# Normal dense subspaces of products of “omega 1” many separable metric factors

Is every normal dense subspace of a product of separable metric spaces collectionwise normal? This question was posed by Arkhangelskii (see Problem I.5.25 in [2]). One partial positive answer is a theorem attributed to Corson: if $Y$ is a normal dense subspace of a product of separable spaces such that $Y \times Y$ is normal, then $Y$ is collectionwise normal. Another partial positive answer: assuming $2^\omega<2^{\omega_1}$, any normal dense subspace of the product space of continuum many separable metric factors is collectionwise normal (see Corollary 4 in this previous post). Another partial positive answer to Arkhangelskii’s question is the theorem due to Reznichenko: If $C_p(X)$, which is a dense subspace of the product space $\mathbb{R}^X$, is normal, then it is collectionwise normal (see Theorem I.5.12 in [2]). In this post, we highlight another partial positive answer to the question posted in [2]. Specifically, we prove the following theorem:

Theorem 1

Let $X=\prod_{\alpha<\omega_1} X_\alpha$ be a product space where each factor $X_\alpha$ is a separable metric space. Let $Y$ be a dense subspace of $X$. Then if $Y$ is normal, then $Y$ is collectionwise normal.

Since any normal space with countable extent is collectionwise normal (see Theorem 2 in this previous post), it suffices to prove the following theorem:

Theorem 1a

Let $X=\prod_{\alpha<\omega_1} X_\alpha$ be a product space where each factor $X_\alpha$ is a separable metric space. Let $Y$ be a dense subspace of $X$. Then if $Y$ is normal, then every closed and discrete subspace of $Y$ is countable, i.e., $Y$ has countable extent.

Arkhangelskii’s question was studied by the author of [3] and [4]. Theorem 1 as presented in this post is essentially the Theorem 1 found in [3]. The proof given in [3] is a beautiful proof. The proof in this post is modeled on the proof in [3] with the exception that all the crucial details are filled in. Theorem 1a (as stated above) is used in [1] to show that the function space $C_p(\omega_1+1)$ contains no dense normal subspace.

It is natural to wonder if Theorem 1 can be generalized to product space of $\tau$ many separable metric factors where $\tau$ is an arbitrary uncountable cardinal. The work of [4] shows that the question at the beginning of this post cannot be answered positively in ZFC. Recall the above mentioned result that assuming $2^\omega<2^{\omega_1}$, any normal dense subspace of the product space of continuum many separable metric factors is collectionwise normal (see Corollary 4 in this previous post). A theorem in [4] implies that assuming $2^\omega=2^{\omega_1}$, for any separable metric space $M$ with at least 2 points, the product of continuum many copies of $M$ contains a normal dense subspace $Y$ that is not collectionwise normal. A side note: for this normal subspace $Y$, $Y \times Y$ is necessarily not normal (according to Corson’s theorem). Thus [3] and [4] collectively show that Arkhangelskii’s question stated here at the beginning of the post is answered positively (in ZFC) among product spaces of $\omega_1$ many separable metric factors and that outside of the $\omega_1$ case, it is impossible to answer the question positively in ZFC.

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Proving Theorem 1a

We use the following lemma. For a proof of this lemma, see the proof for Lemma 1 in this previous post.

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. $Y$ 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 $\overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$.

For any $B \subset \omega_1$, let $\pi_B$ be the natural projection from the product space $X=\prod_{\alpha<\omega_1} X_\alpha$ into the subproduct space $\prod_{\alpha \in B} X_\alpha$.

Proof of Theorem 1a
Let $Y$ be a dense subspace of the product space $X=\prod_{\alpha<\omega_1} X_\alpha$ where each factor $X_\alpha$ has a countable base. Suppose that $D$ is an uncountable closed and discrete subset of $Y$. We then construct a pair of disjoint closed subsets $H$ and $K$ of $Y$ such that for all countable $B \subset \omega_1$, $\pi_B(H)$ and $\pi_B(K)$ are not separated, specifically $\pi_B(H) \cap \overline{\pi_B(K)}\ne \varnothing$. Here the closure is taken in the space $\pi_B(Y)$. By Lemma 2, the dense subspace $Y$ of $X$ is not normal.

For each $\alpha<\omega_1$, let $\mathcal{B}_\alpha$ be a countable base for the space $X_\alpha$. The standard basic open sets in the product space $X$ are of the form $O=\prod_{\alpha<\omega_1} O_\alpha$ such that

• each $O_\alpha$ is an open subset of $X_\alpha$,
• if $O_\alpha \ne X_\alpha$, then $O_\alpha \in \mathcal{B}_\alpha$,
• $O_\alpha=X_\alpha$ for all but finitely many $\alpha<\omega_1$.

We use $supp(O)$ to denote the finite set of $\alpha$ such that $O_\alpha \ne X_\alpha$. Technically we should be working with standard basic open subsets of $Y$, i.e., sets of the form $O \cap Y$ where $O$ is a standard basic open set as described above. Since $Y$ is dense in the product space, every standard open set contains points of $Y$. Thus we can simply work with standard basic open sets in the product space as long as we are working with points of $Y$ in the construction.

Let $\mathcal{M}$ be the collection of all standard basic open sets as described above. Since there are only $\omega_1$ many factors in the product space, $\lvert \mathcal{M} \lvert=\omega_1$. Recall that $D$ is an uncountable closed and discrete subset of $Y$. Let $\mathcal{M}^*$ be the following:

$\mathcal{M}^*=\left\{U \in \mathcal{M}: U \cap D \text{ is uncountable } \right\}$

Claim 1. $\lvert \mathcal{M}^* \lvert=\omega_1$.

First we show that $\mathcal{M}^* \ne \varnothing$. Let $B \subset \omega_1$ be countable. Consider these two cases: Case 1. $\pi_B(D)$ is an uncountable subset of $\prod_{\alpha \in B} X_\alpha$; Case 2. $\pi_B(D)$ is countable.

Suppose Case 1 is true. Since $\prod_{\alpha \in B} X_\alpha$ is a product of countably many separable metric spaces, it is hereditarily Lindelof. Then there exists a point $y \in \pi_B(D)$ such that every open neighborhood of $y$ (open in $\prod_{\alpha \in B} X_\alpha$) contains uncountably many points of $\pi_B(D)$. Thus every standard basic open set $U=\prod_{\alpha \in B} U_\alpha$, with $y \in U$, contains uncountably many points of $\pi_B(D)$. Suppose Case 2 is true. There exists one point $y \in \pi_B(D)$ such that $y=\pi_B(t)$ for uncountably many $t \in D$. Then in either case, every standard basic open set $V=\prod_{\alpha<\omega_1} V_\alpha$, with $supp(V) \subset B$ and $y \in \pi_B(V)$, contains uncountably many points of $D$. Any one such $V$ is a member of $\mathcal{M}^*$.

We can partition the index set $\omega_1$ into $\omega_1$ many disjoint countable sets $B$. Then for each such $B$, obtain a $V \in \mathcal{M}^*$ in either Case 1 or Case 2. Since $supp(V) \subset B$, all such open sets $V$ are distinct. Thus Claim 1 is established.

Claim 2.
There exists an uncountable $H \subset D$ such that for each $U \in \mathcal{M}^*$, $U \cap H \ne \varnothing$ and $U \cap (D-H) \ne \varnothing$.

Enumerate $\mathcal{M}^*=\left\{U_\gamma: \gamma<\omega_1 \right\}$. Choose $h_0,k_0 \in U_0 \cap D$ with $h_0 \ne k_0$. Suppose that for all $\beta<\gamma$, two points $h_\beta,k_\beta$ are chosen such that $h_\beta,k_\beta \in U_\beta \cap D$, $h_\beta \ne k_\beta$ and such that $h_\beta \notin L_\beta$ and $k_\beta \notin L_\beta$ where $L_\beta=\left\{h_\rho: \rho<\beta \right\} \cup \left\{k_\rho: \rho<\beta \right\}$. Then choose $h_\gamma,k_\gamma$ with $h_\gamma \ne k_\gamma$ such that $h_\gamma,k_\gamma \in U_\gamma \cap D$ and $h_\gamma \notin L_\gamma$ and $k_\gamma \notin L_\gamma$ where $L_\gamma=\left\{h_\rho: \rho<\gamma \right\} \cup \left\{k_\rho: \rho<\gamma \right\}$.

Let $H=\left\{h_\gamma: \gamma<\omega_1 \right\}$ and let $K=D-H$. Note that $K_0=\left\{k_\gamma: \gamma<\omega_1 \right\} \subset K$. Based on the inductive process that is used to obtain $H$ and $K_0$, it is clear that $H$ satisfies Claim 2.

Claim 3.
For each countable $B \subset \omega_1$, the sets $\pi_B(H)$ and $\pi_B(K)$ are not separated in the space $\pi_B(Y)$.

Let $B \subset \omega_1$ be countable. Consider the two cases: Case 1. $\pi_B(H)$ is uncountable; Case 2. $\pi_B(H)$ is countable. Suppose Case 1 is true. Since $\prod_{\alpha \in B} X_\alpha$ is a product of countably many separable metric spaces, it is hereditarily Lindelof. Then there exists a point $p \in \pi_B(H)$ such that every open neighborhood of $p$ (open in $\prod_{\alpha \in B} X_\alpha$) contains uncountably many points of $\pi_B(H)$. Choose $h \in H$ such that $p=\pi_B(h)$. Then the following statement holds:

1. For every basic open set $U=\prod_{\alpha<\omega_1} U_\alpha$ with $h \in U$ such that $supp(U) \subset B$, the open set $U$ contains uncountably many points of $H$.

Suppose Case 2 is true. There exists some $p \in \pi_B(H)$ such that $p=\pi_B(t)$ for uncountably many $t \in H$. Choose $h \in H$ such that $p=\pi_B(h)$. Then statement 1 also holds.

In either case, there exists $h \in H$ such that statement 1 holds. The open sets $U$ described in statement 1 are members of $\mathcal{M}^*$. By Claim 2, the open sets described in statement 1 also contain points of $K$. Since the open sets described in statement 1 have supports $\subset B$, the following statement holds:

1. For every basic open set $V=\prod_{\alpha \in B} V_\alpha$ with $\pi_B(h) \in V$, the open set $V$ contains points of $\pi_B(K)$.

Statement 2 indicates that $\pi_B(h) \in \overline{\pi_B(K)}$. Thus $\pi_B(h) \in \pi_B(H) \cap \overline{\pi_B(K)}$. The closure here can be taken in either $\prod_{\alpha \in B} X_\alpha$ or $\pi_B(Y)$ (to apply Lemma 2, we only need the latter). Thus Claim 3 is established.

Claim 3 is the negation of condition 3 of Lemma 2. Therefore $Y$ is not normal. $\blacksquare$

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Remark

The proof of Theorem 1a, though a proof in ZFC only, clearly relies on the fact that the product space is a product of $\omega_1$ many factors. For example, in the inductive step in the proof of Claim 2, it is always possible to pick a pair of points not chosen previously. This is because the previously chosen points form a countable set and each open set in $\mathcal{M}^*$ contains $\omega_1$ many points of the closed and discrete set $D$. With the “$\omega$ versus $\omega_1$” situation, at each step, there are always points not previously chosen. When more than $\omega_1$ many factors are involved, there may be no such guarantee in the inductive process.

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Reference

1. Arkhangelskii, A. V., Normality and dense subspaces, Proc. Amer. Math. Soc., 130 (1), 283-291, 2001.
2. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
3. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
4. Baturov, D. P., On perfectly normal dense subspaces of products, Topology Appl., 154, 374-383, 2007.
5. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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

# Normal dense subspaces of a product of “continuum” many separable metric factors

Is every normal dense subspace of a product of separable metric spaces collectionwise normal? This question was posed by Arkhangelskii in [1] (see Problem I.5.25). A partial positive answer is provided by a theorem that is usually attributed to Corson: If $Y$ is a normal dense subspace of a product of separable metric spaces and if $Y \times Y$ is also normal, then $Y$ is collectionwise normal. In this post, using a simple combinatorial argument, we show that any normal dense subspace of a product of continuum many separable metric space is collectionwise normal (see Corollary 4 below), which is a corollary of the following theorem.

Theorem 1
Let $X$ be a normal space with character $\le 2^\omega$. If $2^\omega<2^{\omega_1}$, then the following holds:

• If $Y$ is a closed and discrete subspace of $X$ with $\lvert Y \lvert=\omega_1$, then $Y$ contains a separated subset of cardinality $\omega_1$.

Theorem 1 gives the corollary indicated at the beginning and several other interesting results. The statement $2^\omega<2^{\omega_1}$ means that the cardinality of the power set (the set of all subsets) of $\omega$ is strictly less than the cardinality of the power set of $\omega_1$. Note that the statement $2^\omega<2^{\omega_1}$ follows from the continuum hypothesis (CH), the statement that $2^\omega=\omega_1$. With the assumption $2^\omega<2^{\omega_1}$, Theorem 1 is a theorem that goes beyond ZFC. We also present an alternative to Theorem 1 that removes the assumption $2^\omega<2^{\omega_1}$ (see Theorem 6 below).

A subset $T$ of a space $S$ is a separated set (in $S$) if for each $t \in T$, there is an open subset $O_t$ of $S$ with $t \in O_t$ such that $\left\{O_t: t \in T \right\}$ is a pairwise disjoint collection. First we prove Theorem 1 and then discuss the corollaries.

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Proof of Theorem 1

Suppose $Y$ is a closed and discrete subset of $X$ with $\lvert Y \lvert=\omega_1$ such that no subset of $Y$ of cardinality $\omega_1$ can be separated. We then show that $2^{\omega_1} \le 2^{\omega}$.

For each $y \in Y$, let $\mathcal{B}_y$ be a local base at the point $y$ such that $\lvert \mathcal{B}_y \lvert \le 2^\omega$. Let $\mathcal{B}=\bigcup_{y \in Y} \mathcal{B}_y$. Thus $\lvert \mathcal{B} \lvert \le 2^\omega$. By normality, for each $W \subset Y$, let $U_W$ be an open subset of $X$ such that $W \subset U_W$ and $\overline{U_W} \cap (Y-W)=\varnothing$. For each $W \subset Y$, consider the following collection of open sets:

$\mathcal{G}_W=\left\{V \in \mathcal{B}_y: y \in W \text{ and } V \subset U_W \right\}$

For each $W \subset Y$, choose a maximal disjoint collection $\mathcal{M}_W$ of open sets in $\mathcal{G}_W$. Because no subset of $Y$ of cardinality $\omega_1$ can be separated, each $\mathcal{M}_W$ is countable. If $W_1 \ne W_2$, then $\mathcal{M}_{W_1} \ne \mathcal{M}_{W_2}$.

Let $\mathcal{P}(Y)$ be the power set (i.e. the set of all subsets) of $Y$. Let $\mathcal{P}_\omega(\mathcal{B})$ be the set of all countable subsets of $\mathcal{B}$. Then the mapping $W \mapsto \mathcal{M}_W$ is a one-to-one map from $\mathcal{P}(Y)$ into $\mathcal{P}_\omega(\mathcal{B})$. Note that $\lvert \mathcal{P}(Y) \lvert=2^{\omega_1}$. Also note that since $\lvert \mathcal{B} \lvert \le 2^\omega$, $\lvert \mathcal{P}_\omega(\mathcal{B}) \lvert \le 2^\omega$. Thus $2^{\omega_1} \le 2^{\omega}$. $\blacksquare$

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Some Corollaries of Theorem 1

Here’s some corollaries that follow easily from Theorem 1. A space $X$ has the countable chain condition (CCC) if every pairwise disjoint collection of non-empty open subset of $X$ is countable. For convenience, if $X$ has the CCC, we say $X$ is CCC. The following corollaries make use of the fact that any normal space with countable extent is collectionwise normal (see Theorem 2 in this previous post).

Corollary 2
Let $X$ be a CCC space with character $\le 2^\omega$. If $2^\omega<2^{\omega_1}$, then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ is countable, i.e., $X$ has countable extent.
• If $X$ is normal, then $X$ is collectionwise normal.

Corollary 3
Let $X$ be a CCC space with character $\le 2^\omega$. If CH holds, then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ is countable, i.e., $X$ has countable extent.
• If $X$ is normal, then $X$ is collectionwise normal.

Corollary 4
Let $X=\prod_{\alpha<2^\omega} X_\alpha$ be a product where each factor $X_\alpha$ is a separable metric space. If $2^\omega<2^{\omega_1}$, then the following conditions hold:

• If $Y$ is a normal dense subspace of $X$, then $Y$ has countable extent.
• If $Y$ is a normal dense subspace of $X$, then $Y$ is collectionwise normal.

Corollary 4 is the result indicated in the title of the post. The product of separable spaces has the CCC. Thus the product space $X$ and any dense subspace of $X$ have the CCC. Because $X$ is a product of continuum many separable metric spaces, $X$ and any subspace of $X$ have characters $\le 2^\omega$. Then Corollary 4 follows from Corollary 2.

When dealing with the topic of normal versus collectionwise normal, it is hard to avoid the connection with the normal Moore space conjecture. Theorem 1 gives the result of F. B. Jones from 1937 (see [3]). We have the following theorem.

Theorem 5
If $2^\omega<2^{\omega_1}$, then every separable normal Moore space is metrizable.

Though this was not how Jones proved it in [3], Theorem 5 is a corollary of Corollary 2. By Corollary 2, any separable normal Moore space is collectionwise normal. It is well known that collectionwise normal Moore space is metrizable (Bing’s metrization theorem, see Theorem 5.4.1 in [2]).

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A ZFC Theorem

We now prove a result that is similar to Corollary 2 but uses no set-theory beyond the Zermelo–Fraenkel set theory plus axiom of choice (abbreviated by ZFC). Of course the conclusion is not as strong. Even though the assumption $2^\omega<2^{\omega_1}$ is removed in Theorem 6, note the similarity between the proof of Theorem 1 and the proof of Theorem 6.

Theorem 6
Let $X$ be a CCC space with character $\le 2^\omega$. Then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ has cardinality less than continuum.

Proof of Theorem 6
Let $X$ be a normal CCC space with character $\le 2^\omega$. Let $Y$ be a closed and discrete subset of $X$. We show that $\lvert Y \lvert < 2^\omega$. Suppose that $\lvert Y \lvert = 2^\omega$.

For each $y \in Y$, let $\mathcal{B}_y$ be a local base at the point $y$ such that $\lvert \mathcal{B}_y \lvert \le 2^\omega$. Let $\mathcal{B}=\bigcup_{y \in Y} \mathcal{B}_y$. Thus $\lvert \mathcal{B} \lvert = 2^\omega$. By normality, for each $W \subset Y$, let $U_W$ be an open subset of $X$ such that $W \subset U_W$ and $\overline{U_W} \cap (Y-W)=\varnothing$. For each $W \subset Y$, consider the following collection of open sets:

$\mathcal{G}_W=\left\{V \in \mathcal{B}_y: y \in W \text{ and } V \subset U_W \right\}$

For each $W \subset Y$, choose $\mathcal{M}_W \subset \mathcal{G}_W$ such that $\mathcal{M}_W$ is a maximal disjoint collection. Since $X$ is CCC, $\mathcal{M}_W$ is countable. It is clear that if $W_1 \ne W_2$, then $\mathcal{M}_{W_1} \ne \mathcal{M}_{W_2}$.

Let $\mathcal{P}(Y)$ be the power set (i.e. the set of all subsets) of $Y$. Let $\mathcal{P}_\omega(\mathcal{B})$ be the set of all countable subsets of $\mathcal{B}$. Then the mapping $W \mapsto \mathcal{M}_W$ is a one-to-one map from $\mathcal{P}(Y)$ into $\mathcal{P}_\omega(\mathcal{B})$. Note that since $\lvert \mathcal{B} \lvert = 2^\omega$, $\lvert \mathcal{P}_\omega(\mathcal{B}) \lvert = 2^\omega$. Thus $\lvert \mathcal{P}(Y) \lvert \le 2^{\omega}$. However, $Y$ is assumed to be of cardinality continuum. Then $\lvert \mathcal{P}(Y) \lvert>2^{\omega_1}$, leading to a contradiction. Thus it must be the case that $\lvert Y \lvert < 2^\omega$. $\blacksquare$

With Theorem 6, Corollary 3 still holds. Theorem 6 removes the set-theoretic assumption of $2^\omega<2^{\omega_1}$. As a result, the upper bound for cardinalities of closed and discrete sets is (at least potentially) higher.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Jones, F. B., Concerning normal and completely normal spaces, Bull. Amer. Math. Soc., 43, 671-677, 1937.

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

# Looking for another closed and discrete subspace of a product space

Let $\omega_1$ be the first uncountable ordinal. In a previous post called Looking for a closed and discrete subspace of a product space, it was shown that the product space $\mathbb{R}^c$, the product of continuum many copies of the real line $\mathbb{R}$, contains a closed and discrete subset of cardinality continuum. This example shows that a product space of uncountably many copies of a “nice” space is “big and wide” enough to hide uncountable closed and discrete sets even when the product space is separable. This post reinforces this same fact by showing that $\mathbb{R}^{\omega_1}$ contains a closed and discrete subset of cardinality $\omega_1$. It follows that for any uncountable cardinal $\tau$, the product space $\mathbb{R}^\tau$ contains an uncountable closed and discrete subset, i.e., the product of uncountably many copies of the real line $\mathbb{R}$ has uncountable extent.

Let $\omega$ be the first infinite ordinal, i.e., the set of all nonnegative integers. Consider $\omega^{\omega_1}$, the product of $\omega_1$ many copies of $\omega$ with the discrete topology. Since $\omega^{\omega_1}$ is a closed subspace of $\mathbb{R}^{\omega_1}$, it suffices to show that $\omega^{\omega_1}$ has an uncountable closed and discrete subset.

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The Construction

We now construct an uncountable closed and discrete subset of $\omega^{\omega_1}$. Let $\delta$ be an infinite ordinal such that $\omega<\delta<\omega_1$. Let $W=\left\{\alpha: \delta \le \alpha<\omega_1 \right\}$. For each $\alpha \in W$, let $Y_\alpha=\left\{\beta<\omega_1: \beta<\alpha \right\}$. We can also use interval notations: $W=[\delta,\omega_1)$ and $Y_\alpha=[0,\alpha)$. Consider $Y_\alpha$ as a space with the discrete topology. Then it is clear that $\omega^{\omega_1}$ is homeomorphic to the product space $\prod_{\alpha \in W} Y_\alpha$. Thus the focus is now on finding an uncountable closed and discrete subset of $\prod_{\alpha \in W} Y_\alpha$.

One interesting fact about the space $\prod_{\alpha \in W} Y_\alpha$ is that every function $f \in \prod_{\alpha \in W} Y_\alpha$ is a pressing down function. That is, for every $f \in \prod_{\alpha \in W} Y_\alpha$, $f(\alpha)<\alpha$ for all $\alpha \in W$. Note that $f$ is defined on $W$, a closed and unbounded subset of $\omega_1$ (hence a stationary set). It follows that for each $f \in \prod_{\alpha \in W} Y_\alpha$, there is a stationary set $S \subset W$ and there exists $\rho<\omega_1$ such that $f(\alpha)=\rho$ for all $\alpha \in S$. This fact is called the pressing down lemma and will be used below. See this post for more information about the pressing down lemma.

For each $\gamma \in W$, let $h_\gamma: Y_{\gamma+1} \rightarrow \delta$ be a one-to-one function. For each $\gamma \in W$, define $t_\gamma \in \prod_{\alpha \in W} Y_\alpha$ as follows:

$t_\gamma(\alpha) = \begin{cases} h_\gamma(\alpha) & \mbox{if } \delta \le \alpha \le \gamma \\ \gamma & \mbox{if } \gamma < \alpha <\omega_1 \end{cases}$

Note that each $t_\gamma$ is a pressing down function. Thus each $t_\gamma \in \prod_{\alpha \in W} Y_\alpha$. Let $T=\left\{t_\gamma: \gamma \in W \right\}$. Clearly $t_\gamma \ne t_\mu$ if $\gamma \ne \mu$. Thus $T$ has cardinality $\omega_1$. We claim that $T$ is a closed and discrete subset of $\prod_{\alpha \in W} Y_\alpha$. It suffices to show that for each $f \in \prod_{\alpha \in W} Y_\alpha$, there exists an open set $V$ with $f \in V$ such that $V$ contains at most one $t_\gamma$.

Let $f \in \prod_{\alpha \in W} Y_\alpha$. As discussed above, there is a stationary set $S \subset W$ and there exists $\rho<\omega_1$ such that $f(\alpha)=\rho$ for all $\alpha \in S$. In particular, choose $\mu, \lambda \in S$ such that $\mu \ne \lambda$. Thus $f(\mu)=f(\lambda)=\rho$. Let $V$ be the open set defined by:

$V=\left\{g \in \prod_{\alpha \in W} Y_\alpha: g(\mu)=g(\lambda)=\rho \right\}$

Clearly, $f \in V$. We show that if $t_\gamma \in V$, then $\gamma=\rho$. Suppose $t_\gamma \in V$. Then $t_\gamma(\mu)=t_\gamma(\lambda)=\rho$. Consider two cases: Case 1: $\delta \le \mu, \lambda \le \gamma$; Case 2: one of $\mu$ and $\lambda>\gamma$. The definition of $t_\gamma$ indicates that $t_\gamma=h_\gamma$ on the interval $[\delta, \gamma]$. Note that $h_\gamma$ is a one-to-one function. Since $t_\gamma(\mu)=t_\gamma(\lambda)$, it cannot be that $\mu, \lambda \in [\delta, \gamma]$, i.e., Case 1 is not possible. Thus Case 2 holds, say $\mu>\gamma$. Then by definition, $t_\gamma(\mu)=\gamma$. Putting everything together, $\gamma=t_\gamma(\mu)=t_\gamma(\lambda)=\rho$. Thus $V \cap T \subset \left\{t_\rho \right\}$. This concludes the proof that the set $T$ is a closed and discrete subset of $\prod_{\alpha \in W} Y_\alpha$. $\blacksquare$

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