap_{\beta<\omega_1} \overline{C_\beta} \ne \varnothing. Let p \in \bigcap_{\beta<\omega_1} \overline{C_\beta}. Consider the following closed subsets of X=\omega_1 \times Y.

    H=\overline{\left\{(\alpha,y_\alpha): \alpha<\omega_1 \right\}}
    K=\left\{(\alpha,p): \alpha<\omega_1 \right\}

We claim that H \cap K=\varnothing. Suppose that (\alpha,p) \in H \cap K. Either p \in \overline{\left\{y_\delta: \delta< \alpha+1 \right\}} or p \in \overline{\left\{y_\delta: \delta \ge \alpha+1 \right\}}. The latter case is not possible. Note that [0,\alpha] \times Y is an open set containing (\alpha,p). This open set cannot contain points of the form (\delta,p) where \delta \ge \alpha+1. So the first case p \in \overline{\left\{y_\delta: \delta< \alpha+1 \right\}} must hold. Since p \in \bigcap_{\beta<\omega_1} \overline{C_\beta}, p \in \overline{C_\alpha}=\overline{\left\{y_\delta: \delta \ge \alpha+1 \right\}}, a contradiction. So H and K are disjoint closed subsets of X=\omega_1 \times Y.

Now consider \beta X, the Stone-Cech compactification of X=\omega_1 \times Y. By Lemma 3, \beta X=\beta (\omega_1 \times Y)=(\omega_1+1) \times Y. Let H^*=\overline{H} and K^*=\overline{K} (closures in \beta X). We claim that (\omega_1,p) \in H^* \cap K^*. Let O=(\theta,\omega_1] \times V be an open set in \beta X with (\omega_1,p) \in O. Note that p \in \overline{C_\theta}=\left\{y_\delta: \delta>\theta \right\}. Thus V \cap \overline{C_\theta} \ne \varnothing. Choose \delta>\theta such that y_\delta \in V. We have (\delta,y_\delta) \in (\theta,\omega_1] \times V and (\delta,y_\delta) \in H^*. On the other hand, (\delta,p) \in K^*. Thus (\omega_1,p) \in H^* \cap K^*, a contradiction. Since X=\omega_1 \times Y is normal, Lemma 4 indicates that H and K should have disjoint closures in \beta X=(\omega_1+1) \times Y. Thus Y has countable tightness.

2 \Longrightarrow 1
Suppose t(Y)=\omega. Let H and K be disjoint closed subsets of \omega_1 \times Y. The following series of claims will complete the proof:

Claim 1
For each y \in Y, there exists an \alpha<\omega_1 such that either W_{H,y} \subset \alpha+1 or W_{K,y} \subset \alpha+1 where

    W_{H,y}=\left\{\delta<\omega_1: (\delta,y) \in H \right\}
    W_{K,y}=\left\{\delta<\omega_1: (\delta,y) \in K \right\}

Proof of Claim 1
Let y \in Y. The set V=\omega_1 \times \left\{y \right\} is a copy of \omega_1. It is a known fact that in \omega_1, there cannot be two disjoint closed and unbounded sets. Let V_H=V \cap H and V_K=V \cap K. If V_H \ne \varnothing and V_K \ne \varnothing, they cannot be both unbounded in V. Thus the claim follows if both V_H \ne \varnothing and V_K \ne \varnothing. Now suppose only one of V_H and V_K is non-empty. If the one that is non-empty is bounded, then the claim follows. Suppose the one that is non-empty is unbounded, say V_K. Then W_{H,y}=\varnothing and the claim follows.

Claim 2
For each y \in Y, there exists an \alpha<\omega_1 and there exists an open set O_y \subset Y with y \in O_y such that one and only one of the following holds:

    H \cap (\omega_1 \times \overline{O_y}) \subset (\alpha+1) \times \overline{O_y} \ \ \ \ \ \ \ \ (1)
    K \cap (\omega_1 \times \overline{O_y}) \subset (\alpha+1) \times \overline{O_y} \ \ \ \ \ \ \ \ (2)

Proof of Claim 2
Let y \in Y. Let \alpha<\omega_1 be as in Claim 1. Assume that W_{H,y} \subset \alpha+1. We want to show that there exists an open set O_y \subset Y with y \in O_y such that (1) holds. Suppose that for each open O \subset Y with y \in O, there is a q \in \overline{O} and there exists \delta_q>\alpha such that (\delta_q,q) \in H. Let S be the set of all such points q. Then y \in \overline{S}. Since Y has countable tightness, there exists countable T \subset S such that y \in \overline{T}. Since T is countable, choose \gamma >\omega_1 such that \alpha<\delta_q<\gamma for all q \in T. Note that [\alpha,\gamma] \times \left\{y \right\} does not contain points of H since W_{H,y} \subset \alpha+1. For each \theta \in [\alpha,\gamma], the point (\theta,y) has an open neighborhood that contains no point of H. Since [\alpha,\gamma] \times \left\{y \right\} is compact, finitely many of these neighborhoods cover [\alpha,\gamma] \times \left\{y \right\}. Let these finitely many open neighborhoods be M_i \times N_i where i=1,\cdots,m. Let N=\bigcap_{i=1}^m N_i. Then y \in N and N would contain a point of T, say q. Then (\delta_q,q) \in M_i \times N_i for some i, a contradiction. Note that (\delta_q,q) is a point of H. Thus there exists an open O_y \subset Y with y \in O_y such that (1) holds. This completes the proof of Claim 2.

Claim 3
For each y \in Y, there exists an \alpha<\omega_1 and there exists an open set O_y \subset Y with y \in O_y such that there are disjoint open subsets Q_H and Q_K of \omega_1 \times \overline{O_y} with H \cap (\omega_1 \times \overline{O_y}) \subset Q_H and K \cap (\omega_1 \times \overline{O_y}) \subset Q_K.

Proof of Claim 3
Let y \in Y. Let \alpha and O_y be as in Claim 2. Assume (1) in the statement of Claim 2 holds. Note that (\alpha+1) \times \overline{O_y} is a product of two compact spaces and is thus compact (and normal). Let R_{H,y} and R_{K,y} be disjoint open sets in (\alpha+1) \times \overline{O_y} such that H \cap (\alpha+1) \times \overline{O_y} \subset R_{H,y} and K \cap (\alpha+1) \times \overline{O_y} \subset R_{K,y}. Note that [\alpha+1,\omega_1) \times \overline{O_y} contains no points of H. Then Q_{H,y}=R_{H,y} and Q_{K,y}=R_{K,y} \cup [\alpha+1,\omega_1) \times \overline{O_y} are the desired open sets. This completes the proof of Claim 3.

To make the rest of the proof easier to see, we prove the following claim , which is a general fact that is cleaner to work with. Claim 4 describes precisely (in a topological way) what is happening at this point in the proof.

Claim 4
Let Z be a space. Let C and D be disjoint closed subsets of Z. Suppose that \left\{U_1,U_2,\cdots,U_m \right\} is a collection of open subsets of Z covering C \cup D such that for each i=1,2,\cdots,m, only one of the following holds:

    C \cap \overline{U_i} \ne \varnothing \text{ and } D \cap \overline{U_i}=\varnothing
    C \cap \overline{U_i} = \varnothing \text{ and } D \cap \overline{U_i} \ne \varnothing

Then there exist disjoint open subsets of Z separating C and D.

Proof of Claim 4
Let U_C=\cup \left\{U_i: \overline{U_i} \cap C \ne \varnothing \right\} and U_D=\cup \left\{U_i: \overline{U_i} \cap D \ne \varnothing \right\}. Note that \overline{U_C}=\cup \left\{\overline{U_i}: \overline{U_i} \cap C \ne \varnothing \right\}. Likewise, \overline{U_D}=\cup \left\{\overline{U_i}: \overline{U_i} \cap D \ne \varnothing \right\}. Let V_C=U_C-\overline{U_D} and V_D=U_D-\overline{U_C}. Then V_C and V_D are disjoint open sets. Furthermore, C \subset V_C and D \subset V_D. This completes the proof of Claim 4.

Now back to the proof of Theorem 1. For each y \in Y, let O_y, Q_{H,y} and Q_{K,y} be as in Claim 3. Since Y is compact, there exists \left\{y_1,y_2,\cdots,y_n \right\} \subset Y such that \left\{O_{y_1},O_{y_2},\cdots,O_{y_n} \right\} is a cover of Y. For each i=1,\cdots,n, let L_i=Q_{H,y_i} \cap (\omega_1 \times O_y) and M_i=Q_{K,y_i} \cap (\omega_1 \times O_y). Note that both L_i and M_i are open in \omega_1 \times Y. To apply Claim 4, rearrange the open sets L_i and M_i and re-label them as U_1,U_2,\cdots,U_m. By letting Z=\omega_1 \times Y, C=H and D=K, the open sets U_i satisfy Claim 4. Tracing the U_i to L_j or M_j and then to Q_{H,y_j} and Q_{K,y_j}, it is clear that the two conditions in Claim 4 are satisfied:

    H \cap \overline{U_i} \ne \varnothing \text{ and } K \cap \overline{U_i}=\varnothing
    H \cap \overline{U_i} = \varnothing \text{ and } K \cap \overline{U_i} \ne \varnothing

Then by Claim 4, the disjoint closed sets H and K can be separated by two disjoint open subsets of \omega_1 \times Y. \blacksquare

The theorem proved in [4] is essentially the statement that for any compact space Y, the product \kappa^+ \times Y is normal if and only t(Y) \le \kappa. Here \kappa^+ is the first ordinal of the next cardinal that is greater than \kappa.

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Gruenhage, G., Nogura, T., Purisch, S., Normality of X \times \omega_1, Topology and its Appl., 39, 263-275, 1991.
  3. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
  4. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976.

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\copyright \ 2014 - 2015 \text{ by Dan Ma}

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}