# Equivalent conditions for hereditarily Lindelof spaces

A topological space $X$ is Lindelof if every open cover $X$ has a countable subcollection that also is a cover of $X$. A topological space $X$ is hereditarily Lindelof if every subspace of $X$, with respect to the subspace topology, is a Lindelof space. In this post, we prove a theorem that gives two equivalent conditions for the hereditarily Lindelof property. We consider the following theorem.

Theorem 1
Let $X$ be a topological space. The following conditions are equivalent.

1. The space $X$ is a hereditarily Lindelof space.
2. Every open subspace of $X$ is Lindelof.
3. For every uncountable subspace $Y$ of $X$, there exists a point $y \in Y$ such that every open subset of $X$ containing $y$ contains uncountably many points of $Y$.

This is an excellent exercise for the hereditarily Lindelof property and for transfinite induction (for one of the directions). The equivalence $1 \longleftrightarrow 3$ is the exercise 3.12.7(d) on page 224 of [1]. The equivalence of the 3 conditions of Theorem 1 is mentioned on page 182 (chapter d-8) of [2].

Proof of Theorem 1
The direction $1 \longrightarrow 2$ is immediate. The direction $2 \longrightarrow 3$ is straightforward.

$3 \longrightarrow 1$
We show $\text{not } 1 \longrightarrow \text{not } 3$. Suppose $T$ is a non-Lindelof subspace of $X$. Let $\mathcal{U}$ be an open cover of $T$ such that no countable subcollection of $\mathcal{U}$ can cover $T$. By a transfinite inductive process, choose a set of points $\left\{t_\alpha \in T: \alpha < \omega_1 \right\}$ and a collection of open sets $\left\{U_\alpha \in \mathcal{U}: \alpha < \omega_1 \right\}$ such that for each $\alpha < \omega_1$, $t_\alpha \in U_\alpha$ and $t_\alpha \notin \cup \left\{U_\beta: \beta<\alpha \right\}$. The inductive process is possible since no countable subcollection of $\mathcal{U}$ can cover $T$. Now let $Y=\left\{t_\alpha: \alpha<\omega_1 \right\}$. Note that each $U_\alpha$ can at most contain countably many points of $Y$, namely the points in $\left\{t_\beta: \beta \le \alpha \right\}$.

For each $\alpha$, let $V_\alpha$ be an open subset of $X$ such that $U_\alpha=V_\alpha \cap Y$. We can now conclude: for every point $t_\alpha$ of $Y$, there exists an open set $V_\alpha$ containing $t_\alpha$ such that $V_\alpha$ contains only countably many points of $Y$. This is the negation of condition 3. $\blacksquare$

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Remarks

Condition 3 indicates that every uncountable set has a certain special type of limit points. Let $p \in X$. We say $p$ is a limit point of the set $Y \subset X$ if every open set containing $p$ contains a point of $Y$ different from $p$. Being a limit point of $Y$, we only know that each open set containing $p$ contain infinitely many points of $Y$ (assuming a $T_1$ space). Thus the limit points indicated in condition 3 are a special type of limit points. According to the terminology of [1], if $p$ is a limit point of $Y$ satisfying condition 3, then $p$ is said to be a condensation point of $Y$. According to Theorem 1, existence of condensation point in every uncountable set is a strong topological property (being equivalent to the hereditarily property). It is easy to see that of condition 3 holds, all but countably many points of any uncountable set $Y$ is a condensation point of $Y$.

In some situations, we may not need the full strength of condition 3. In such situations, the following corollary may be sufficient.

Corollary 2
If the space $X$ is hereditarily Lindelof, then every uncountable subspace $Y$ of $X$ contains one of its limit points.

As noted earlier, if every uncountable set contains one of its limits, then all but countably many points of any uncountable set are limit points. To contrast the hereditarily Lindelof property with the Lindelof property, consider the following theorem.

Theorem 3
If the space $X$ is Lindelof, then every uncountable subspace $Y$ of $X$ has a limit point.

The condition “every uncountable subspace $Y$ of $X$ has a limit point” has another name. When a space satisfies this condition, it is said to have countable extent. The ideas in Corollary 2 and Theorem 3 are also discussed in this previous post.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.

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

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

# One theorem about normality of Cp(X)

Assuming that the function space $C_p(X)$ is normal, what can be said about the domain space $X$? In this post, we prove a theorem that yields a corollary that for any normal space $X$, if $C_p(X)$ is normal, then $X$ has countable extent (i.e. every closed and discrete subset of $X$ is countable). Thus the normality of the function space limits the size of a closed and discrete subset of the domain space. It then follows that for any metric space $X$, if $C_p(X)$ is normal, $X$ has is second countable (i.e. having a countable base). Another immediate, but slightly less obvious, corollary is that for any $X$ that is a normal Moore space, if $C_p(X)$ is normal, then $X$ is metrizable.

For definitions of basic open sets and other background information on the function space $C_p(X)$, see this previous post.

Let $X$ be a space. Let $Y \subset X$. Let $\pi_Y$ be the natural projection from the product space $\mathbb{R}^X$ into the product space $\mathbb{R}^Y$. Specifically, if $f \in \mathbb{R}^X$, then $\pi_Y(f)=f \upharpoonright Y$, i.e., the function $f$ restricted to $Y$. In the discussion below, $\pi_Y$ is defined just on $C_p(X)$, i.e., $\pi_Y$ is the natural projection from $C_p(X)$ into $C_p(Y)$. It is always the case that $\pi_Y(C_p(X)) \subset C_p(Y)$. It is not necessarily the case that $\pi_Y(C_p(X))=C_p(Y)$. However, if $X$ is a normal space and $Y$ is closed in $X$, then $\pi_Y(C_p(X))=C_p(Y)$ and $\pi_Y$ is the natural projection from $C_p(X)$ onto $C_p(Y)$. We prove the following theorem.

Theorem 1

Suppose that $C_p(X)$ is a normal space. Let $Y$ be a closed subspace of $X$. Then $\pi_Y(C_p(X))$ is a normal space.

Theorem 1 is found in [1] (see Theorem I.6.2). In proving Theorem 1, we need the following lemma.

Lemma 2

Let $T=\prod_{\alpha \in A} T_\alpha$ be a product of separable metrizable spaces. Let $S$ be a dense subspace of $T$. Then the following conditions are equivalent.

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

For a proof of Lemma 2, see Lemma 1 in this previous post.

Proof of Theorem 1
Note that $\pi_Y(C_p(X))$ is a dense subspace of $\mathbb{R}^Y$. Let $H$ and $K$ be disjoint closed subsets of $\pi_Y(C_p(X))$. To show $\pi_Y(C_p(X))$ is normal, by Lemma 2, we only need to produce a countable $B \subset Y$ such that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$. The closure here is taken in $\pi_B(\pi_Y(C_p(X)))$.

Let $H_1=\pi_Y^{-1}(H)$ and $K_1=\pi_Y^{-1}(K)$. Both $H_1$ and $K_1$ are closed subsets of $C_p(X)$. By Lemma 2, there exists some countable $C \subset X$ such that $\overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing$. The closure here is taken in $\pi_C(C_p(X))$. According to the remark at the end of this previous post, for any countable $D \subset X$ such that $C \subset D$, $\overline{\pi_D(H_1)} \cap \overline{\pi_D(K_1)}=\varnothing$. In other words, the countable set $C$ can be enlarged and the conclusion of the lemma still holds. With this observation in mind, we can assume that $C \cap Y \ne \varnothing$. If not, we can always throw countably many points of $Y$ into $C$ and still have $\overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing$.

Let $B=C \cap Y$. We claim that $\overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}=\varnothing$. The closure here is taken $\pi_B(C_p(X))$. Suppose that $\overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)} \ne \varnothing$. Choose $f \in C_p(X)$ such that $f \upharpoonright B \in \overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}$. It follows that $f \upharpoonright C \in \overline{\pi_C(H_1)}$. To see this, let $f \upharpoonright C \in U=\prod_{x \in C} U_x$ where $U$ is a standard basic open set. Let $F$ be the support of $U$, i.e., the finite set of $x \in C$ such that $U_x \ne \mathbb{R}$. Let $F_1=F \cap Y$ and $F_2=F \cap (X-Y)$. Let $U^*=\prod_{x \in B} U_x$. Note that $f \upharpoonright B \in U^*$. Since $f \upharpoonright B \in \overline{\pi_B(H_1)}$, there is some $g \in H_1$ such that $\pi_B(g) \in U^*$. Note that $F_1$ is the support of $U^*$.

Because the space $X$ is completely regular, there is a $h \in C_p(X)$ such that $h(x)=0$ for all $x \in Y$ and $h(x)=f(x)-g(x)$ for all $x \in F_2$. Let $w=h+g$. Since $w \upharpoonright Y=g \upharpoonright Y$, $w \in H_1$. Note that $w=g$ on $Y$, hence on $F_1$ and that $w=f$ on $F_2$. Thus $w \upharpoonright C \in U$. Since $U$ is an arbitrary open set containing $f \upharpoonright C$, it follows that $f \upharpoonright C \in \overline{\pi_C(H_1)}$. By a similar argument, it can be shown that $f \upharpoonright C \in \overline{\pi_C(K_1)}$. This is a contradiction since $\overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing$. Therefore the claim that $\overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}=\varnothing$ is true, with the closure being taken in $\pi_B(C_p(X))$.

Because $B \subset Y$, observe that $\pi_B(H_1)=\pi_B(H)$ and $\pi_B(K_1)=\pi_B(K)$. Furthermore, $\pi_B(\pi_Y(C_p(X)))=\pi_B(C_p(X))$. Thus we can claim that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$, with the closure being taken in $\pi_B(\pi_Y(C_p(X)))$. By Lemma 2, $\pi_Y(C_p(X))$ is normal. $\blacksquare$

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Some Corollaries

Corollary 3

Let $X$ be a normal space. If $C_p(X)$ is normal, then $X$ has countable extent, i.e., every closed and discrete subset of $X$ is countable.

Proof of Corollary 3
Let $Y$ be a closed and discrete subset of $X$. We show that $Y$ must be countable. Since $Y$ is closed and $X$ is normal, $\pi_Y(C_p(X))=C_p(Y)$. By Theorem 1, $C_p(Y)$ is normal. Since $Y$ is discrete, $C_p(Y)=\mathbb{R}^Y$. If $Y$ is uncountable, $\mathbb{R}^Y$ is not normal. Thus $Y$ must be countable. $\blacksquare$

Corollary 4

Let $X$ be a metrizable space. If $C_p(X)$ is normal, then $X$ has a countable base.

Proof of Corollary 4
Note that in any metrizable space, the weight equals the extent. By Corollary 3, $X$ has countable extent and thus has countable base. $\blacksquare$

Corollary 5

Let $X$ be a normal space. If $C_p(X)$ is normal, then $X$ is collectionwise normal.

Proof of Corollary 5
Any normal space with countable extent is collectionwise normal. See Theorem 2 in this previous post. $\blacksquare$

Corollary 6

Let $X$ be a normal Moore space. If $C_p(X)$ is normal, then $X$ is metrizable.

Proof of Corollary 6
Suppose $C_p(X)$ is normal. By Theorem 1, $X$ has countable extent. By Corollary 5, $X$ is collectionwise normal. According to Bing’s metrization theorem, any collectionwise normal Moore space is metrizable (see [2] Theorem 5.4.1 in page 329). $\blacksquare$

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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.

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

# A Note On The Sorgenfrey Line

The Sorgenfrey Line is a topological space whose underlying space is the real line. The topology is generated by the basis of the half open intervals $[a, b)$ where $a$ and $b$ are real numbers. For students of topology, the Sorgenfrey Line is a handy example of (1) “Lindelof x Lindelof” does not have to be Lindelof, (2) “normal x normal” does not have to be normal, (3) “paracompact x paracompact” does not have to be paracompact, (4) “perfectly normal x perfectly normal” does not have to be perfectly normal (does not even have to be normal). In other words, these four properties are not preserved by taking Cartesian product. The goal of this note is to prove these and a few other results about the Sorgenfrey Line. In this note, $S$ is to denote the Sorgenfrey Line.

We will show these results:

A
$S$ is Lindelof (thus is normal and paracompact).
B
$S$ is hereditarily Lindelof.
C
Compact subsets of $S$ are countable. Thus $S$ is an example of a space that is Lindelof and not $\sigma$-compact.
D
$S \times S$ is not Lindelof.
E
$S \times S$ is not normal.
F
$S \times S$ is not paramcompact.
G
$S$ is not second countable, thus not metrizable.
H
$S$ is perfectly normal.

In proving C, we will use the following lemma (Lemma 1), which was proved in a previous post. This is a special countable extent property of the real line. Note that a space $X$ has extent of cardinality $\mathcal{K}$ if $\mathcal{K}$ is the least upper bound on the sizes of all closed and discrete subsets of $X$. In proving F, the Jones Lemma will be used. This lemma is essentially saying that the extent of a separable normal space cannot be the cardinality continuum or greater. A space $X$  is perfectly normal if $X$ is normal and every closed subset is a $G_{\delta}$ set (equivalently every open subset is an $F_{\sigma}$ set). These two lemmas are stated below.

Lemma 1
Every uncountable subset of $\mathbb{R}$ has a two-sided limit point.
Lemma 2 (Jones’ Lemma)
If $X$ is a separable normal space, then it has no closed and discrete subset of cardinality continuum.

See this post for a proof of Jones’ Lemma.

Proof of A. Let $\mathcal{A}$ be an open cover of $S$ consisting of open intervals of the form $[a, b)$.

Let $T=\cup\lbrace{(a, b): [a, b)\epsilon\mathcal{A}}\rbrace$. We claim that $U=S - T$ is a countable set. Suppose that $U$ is uncountable. By Lemma 1, there exists a real number $p$ that is a two-sided limit point of $U$. This means that for every open interval $(s,t)$ (open interval in the usual topology on the real line) with $p \in (s,t)$, the interval $(s,t)$ contains points of $U$ on the left side of $p$ as well as on the right side of $p$.

Choose some $[a,b) \in \mathcal{A}$ such that $p \in [a,b)$. Note that $(a,b)$ is a subset of $T$ and so should not contain points of $U=S-T$. Because $p$ is a two-sided limit point of $U$, $(a,b)$ will contain points of $U$, a contradiction. So $U$ must be countable.

Note that $T$ is an open set in the usual topology, which is Lindelof. So we can find countably many $[c,d) \in \mathcal{A}$ such that the union of all such $(c,d)$ covers $T$. Then find countably many $[c,d) \in \mathcal{A}$ that cover the countably many points in $U=S-T$. Combining both sets of $[c,d)$, we see that $\mathcal{A}$ has a countable subcover. $\blacksquare$

Proof of B. Take any uncountable subspace of $S$, we can apply the same proof as in A.

Proof of C. Let $A\subset{S}$ be a compact subspace that is uncoutable. By Lemma 1, $A$ has a two-sided limit point $y$ (i.e. it is both a left-sided limit point and a right-sided limit point in the usual topology). Let $\lbrace{y_n}\rbrace$ be a sequence of points in $A$ that converges to $y$ from the left. Then $\lbrace{(\infty,y_n)}\rbrace$  and $[y,\infty)$ form an open cover of $A$ that has no finite subcover. Thus all compact subspaces of the Sorgenfrey Line are countable. Furthermore $S$ is an example of a Lindelof but not $\sigma$-compact space.

Proof of D. If $S \times S$ is Lindelof, then it would be a separable normal space and cannot have closed and discrete subset of cardinality continuum or greater (according to Jones’ Lemma). Note that $D=\lbrace{(x,-x): x\epsilon{S}}\rbrace$ is a closed and discrete subset of $S \times S$, which has cardinality continuum. This shows that $S \times S$ cannot be Lindeloff. $\blacksquare$

Proof of E and F. For E, use the same argument as in the proof of D. For F, note that paracompactness implies normality. $\blacksquare$

Proof of G. If $S$ is second countable (has a countable base), then it would be a separable metrizable space and $S \times S$ would also be a separable metrizable space. But $S \times S$ is not even Lindeloff. Thus the Sorgenfrey Line cannot be second countable. $\blacksquare$

Proof of H. Let $W$ be an open subset of $S$. Let $O$ be the interor of $W$ in the usual topology. By a similar argument in the proof of A above, we can show that $W-O$ is countable. Since $O$ is an open set in the usual topology, it is an $F_{\sigma}$ set in the usual topology (and thus in the Sorgenfrey topology). It follows that $O$ plus countably many points would form an $F_{\sigma}$ set. Thus every open subset of the Sorgenfrey line $S$ is a $G_\delta$-set. Since the Sorgenfrey line is Lindelof, it is normal. Thus the Sorgenfrey line is a normal space in which every open set is a $G_\delta$-set (i.e. it is perfectly normal). $\blacksquare$

# A Countable Extent Property Unique to the Real Line

Being a separable metrizable space, the real line with the usual topology has countable extent. A space $X$ has countable extent if all closed and discrete subsets of $X$ are countable. In general, the extent of a space is the least cardinality of a closed and discrete set in that space. In this brief note, we show that the real line has another countable extent property that is unique to the real line. We show that every uncountable subset of the real line has a two-sided limit point. As an application, it follows from this fact that the Sorgenfrey line is Lindelof and that all compact subspaces in the Sorgenfrey Line are countable (see the blog post A Note On The Sorgenfrey Line).

Given a space $X$ and given $p \in {X}$, the point $p$ is a limit point of a subset $A\subset{X}$ if every open set containing $p$ contains a point of $A$ distinct from $p$. Equivalently, we can replace open set in this definition with members of a base (basic open sets). Thus, the limit here is from a topological perspective and not from a metric space perspective. A space $X$ has countable extent if all closed and discrete subsets are countable. This is equivalent to the statement that every uncountable subset has a limit point.

In the real line, we say that the point $p$ is a limit point of $A\subset{\mathbb{R}}$ if every open interval $(a, b)$ containing $p$ contains a point of $A$ distinct from $p$. We say $p$ is a two-sided limit point of $A\subset{\mathbb{R}}$ if every open interval $(a, b)$ containing $p$ contains points of $A$ on both sides of $p$ (i.e. both intervals $(a, p)$ and $(p, b)$ contain points of $A$). Likewise, one-sided (left-sided, right-sided) limit point means every open interval containing $p$ contains points of $A$ on one side (left side, right side, respectively).

Lemma
Every uncountable subset of $\mathbb{R}$ has a two-sided limit point.

Proof. Let $A\subset{\mathbb{R}}$ be uncountable. The set $A$ as a topological space is a Lindelof space. Thus $A$ has a limit point (in fact, has uncountably many limit points).

Suppose that $A$ has no two-sided limit points. Then the uncountably many limit points of $A$ must be either left-sided or right-sided limit points. We assume the case that $A$ has uncountably many right sided limit points. The proof for the left-sided case is similar. Let $B$ be the set of all right-sided limit points of $A$.

Since points in $B$ are right-sided limits of $A$, for each $x \in {B}$, there is a rational number $a_x$ such that $(a_x, x)\cap{A}=\phi$. Matching up the rational numbers with uncountably many points in $B$, there is a rational number $r$ such that the following set $C$ is uncountable.

$C=\lbrace{x \in {B}: r=a_x}\rbrace$

The uncountable set $C$ has a limit point $y$. Note that $y$ is a limit point of $A$. For the rational number $r$ indicated above, it must be the case $r < y$. Note that all points of $C$ are to the right of $r$. If $r \ge y$, then there would no points of $C$ in an open interval of $y$, which goes against the fact that $y$ is a limit point of $C$.

By a similar argument as in the preceding paragraph, there are no points of $C$ on the right side of $y$. If there is some $x \in {C}$ and $y < x$ then $(a_x, x)=(r, x)$ is an open interval containing the point $y$ that contains no points of $A$, contradicting the fact that $y$ is a limit point of $A$. Thus $y$ is only a left-sided limit point of $C$.

Choose two points $w$ and $z$ from $C$ such that $r < w < z < y$. We know that $(a_z, z)=(r, z)$ contains no points of $A$. It follows that $(w, z)$ contains no points of $A$. Since $w$ is a right-sided limit point of $A$, $(w, z)$ would contain points of $A$, a contradiction. Thus the set $A$ must have a two-sided limit point and the lemma is proved. $\blacksquare$