Bing’s Example G

Posted on November 18, 2012 by Dan Ma

Bing’s Example G is an example of a topological space that is normal but not collectionwise normal. It was introduced in an influential paper of R. H. Bing in 1951 (see [1]). This paper has a metrization theorem that is now called Bing’s metrization theorem (any regular space is metrizable if and only if it has a $\sigma$ -discrete base). The paper also introduced the notion of collectionwise normality and discussed the roles it plays in metrization theory (e.g. a Moore space is metrizable if and only if it is collectionwise normal). Example G was an influential example from an influential paper. It became the basis of construction for many other counterexamples (see [5] for one example). Investigations were also conducted by looking at various covering properties among subspaces of Example G (see [2] and [4] are two examples).

In this post we prove some basic results about Bing’s Example G. Some of the results we prove are found in Bing’s 1951 paper. The other results shown here are usually mentioned without proof in various places in the literature.

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Bing’s Example G – Definition

Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$ . Let $F=2^Q$ be the set of all functions $f: Q \rightarrow 2=\left\{0,1 \right\}$ . Another notation for $2^Q$ is the Cartesian product $\prod \limits_{q \in Q} \left\{0,1 \right\}$ . For each $p \in P$ , define the function $f_p: Q \rightarrow 2$ by the following:

$\forall q \in Q$

$f_p(q)=1$

$p \in q$

$f_p(q)=0$

$p \notin q$

Let $F_P=\left\{f_p: p \in P \right\}$ . Let $\tau$ be the set of all open subsets of $2^Q$ in the product topology. We now consider another topology on $2^Q$ generated by the following base:

$\mathcal{B}=\tau \cup \left\{\left\{x \right\}: x \in F-F_P \right\}$

Bing’s Example G is the set $F=2^Q$ with the topology generated by the base $\mathcal{B}$ . In other words, each $x \in F-F_P$ is made an isolated point and points in $F_P$ retain the usual product open sets.

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Bing’s Example G – Initial Discussion

Bing’s Example G, i.e. the space $F$ as defined above, is obtained by altering the topology of the product space of $2^{\lvert \mathcal{K} \lvert}$ many copies of the two-point discrete space where $\mathcal{K}$ is the cardinality of the power set of the uncountable index set $P$ we start with. Out of this product space, a set $F_P$ of points is carefully chosen such that $F_P$ has the same cardinality as $P$ and such that $F_P$ is relatively discrete in the product space. Points in $F_P$ are made to retain the product topology and all points outside of $F_P$ are declared as isolated points.

We now show that the set $F_P$ is a discrete set in the space $F$ . For each $p \in P$ , let $W_p$ be the open set defined by

$W_p=\left\{f \in F: f(\left\{p \right\})=1 \text{ and } f(P-\left\{p \right\})=0 \right\}$ .

It is clear that $f_p$ is the only point of $F_P$ belonging to $W_p$ . Therefore, in the Example G topology, the set $F_P$ is discrete and closed . In the section “Bing’s Example G is not Collectionwise Hausdorff” below, we show below that $F_P$ cannot be separated by any pairwise disjoint collection of open sets.

The character at a point is the minimum cardinality of a local base at that point. The character at a point in $F_P$ in the Example G topology agrees with the product topology. Points in $F_P$ have character $\lvert Q \lvert=2^{\lvert P \lvert}$ . Specifically if the starting $P$ has cardinality $\omega_1$ , then points in $F_P$ have character $2^{\omega_1}$ . Thus Example G has large character and cannot be a Moore space (any Moore space has a countable base at every point).

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Bing’s Example G is Normal

Let $H_1$ and $H_2$ be disjoint closed subsets of $F$ . The easy case is that one of $H_1$ and $H_2$ is a subset of $F-F_P$ , say $H_1 \subset F-F_P$ . Then $H_1$ is a closed and open set in $F$ . Then $H_1$ and $F-H_1$ are disjoint open sets containing $H_1$ and $H_2$ , respectively. So we can assume that both $H_1 \cap F_P \ne \varnothing$ and $H_2 \cap F_P \ne \varnothing$ .

Let $A_1=H_1 \cap F_P$ and $A_2=H_2 \cap F_P$ . Let $q_1=\left\{p \in P: f_p \in A_1 \right\}$ and $q_2=\left\{p \in P: f_p \in A_2 \right\}$ . Define the following open sets:

$U_1=\left\{f \in F: f(q_1)=1 \text{ and } f(q_2)=0 \right\}$

$U_2=\left\{f \in F: f(q_1)=0 \text{ and } f(q_2)=1 \right\}$

Because $H_1 \cap H_2=\varnothing$ , we have $A_1 \subset U_1$ and $A_2 \subset U_2$ . Furthermore, $U_1 \cap U_2=\varnothing$ . Let $B_1=H_1 \cap (F-F_P)$ and $B_2=H_2 \cap (F-F_P)$ , which are open since they consist of isolated points. Then $O_1=(U_1 \cup B_1)-H_2$ and $O_2=(U_2 \cup B_2)-H_1$ are disjoint open subsets of $F$ with $H_1 \subset O_1$ and $H_2 \subset O_2$ .

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Collectionwise Normal Spaces

Let $X$ be a space. Let $\mathcal{A}$ be a collection of subsets of $X$ . We say $\mathcal{A}$ is pairwise disjoint if $A \cap B=\varnothing$ whenever $A,B \in \mathcal{A}$ with $A \ne B$ . We say $\mathcal{A}$ is discrete if for each $x \in X$ , there is an open set $O$ containing $x$ such that $O$ intersects at most one set in $\mathcal{A}$ .

The space $X$ is said to be collectionwise normal if for every discrete collection $\mathcal{D}$ of closed subsets fo $X$ , there is a pairwise disjoint collection $\left\{U_D: D \in \mathcal{D} \right\}$ of open subsets of $X$ such that $D \subset U_D$ for each $D \in \mathcal{D}$ . Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [3]). Thus Bing’s Example G is not paracompact.

When discrete collection of closed sets in the definition of “collectionwise normal” is replaced by discrete collection of singleton sets, the space is said to be collectionwise Hausdorff. Clearly any collectionwise normal space is collectionwise Hausdorff. Bing’s Example is actually not collectionwise Hausdorff.

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Bing’s Example G is not Collectionwise Hausdorff

The discrete set $F_P$ cannot be separated by disjoint open sets. For each $p \in P$ , let $O_p$ be an open subset of $F$ such that $p \in O_p$ . We show that the open sets $O_p$ cannot be pairwise disjoint. For each $p \in P$ , choose an open set $L_p$ in the product topology of $2^Q$ such that $p \in L_p \subset O_p$ . The product space $2^Q$ is a product of separable spaces, hence has the countable chain condition (CCC). Thus the open sets $L_p$ cannot be pairwise disjoint. Thus $L_t \cap L_s \ne \varnothing$ and $O_t \cap O_s \ne \varnothing$ for at least two points $s,t \in P$ .

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Bing’s Example G is Completely Normal

The proof for showing Bing’s Example G is normal can be modified to show that it is completely normal. First some definitions. Let $X$ be a space. Let $A \subset X$ and $B \subset X$ . The sets $A$ and $B$ are separated sets if $A \cap \overline{B}=\varnothing=\overline{A} \cap B$ . Essentially, any two disjoint sets are separated sets if and only if none of them contains limit points (i.e. accumulation points) of the other set. A space $X$ is said to be completely normal if for every two separated sets $A$ and $B$ in $X$ , there exist disjoint open subsets $U$ and $V$ of $X$ such that $A \subset U$ and $B \subset V$ . Any two disjoint closed sets are separated sets. Thus any completely normal space is normal. It is well known that for any regular space $X$ , $X$ is completely normal if and only if $X$ is hereditarily normal. For more about completely normality, see [3] and [6].

Let $H_1 \subset F$ and $H_2 \subset F$ such that $H_1 \cap \overline{H_2}=\varnothing=\overline{H_1} \cap H_2$ . We consider two cases. One is that one of $H_1$ and $H_2$ is a subset of $F-F_P$ . The other is that both $H_1 \cap F_P \ne \varnothing$ and $H_2 \cap F_P \ne \varnothing$ .

The first case. Suppose $H_1 \subset F-F_P$ . Then $H_1$ consists of isolated points and is an open subset of $F$ . For each $x \in H_2 \cap F_P$ , choose an open subset $V_x$ of $F$ such that $x \in V_x$ and $V_x$ contains no points of $F_P-\left\{ x \right\}$ and $V_x \cap \overline{H_1}=\varnothing$ . For each $x \in H_2 \cap (F-F_P)$ , let $V_x=\left\{x \right\}$ . Let $V$ be the union of all $V_x$ where $x \in H_2$ . Let $U=H_1$ . Then $U$ and $V$ are disjoint open sets with $H_1 \subset U$ and $H_2 \subset V$ .

The second case. Suppose $A_1=H_1 \cap F_P \ne \varnothing$ and $A_2=H_2 \cap F_P \ne \varnothing$ . Let $q_1=\left\{p \in P: f_p \in A_1 \right\}$ and $q_2=\left\{p \in P: f_p \in A_2 \right\}$ . Define the following open sets:

$U_1=\left\{f \in F: f(q_1)=1 \text{ and } f(q_2)=0 \right\}$

$U_2=\left\{f \in F: f(q_1)=0 \text{ and } f(q_2)=1 \right\}$

Because $H_1 \cap H_2=\varnothing$ , we have $A_1 \subset U_1$ and $A_2 \subset U_2$ . Furthermore, $U_1 \cap U_2=\varnothing$ . Let $B_1=H_1 \cap (F-F_P)$ and $B_2=H_2 \cap (F-F_P)$ , which are open since they consist of isolated points. Then $O_1=(U_1 \cup B_1)-\overline{H_2}$ and $O_2=(U_2 \cup B_2)-\overline{H_1}$ are disjoint open subsets of $F$ with $H_1 \subset O_1$ and $H_2 \subset O_2$ .

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Bing’s Example G is not Perfectly Normal

A space is perfectly normal if it is normal and that every closed subset is $G_\delta$ (i.e. the intersection of countably many open subsets). The set $F_P$ of non-isolated points is a closed set in $F$ . We show that $F_P$ cannot be a $G_\delta$ -set. Before we do so, we need to appeal to a fact about the product space $2^Q$ .

According to the Tychonoff theorem, the product space $2^Q$ is a compact space since it is a product of compact spaces. On the other hand, $2^Q$ is a product of uncountably many factors and is thus not first countable. It is a well known fact that in a compact Hausdorff space, if a point is a $G_\delta$ -point, then there is a countable local base at that point (i.e. the space is first countable at that point). Thus no point of the compact product space $2^Q$ can be a $G_\delta$ -point. Since points of $F_P$ retain the open sets of the product topology, no point of $F_P$ can be a $G_\delta$ -point in the Bing’s Example G topology.

For each $p \in P$ , let $W_p$ be open in $F$ such that $f_p \in W_p$ and $W_p$ contains no points $F_P-\left\{f_p \right\}$ . For example, we can define $W_p$ as in the above section “Bing’s Example G – Initial Discussion”.

Suppose that $F_P$ is a $G_\delta$ -set. Then $F_P=\bigcap \limits_{i=1}^\infty O_i$ where each $O_i$ is an open subset of $F$ . Now for each $p \in P$ , we have $\left\{f_p \right\}=\bigcap \limits_{i=1}^\infty (O_i \cap W_p)$ , contradicting the fact that the point $f_p$ cannot be a $G_\delta$ -point in the space $F$ (and in the product space $2^Q$ ). Thus $F_P$ is not a $G_\delta$ -set in the space $F$ , leading to the conclusion that Bing’s Example G is not perfectly normal.

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Bing’s Example G is not Metacompact

A space $M$ is said to have caliber $\omega_1$ if for every uncountable collection $\left\{U_\alpha: \alpha < \omega_1 \right\}$ of non-empty open subsets of $M$ , there is an uncountable $A \subset \omega_1$ such that $\bigcap \left\{U_\alpha: \alpha \in A \right\} \ne \varnothing$ . Any product of separable spaces has this property (see Topological Spaces with Caliber Omega 1). Thus the product space $2^Q$ has caliber $\omega_1$ . Thus in the product space $2^Q$ , no collection of uncountably many non-empty open sets can be a point-finite collection (in fact cannot even be point-countable).

To see that the Example G is not metacompact, let $\mathcal{W}=\left\{W_p: p \in P \right\}$ be a collection of open sets such that for $p \in P$ , $f_p \in W_p$ , $W_p$ is open in the product topology of $2^Q$ and $W_p$ contains no points $F_P-\left\{f_p \right\}$ . For example, we can define $W_p$ as in the above section “Bing’s Example G – Initial Discussion”.

Let $W=\bigcup \mathcal{W}$ . Let $\mathcal{V}=\mathcal{W} \cup \left\{\left\{ x \right\}: x \in F-W \right\}$ . Any open refinement of $\mathcal{V}$ would contain uncountably many open sets in the product topology and thus cannot be point-finite. Thus the space $F$ cannot be metacompact.

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Reference

Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
Burke, D. K., A note on R. H. Bing’s example G, Top. Conf. VPI, Lectures Notes in Mathematics, 375, Springer Verlag, New York, 47-52, 1974.
Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
Lewis, I. W., On covering properties of subspaces of R. H. Bing’s Example G, Gen. Topology Appl., 7, 109-122, 1977.
Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

Topological Spaces with Caliber Omega 1

Posted on November 17, 2012 by Dan Ma

Let $\omega_1$ be the first uncountable ordinal. A separable space is one that has a countable dense subset. Any separable space $X$ satisfies this property:

$\left\{U_\alpha: \alpha < \omega_1 \right\}$

$X$

$A \subset \omega_1$

$\bigcap \limits_{\alpha \in A} U_\alpha \ne \varnothing$

Any space that has this property is said to have caliber $\omega_1$ . Spaces that have caliber $\omega_1$ have the countable chain condition (abbreviated by CCC, which means that there is no uncountable pairwise disjoint collection of open subsets of the space). So we have the following implications:

$\text{separable} \Longrightarrow \text{caliber } \omega_1 \Longrightarrow \text{CCC}$

The chain condition “caliber $\omega_1$ ” is an interesting one. Some of the spaces that have the CCC actually have caliber $\omega_1$ . For example, separable spaces and products of separable spaces have caliber $\omega_1$ . Thus the product space $\left\{0,1 \right\}^{\mathcal{K}}$ (for any uncountable cardinal $\mathcal{K}$ ) not only has the countable chain condition. It has the stronger property of having caliber $\omega_1$ , which may make certain proof easier to do. In this post we also provide examples to show that none of the above implications is reversible.

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Delta System Lemma

In proving that product of separable spaces has caliber $\omega_1$ , the Delta system lemma is used. A collection $\mathcal{D}$ of sets is said to be a Delta-system (or $\Delta$ -system) if there is a set $D$ such that for every $A,B \in \mathcal{D}$ with $A \ne B$ , we have $A \cap B = D$ . When such set $D$ exists, it is called the root of the Delta-system $\mathcal{D}$ . The following is the statement of Delta-system lemma.

Lemma 1 – Delta-System Lemma
For every uncountable collection $\mathcal{A}$ of finite sets, there is an uncountable $\mathcal{D} \subset \mathcal{A}$ such that $\mathcal{D}$ is a $\Delta$ -system.

The statement of Delta-system lemma presented here is a special case for a general version (see [2], page 49).

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Products of Spaces with Caliber $\omega_1$

Theorem 2 below shows that whenever “caliber $\omega_1$ ” is preserved by taking product with any finite number of factors, “caliber $\omega_1$ ” is preserved by taking the product of any number of factors. As a corollary, we have the result that the product of any number of separable spaces has caliber $\omega_1$ . The proof of Theorem 2 is similar to the one stating that whenever CCC is preserved by taking product with any finite number of factors, CCC is preserved by taking the product of any number of factors (see the previous post Product of Spaces with Countable Chain Condition).

Theorem 2
Suppose that $\left\{X_\alpha: \alpha \in T \right\}$ is a family of spaces such that $\prod \limits_{\alpha \in F} X_\alpha$ has caliber $\omega_1$ for every finite $F \subset T$ . Then $\prod \limits_{\alpha \in T} X_\alpha$ has caliber $\omega_1$ .

Proof
In proving the product space $\prod \limits_{\alpha \in T} X_\alpha$ having caliber $\omega_1$ , it suffices to work with basic open sets of the form $\prod \limits_{\alpha \in T} O_\alpha$ where $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in T$ . Let $\mathcal{U}=\left\{U_\beta: \beta < \omega_1 \right\}$ be an uncountable collection of such non-empty open sets. Our plan is to find an uncountable $W_0 \subset \omega_1$ such that $\bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing$ .

$\text{ }$

For each $U_\beta=\prod \limits_{\alpha \in T} U_{\beta,\alpha}$ , let $F_\beta \subset T$ be the finite set such that $\alpha \in F_\beta$ if and only if $U_{\beta, \alpha} \ne X_\alpha$ . Consider $\mathcal{A}=\left\{F_\beta: \beta < \omega_1 \right\}$ . By the Delta-system lemma, there is an uncountable $W \subset \omega_1$ such that $\mathcal{F}=\left\{F_\beta: \beta \in W \right\}$ is a Delta-system. Let $F$ be the root of this Delta-system.

$\text{ }$

Consider the case that the root of the Delta-system is empty. Then for any $F_{\beta_1} \in \mathcal{F}$ and $F_{\beta_2} \in \mathcal{F}$ where $F_{\beta_1} \ne F_{\beta_2}$ , we have $F_{\beta_1} \cap F_{\beta_2}=\varnothing$ . For each $\beta \in W$ , we have $U_\beta=\prod \limits_{\alpha \in T} U_{\beta,\alpha}$ and we choose $h_\beta$ in $\prod \limits_{\alpha \in F_\beta} U_{\beta,\alpha}$ . Then we can define an $h$ in $\prod \limits_{\alpha \in T} X_\alpha$ such that $h$ extends $h_\beta$ for all $\beta \in W$ . Then in this case let $W_0=W$ and we have $\bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing$ . So we move onto the case that $F \ne \varnothing$ .

$\text{ }$

Now assume $F \ne \varnothing$ . Let $\mathcal{U}^*=\left\{U_\beta: \beta \in W \right\}$ . For each $U_\beta \in \mathcal{U}^*$ where $U_\beta=\prod \limits_{\alpha \in T} U_{\beta,\alpha}$ , let $p(U_\beta)$ be $\prod \limits_{\alpha \in F} U_{\beta,\alpha}$ (i.e. $p$ is a projection map). Let $\mathcal{U}^{**}=\left\{p(U_\beta): \beta \in W \right\}$ .

$\text{ }$

Consider two cases. One is that $\mathcal{U}^{**}$ is countable. The second is that $\mathcal{U}^{**}$ is uncountable. Suppose $\mathcal{U}^{**}$ is countable. Then there is an uncountable $W_0 \subset W$ such that $p(U_\gamma)=p(U_\mu)$ for all $\gamma,\mu \in W_0$ . Then fix $\gamma \in W_0$ and choose $g^* \in p(U_\gamma)$ . Then $g^*$ is extendable to some $g$ in $\prod \limits_{\alpha \in T} X_\alpha$ such that $g \in U_\beta$ for all $\beta \in W_0$ . Thus we have $\bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing$ .

$\text{ }$

Now assume $\mathcal{U}^{**}$ is uncountable. By assumption $\prod \limits_{\alpha \in F} X_\alpha$ has caliber $\omega_1$ . Then there is an uncountable $W_0 \subset W$ such that $\bigcap \limits_{\beta \in W_0} p(U_\beta) \ne \varnothing$ . Choose $g^* \in \bigcap \limits_{\beta \in W_0} p(U_\beta)$ . As in the previous case, $g^*$ is extendable to some $g$ in $\prod \limits_{\alpha \in T} X_\alpha$ such that $g \in U_\beta$ for all $\beta \in W_0$ . Thus we have $\bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing$ . $\blacksquare$

Corollary 3
Suppose that $\left\{X_\alpha: \alpha \in T \right\}$ is a family of separable spaces. Then $\prod \limits_{\alpha \in T} X_\alpha$ has caliber $\omega_1$ .

Proof
This follows directly from Theorem 2. Note that the product of finitely many separable is separable (hence has caliber $\omega_1$ ). $\blacksquare$

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Examples

We now show that the following implications are not reversible.

$\text{ }$

$\text{separable} \Longrightarrow \text{caliber } \omega_1 \Longrightarrow \text{CCC}$

$\text{ }$

To get a space with caliber $\omega_1$ that is not separable, consider the product of $\mathcal{K}$ many copies of $\left\{0,1 \right\}$ where $\mathcal{K}$ is a cardinal number greater than continuum. Since $\left\{0,1 \right\}^{\mathcal{K}}$ is a product of separable spaces, it has caliber $\omega_1$ according to Corollary 3. It is well known that the product of more than continuum many separable spaces is not separable (see Product of Separable Spaces).

To get a space with the CCC that does not have caliber $\omega_1$ . Consider the subspace $S$ of the product space $H=\left\{0,1 \right\}^{\omega_1}$ ( $\omega_1$ many copies of $\left\{0,1 \right\}$ ) where $S$ is the set of all $h \in H$ such that $h(\alpha) \ne 0$ for at most countably many $\alpha < \omega_1$ . Note that $H$ is a space with the CCC since it is a product of separable spaces. Furthermore $S$ is a dense subspace of $H$ . The property of CCC is hereditary with respect to dense subsets. Thus $S$ has the countable chain condition. Here's a discussion of the countable chain condition.

To see that $S$ does not have caliber $\omega_1$ , look at the collection of open sets $\left\{V_\alpha: \alpha < \omega_1 \right\}$ where each $V_\alpha=\left\{h \in S: h(\alpha)=1 \right\}$ . Note that each $h \in S$ belongs to at most countably many $V_\alpha$ . Thus for any uncountable $A \subset \omega_1$ , $\bigcap \limits_{\alpha \in A} V_\alpha = \varnothing$ .

The example $S$ shows that dense subspace of a space with caliber $\omega_1$ need not have caliber $\omega_1$ .

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Reference

Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

Cartesian Products of Two Paracompact Spaces – Continued

Posted on November 11, 2012 by Dan Ma

Consider the real line $\mathbb{R}$ with a topology finer than the usual topology obtained by isolating each point in $\mathbb{P}$ where $\mathbb{P}$ is the set of all irrational numbers. The real line with this finer topology is called the Michael line and we use $\mathbb{M}$ to denote this topological space. It is a classic result that $\mathbb{M} \times \mathbb{P}$ is not normal (see “Michael Line Basics”). Even though the Michael line $\mathbb{M}$ is paracompact (it is in fact hereditarily paracompact), $\mathbb{M}$ is not perfectly normal. Result 3 below will imply that the Michael line cannot be perfectly normal. Otherwise $\mathbb{M} \times \mathbb{P}$ would be paracompact (hence normal). Result 3 is the statement that if $X$ is paracompact and perfectly normal and Y is a metric space then $X \times Y$ is paracompact and perfectly normal. We also use this result to show that if $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof (see Result 4 below).

This post is a continuation of the post “Cartesian Products of Two Paracompact Spaces”. In that post, four results are listed. They are:

Result 1

$X$

$Y$

$X \times Y$

Result 2

$X$

$Y$

$\sigma$

$X \times Y$

Result 3

$X$

$Y$

$X \times Y$

Result 4

$X$

$Y$

$X \times Y$

Result 1 and Result 2 are proved in the previous post “Cartesian Products of Two Paracompact Spaces”. Result 3 and Result 4 are proved in this post. All spaces are assumed to be regular.

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Paracompact Spaces, Lindelof Spaces and Other Information

A paracompact space is one in which every open cover has a locally finite open refinement. The previous post “Cartesian Products of Two Paracompact Spaces” has a basic discussion on paracompact spaces. For the sake of completeness, we repeat here some of the results discussed in that post. A proof of Proposition 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [2] (Theorem 20.7 in page 146).. For a proof of Proposition 2, see Theorem 3 in the previous post “Cartesian Products of Two Paracompact Spaces”. We provide a proof for Proposition 3.

Proposition 1
Let $X$ be a regular space. Then $X$ is paracompact if and only if every open cover $\mathcal{U}$ of $X$ has a $\sigma$ -locally finite open refinement.

Proposition 2
Every $F_\sigma$ -subset of a paracompact space is paracompact.

Proposition 3
Any paracompact space with a dense Lindelof subspace is Lindelof.

Proof of Proposition 3
Let $L$ be a paracompact space. Let $M \subset L$ be a dense Lindelof subspace. Let $\mathcal{U}$ be an open cover of $L$ . Since we are working with a regular space, let $\mathcal{V}$ be an open cover of $L$ such that $\left\{\overline{V}: V \in \mathcal{V} \right\}$ refines $\mathcal{U}$ . Let $\mathcal{W}$ be a locally finite open refinement of $\mathcal{V}$ . Choose $\left\{W_1,W_2,W_3,\cdots \right\} \subset \mathcal{W}$ such that it is a cover of $M$ . Since $M \subset \bigcup \limits_{i=1}^\infty W_i$ , $\overline{\bigcup \limits_{i=1}^\infty W_i}=L$ .

Since the sets $W_i$ come from a locally finite collection, they are closure preserving. Hence we have:

$\overline{\bigcup \limits_{i=1}^\infty W_i}=\bigcup \limits_{i=1}^\infty \overline{W_i}=L$

For each $i$ , choose some $U_i \in \mathcal{U}$ such that $\overline{W_i} \subset U_i$ . Then $\left\{U_1,U_2,U_3,\cdots \right\}$ is a countable subcollection of $\mathcal{U}$ covering the space $L$ . $\blacksquare$

A space is said to be a perfectly normal if it is a normal space with the additional property that every closed subset is a $G_\delta$ -set in the space (equivalently every open subset is an $F_\sigma$ -set). We need two basic results about hereditarily Lindelof spaces. A space is Lindelof if every open cover of that space has a countable subcover. A space is hereditarily Lindelof if every subspace of that space is Lindelof. Proposition 4 below, stated without proof, shows that to prove a space is hereditarily Lindelof, we only need to show that every open subspace is Lindelof.

Proposition 4
Let $L$ be a space. Then $L$ is hereditarily Lindelof if and only if every open subspace of $L$ is Lindelof.

Proposition 5
Let $L$ be a Lindelof space. Then $L$ is hereditarily Lindelof if and only if $L$ is perfectly normal.

Proof of Proposition 5
$\Rightarrow$ Suppose $L$ is hereditarily Lindelof. It is well known that regular Lindelof space is normal. Thus $L$ is normal. It remains to show that every open subset of $L$ is $F_\sigma$ . Let $U \subset L$ be an non-empty open set. For each $x \in U$ , let $V_x$ be open such that $x \in V_x$ and $\overline{V_x} \subset U$ (the space is assumed to be regular). By assumption, the open set $U$ is Lindelof. The open sets $V_x$ form an open cover of $U$ . Thus $U$ is the union of countably many $\overline{V}_x$ .

$\Leftarrow$ Suppose $L$ is perfectly normal. To show that $L$ is hereditarily Lindelof, it suffices to show that every open subset of $L$ is Lindelof (by Proposition 4). Let $U \subset L$ be non-empty open. By assumption, $U=\bigcup \limits_{i=1}^\infty F_i$ where each $F_i$ is a closed set in $L$ . Since the Lindelof property is hereditary with respect to closed subsets, $U$ is Lindelof. $\blacksquare$

Another important piece of information that we need is the following metrization theorem. It shows that being a metrizable space is equivalent to have a base that is $\sigma$ -locally finite. In proving Result 3, we will assume that the metric factor has such a base. This is a classic metrization theorem (see [1] or [2] or any other standard topology text).

Theorem 6
Let $X$ be a space. Then $X$ is metrizable if and only if $X$ has a $\sigma$ -locally finite base.

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Result 3

Result 3 is the statement that:

$X$

$X \times Y$

Result 3 follows from the following two lemmas.

Lemma 7
If the following two conditions hold:

every open subset of $X$ is an $F_\sigma$ -set in $X$ ,
$Y$ is a metric space,

then every open subset of $X \times Y$ is an $F_\sigma$ -set in $X \times Y$ .

Proof of Lemma 7
Let $U$ be a open subset of $X \times Y$ . If $U=\varnothing$ , then $U$ is certainly the union of countably many closed sets. So assume $U \ne \varnothing$ . Let $\mathcal{B}=\bigcup \limits_{i=1}^\infty \mathcal{B}_i$ be a base for $Y$ such that each $\mathcal{B}_i$ is locally finite in $Y$ (by Theorem 6, such a base exists since $Y$ is metrizable).

Consider all non-empty $B \in \mathcal{B}$ such that we can choose nonempty open set $W_B \subset X$ with $W_B \times \overline{B} \subset U$ . Since $U$ is non-empty open, such pairs $(B, W_B)$ exist. Let $\mathcal{B}^*$ be the collection of all non-empty $B \in \mathcal{B}$ for which there is a matching non-empty $W_B$ . For each $i$ , let $\mathcal{B}_i^*=\mathcal{B}^* \cap \mathcal{B}_i$ . Of course, each $\mathcal{B}_i^*$ is still locally finite.

Since every open subset of $X$ is an $F_\sigma$ -set in $X$ , for each $W_B$ , we can write $W_B$ as

$W_B=\bigcup \limits_{j=1}^\infty W_{B,j}$

where each $W_{B,i}$ is closed in $X$ .

For each $i=1,2,3,\cdots$ and each $j=1,2,3,\cdots$ , consider the following collection:

$\mathcal{V}_{i,j}=\left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}$

Each element of $\mathcal{V}_{i,j}$ is a closed set in $X \times Y$ . Since $\mathcal{B}_i^*$ is a locally finite collection in $Y$ , $\mathcal{V}_{i,j}$ is a locally finite collection in $X \times Y$ . Define $V_{i,j}=\bigcup \mathcal{V}_{i,j}$ . The set $V_{i,j}$ is a union of closed sets. In general, the union of closed sets needs not be closed. However, $V_{i,j}$ is still a closed set in $X \times Y$ since $\mathcal{V}_{i,j}$ is a locally finite collection of closed sets. This is because a locally finite collection of sets is closure preserving. Note the following:

$\overline{V_{i,j}}=\overline{\bigcup \mathcal{V}_{i,j}}=\overline{\bigcup \left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}}=\bigcup \left\{\overline{W_{B,j} \times \overline{B}}: B \in \mathcal{B}_i^* \right\}$

$=\bigcup \left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}=V_{i,j}$

Finally, we have $U=\bigcup \limits_{i=1}^\infty \bigcup \limits_{j=1}^\infty V_{i,j}$ , which is the union of countably many closed sets. $\blacksquare$

Lemma 8
If $X$ is a paracompact space satisfying the following two conditions:

every open subset of $X$ is an $F_\sigma$ -set in $X$ ,
$Y$ is a metric space,

then $X \times Y$ is paracompact.

Proof of Lemma 8
As in the proof of the above lemma, let $\mathcal{B}=\bigcup \limits_{i=1}^\infty \mathcal{B}_i$ be a base for $Y$ such that each $\mathcal{B}_i$ is locally finite in $Y$ . Let $\mathcal{U}$ be an open cover of $X \times Y$ . Assume that elements of $\mathcal{U}$ are of the form $A \times B$ where $A$ is open in $X$ and $B \in \mathcal{B}$ .

For each $B \in \mathcal{B}$ , consider the following two items:

$\mathcal{W}_B=\left\{A: A \times B \in \mathcal{U} \right\}$

$W_B=\bigcup \mathcal{W}_B$

To simplify matter, we only consider $B \in \mathcal{B}$ such that $\mathcal{W}_B \ne \varnothing$ . Each $W_B$ is open in $X$ and hence by assumption an $F_\sigma$ -set in $X$ . Thus by Proposition 2, each $W_B$ is paracompact. Note that $\mathcal{W}_B$ is an open cover of $W_B$ . Let $\mathcal{H}_B$ be a locally finite open refinement of $\mathcal{W}_B$ . Consider the following two items:

$j=1,2,3,\cdots$

$\mathcal{V}_j=\left\{A \times B: A \in \mathcal{H}_B \text{ and } B \in \mathcal{B}_j \right\}$

$\mathcal{V}=\bigcup \limits_{j=1}^\infty \mathcal{V}_j$

We observe that $\mathcal{V}$ is an open cover of $X \times Y$ and that $\mathcal{V}$ refines $\mathcal{U}$ . Furthermore each $\mathcal{V}_j$ is a locally finite collection. The open cover $\mathcal{U}$ we start with has a $\sigma$ -locally finite open refinement. Thus $X \times Y$ is paracompact. $\blacksquare$

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Result 4

Result 4 is the statement that:

$X$

$Y$

$X \times Y$

Proof of Result 4
Suppose $X$ is hereditarily Lindelof and that $Y$ is a separable metric space. It is well known that regular Lindelof spaces are paracompact. Thus $X$ is paracompact. By Proposition 5, $X$ is perfectly normal. By Result 3, $X \times Y$ is paracompact and perfectly normal.

Let $D$ be a countable dense subset of $Y$ . We can think of $D$ as a $\sigma$ -compact space. The product of any Lindelof space with a $\sigma$ -compact space is Lindelof (see Corollary 3 in the post “The Tube Lemma”). Thus $X \times D$ is Lindelof. Furthermore $X \times D$ is a dense Lindelof subspace of $X \times Y$ . By Proposition 3, $X \times Y$ is Lindelof. By Proposition 5, $X \times Y$ is hereditarily Lindelof. $\blacksquare$

Remark
In the previous post “Bernstein Sets and the Michael Line”, a non-normal product space where one factor is Lindelof and the other factor is a separable metric space is presented. That Lindelof space is not hereditarily Lindelof (it has uncountably many isolated points). Note that by Result 4, for any such non-normal product space, the Lindelof factor cannot be hereditarily Lindelof.

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Reference

Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

Cartesian Products of Two Paracompact Spaces

Posted on November 8, 2012 by Dan Ma

In some previous posts we discuss examples surrounding the Michael line showing that the product of a paracompact space and a complete metric space needs not be normal (see “Michael Line Basics”) and that the product of a Lindelof space and a separable metric space need not be normal (see “Bernstein Sets and the Michael Line”). These examples are classic counterexamples demonstrating that both paracompactness and Lindelofness are not preserved by taking two-factor cartesian products even when one of the factors is nice (complete metric space in the first example and separable metric space in the second example). We now show some positive results. Of course, these results require additional conditions on one or both of the factors. We prove the following results.

Result 1

$X$

$Y$

$X \times Y$

Result 2

$X$

$Y$

$\sigma$

$X \times Y$

Result 3

$X$

$Y$

$X \times Y$

Result 4

$X$

$Y$

$X \times Y$

With Results 1 and 2, compact spaces and $\sigma$ -compact spaces can be called productively paracompact since the product of each of these spaces with any paracompact space is paracompact. We prove Result 1 and Result 2 below.

Result 3 and Result 4 are proved in another post Cartesian Products of Two Paracompact Spaces – Continued.

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Paracompact Spaces

First, recall some definitions. All spaces are at least regular (to us regular implies Hausdorff). Let $X$ be a space. A collection $\mathcal{A}$ of subsets of $X$ is said to be a cover of $X$ if $X=\bigcup \mathcal{A}$ (in words every point of the space belongs to one set in the collection). Furthermore, $\mathcal{A}$ is an open cover of $X$ is it is a cover of $X$ consisting of open subsets of $X$ .

Let $\mathcal{A}$ and $\mathcal{B}$ be covers of the space $X$ . The cover $\mathcal{B}$ is said to be a refinement of $\mathcal{A}$ ( $\mathcal{B}$ is said to refine $\mathcal{A}$ ) if for every $B \in \mathcal{B}$ , there is some $A \in \mathcal{A}$ such that $B \subset A$ . The cover $\mathcal{B}$ is said to be an open refinement of $\mathcal{A}$ if $\mathcal{B}$ refines $\mathcal{A}$ and $\mathcal{B}$ is an open cover.

A collection $\mathcal{A}$ of subsets of $X$ is said to be a locally finite collection if for each point $x \in X$ , there is a non-empty open subset $V$ of $X$ such that $x \in V$ and $V$ has non-empty intersection with at most finitely many sets in $\mathcal{A}$ . An open cover $\mathcal{A}$ of $X$ is said to have a locally finite open refinement if there exists an open cover $\mathcal{C}$ of $X$ such that $\mathcal{C}$ refines $\mathcal{A}$ and $\mathcal{C}$ is a locally finite collection. We have the following definition.

Definition

$X$

A collection $\mathcal{U}$ of subsets of the space $X$ is said to be a $\sigma$ -locally finite collection if $\mathcal{U}=\bigcup \limits_{i=1}^\infty \mathcal{U}_i$ such that each $\mathcal{U}_i$ is a locally finite collection of subsets of $X$ . Consider the property that every open cover of $X$ has a $\sigma$ -locally finite open refinement. This on the surface is a stronger property than paracompactness. However, Theorem 1 below shows that it is actually equivalent to paracompactness. The proof of Theorem 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [2] (Theorem 20.7 in page 146).

Theorem 1
Let $X$ be a regular space. Then $X$ is paracompact if and only if every open cover $\mathcal{U}$ of $X$ has a $\sigma$ -locally finite open refinement.

Theorem 2 below is another characterization of paracompactness that is useful. For a proof of Theorem 2, see “Finite and Countable Products of the Michael Line”.

Theorem 2
Let $X$ be a regular space. Then $X$ is paracompact if and only if the following holds:

$\left\{U_t: t \in T \right\}$

$X$

$\left\{V_t: t \in T \right\}$

$\overline{V_t} \subset U_t$

$t \in T$

Theorem 3 below shows that paracompactness is hereditary with respect to $F_\sigma$ -subsets.

Theorem 3
Every $F_\sigma$ -subset of a paracompact space is paracompact.

Proof of Theorem 3
Let $X$ be paracompact. Let $Y \subset X$ such that $Y=\bigcup \limits_{i=1}^\infty Y_i$ where each $Y_i$ is a closed subset of $X$ . Let $\mathcal{U}$ be an open cover of $Y$ . For each $U \in \mathcal{U}$ , let $U^*$ be open in $X$ such that $U^* \cap Y=U$ .

For each $i$ , let $\mathcal{U}_i^*$ be the set of all $U^*$ such that $U \cap Y_i \ne \varnothing$ . Let $\mathcal{V}_i^*$ be a locally finite refinement of $\mathcal{U}_i^* \cup \left\{X-Y_i \right\}$ . Let $\mathcal{V}_i$ be the following:

$\mathcal{V}_i=\left\{V \cap Y: V \in \mathcal{V}_i^* \text{ and } V \cap Y_i \ne \varnothing \right\}$

It is clear that each $\mathcal{V}_i$ is a locally finite collection of open set in $Y$ covering $Y_i$ . All the $\mathcal{V}_i$ together form a refinement of $\mathcal{U}$ . Thus $\mathcal{V}=\bigcup \limits_{i=1}^\infty \mathcal{V}_i$ is a $\sigma$ -locally finite open refinement of $\mathcal{U}$ . By Theorem 1, the $F_\sigma$ -set $Y$ is paracompact. $\blacksquare$
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Result 1

Result 1 is the statement that:

$X$

$Y$

$X \times Y$

To prove Result 1, we use the Tube lemma (for a proof, see “The Tube Lemma”).

The Tube Lemma
Let $X$ be any space and $Y$ be compact. For each $x \in X$ and for each open set $U \subset X \times Y$ such that $\left\{x \right\} \times Y \subset U$ , there is an open set $O \subset X$ such that $\left\{x \right\} \times Y \subset O \times Y \subset U$ .

Proof of Result 1
Let $\mathcal{U}$ be an open cover of $X \times Y$ . For each $x \in X$ , choose a finite $\mathcal{U}_x \subset \mathcal{U}$ such that $\mathcal{U}_x$ is a cover of $\left\{x \right\} \times Y$ . By the Tube Lemma, for each $x \in X$ , there is an open set $O_x \subset X$ such that $\left\{x \right\} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x$ . Since $X$ is paracompact, by Theorem 2, let $\left\{W_x: x \in X \right\}$ be a locally finite open refinement of $\left\{O_x: x \in X \right\}$ such that $W_x \subset O_x$ for each $x \in X$ .

Let $\mathcal{W}=\left\{(W_x \times Y) \cap U: x \in X, U \in \mathcal{U}_x \right\}$ . We claim that $\mathcal{W}$ is a locally finite open refinement of $\mathcal{U}$ . First, this is an open cover of $X \times Y$ . To see this, let $(a,b) \in X \times Y$ . Then $a \in W_x$ for some $x \in X$ . Furthermore, $a \in O_x$ and $(a,b) \in \cup \mathcal{U}_x$ . Thus, $(a,b) \in (W_x \times Y) \cap U$ for some $U \in \mathcal{U}_x$ . Secondly, it is clear that $\mathcal{W}$ is a refinement of the original cover $\mathcal{U}$ .

It remains to show that $\mathcal{W}$ is locally finite. To see this, let $(a,b) \in X \times Y$ . Then there is an open $V$ in $X$ such that $x \in V$ and $V$ can meets only finitely many $W_x$ . Then $V \times Y$ can meet only finitely many sets in $\mathcal{W}$ . $\blacksquare$

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Result 2

Result 2 is the statement that:

$X$

$Y$

$\sigma$

$X \times Y$

Proof of Result 2
Note that the $\sigma$ -compact space $Y$ is Lindelof. Since regular Lindelof are normal, $Y$ is normal and is thus completely regular. So we can embed $Y$ into a compact space $K$ . For example, we can let $K=\beta Y$ , which is the Stone-Cech compactification of $Y$ (see “Embedding Completely Regular Spaces into a Cube”). For our purpose here, any compact space containing $Y$ will do. By Result 1, $X \times K$ is paracompact. Note that $X \times Y$ can be regarded as a subspace of $X \times K$ .

Let $Y=\bigcup \limits_{i=1}^\infty Y_i$ where each $Y_i$ is compact in $Y$ . Note that $X \times Y=\bigcup \limits_{i=1}^\infty X \times Y_i$ and each $X \times Y_i$ is a closed subset of $X \times K$ . Thus the product $X \times Y$ is an $F_\sigma$ -subset of $X \times K$ . According to Theorem 3, $F_\sigma$ -subsets of any paracompact space is paracompact space. Thus $X \times Y$ is paracompact. $\blacksquare$

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Reference

Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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The Michael Line and the Continuum Hypothesis

Posted on November 4, 2012 by Dan Ma

There exist a Lindelof space and a separable metric space such that their Cartesian product is not normal (discussed in the post “Bernstein Sets and the Michael Line”). The separable metric space is a Bernstein set, a subspace of the real line that is far from being a complete metric space. However, this example is constructed without using any additional set theory axiom beyond the Zermelo-Fraenkel axioms plus the axiom of choice (abbreviated ZFC). A natural question is whether there exists a Lindelof space and a complete metric space such that their product is not normal. In particular, does there exist a Lindelof space $L$ such that the product of $L$ with the space of all irrational numbers is not normal? As of the writing of this post, it is still unknown that such a Lindelof space can exist in just ZFC alone without applying additional set theory axiom. However, such a Lindelof space can be constructed from various additional axioms (e.g. continuum hypothesis or Martin’s axiom). In this post, we present an example of such construction using the continuum hypothesis (the statement that the cardinality of the real line is the same as the first uncountable cardinal $\aleph_1$ ).

Let $\mathbb{M}$ be the Michael line. Let $\mathbb{P}$ be the set of irrational numbers with the usual topology inherited from the real line. It is a classical result that the product $\mathbb{M} \times \mathbb{P}$ is not normal (see “Michael Line Basics”). The Lindelof example we wish to discuss is an uncountable Lindelof subspace $L$ of $\mathbb{M}$ such that $L$ contains the set $\mathbb{Q}$ of rational numbers. The same proof that $\mathbb{M} \times \mathbb{P}$ is not normal will show that $L \times \mathbb{P}$ is not normal.

See the following posts for a basic discussion of the Michael line:

“Michael Line Basics”

“Finite and Countable Products of the Michael Line”

“Bernstein Sets and the Michael Line”

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Luzin Sets

The Lindelof space $X$ we want to find is a subset of the real line that is called a Luzin set. Before defining Luzin sets, recall some definitions. Let $Y$ be a space. Let $A \subset Y$ . The set $A$ is said to be nowhere dense in $Y$ if for every non-empty open subset $U$ of $Y$ , there is a non-empty open subset $V$ of $Y$ such that $V \subset U$ and $V$ misses $A$ (equivalently, the closure of $A$ has no interior). The set $A$ is of first category in $Y$ if it is the union of countably many nowhere dense sets.

To define Luzin sets, we focus on the Euclidean space $\mathbb{R}$ . Let $A \subset \mathbb{R}$ . The set $A$ is said to be a Luzin set if for every set $W \subset \mathbb{R}$ that is of first category in the real line, $A \cap W$ is at most countable. The Russian mathematician Luzin in 1914 constructed such an uncountable Luzin set using continuum hypothesis (CH). A good reference for Luzin sets is [4]. We have the following theorem.

Theorem 1
Assume CH. There exists an uncountable Luzin set.

Proof of Theorem 1
There are continuum many closed nowhere dense subsets of the real line. Since we assume the continuum hypothesis, we can enumerate these sets in a sequence of length $\omega_1$ . Let $\left\{F_\alpha: \alpha < \omega_1 \right\}$ be the set of all closed nowhere dense sets in the real line. Choose a real number $x_0 \notin F_0$ to start. For each $\alpha$ with $0 < \alpha <\omega_1$ , choose a real number $x_\alpha$ not in the following set:

$\left\{x_\beta: \beta<\alpha \right\} \cup \bigcup \limits_{\beta<\alpha} F_\beta$

The above set is a countable union of closed nowhere dense sets of the real line. As a complete metric space, the real line cannot be of first category. In fact, according to the Baire category theorem, the complement of a set of first category (such as the one described above) is dense in the real line. So such an $x_\alpha$ can always be selected at each $\alpha<\omega_1$ . Then $X=\left\{x_\alpha: \alpha<\omega_1 \right\}$ is a Luzin set. $\blacksquare$

Now that we have a way of constructing an uncountable Luzin sets, the following observations provide some useful facts for our problem at hand.

Nowhere dense sets and sets of first category are "thin" sets. Any "thin" set can intersect with a Luzin set with only countably many points. Thus any "co-thin" set contains all but countably many points of a Luzin set. For example, let $A$ be an uncountable Luzin set. Then if $F$ is a closed nowhere dense set in the real line, then $\mathbb{R}-F$ contains all but countably many points of $A$ . Furthermore, if $F_1,F_2,F_3,\cdots,$ are closed nowhere dense subsets of the real line, then $\mathbb{R}- \bigcup \limits_{i=1}^\infty F_i$ contains all but countably many points of the Luzin set $A$ .

Note that the set $\mathbb{R}-F$ in the preceding paragraph is a dense open set. Thus the complement of a closed nowhere dense set is a dense open set. Note that the set $\mathbb{R}- \bigcup \limits_{i=1}^\infty F_i$ in the preceding paragraph is a dense $G_\delta$ -set. Thus the complement of the union of countably many closed nowhere dense sets is a dense $G_\delta$ -set. Thus the observation in the preceding paragraph gives the following proposition:

Proposition 2
Given an uncountable Luzin set $A$ and given a dense $G_\delta$ subset $H$ of the real line, $H$ contains all but countably many points of $A$ .

In fact, Proposition 2 not only hold in the real line, it also holds in any uncountable dense subset of the real line.

Proposition 3
Let $A$ be an uncountable Luzin set. Let $Y \subset \mathbb{R}$ be uncountable and dense in the real line such that $A \cap Y$ is uncountable. Given a dense $G_\delta$ subset $H$ of $Y$ , $H$ contains all but countably many points of $A \cap Y$ .

Proof of Proposition 3
We want to show that $Y-H$ can only contain countably many points of $A$ . Let $H=\bigcap \limits_{i=1}^\infty O_i$ where each $O_i$ is open and dense in $Y$ . Then for each $i$ , let $U_i$ be open in the real line such that $U_i \cap Y=O_i$ . Each $U_i$ is open and dense in the real line. Thus $H^*=\bigcap \limits_{i=1}^\infty U_i$ contains all but countably many points of the Luzin set $A$ . Note the following set inclusion:

$H=\bigcap \limits_{i=1}^\infty U_i \cap Y=\bigcap \limits_{i=1}^\infty O_i \subset \bigcap \limits_{i=1}^\infty U_i=H^*$

Suppose that $Y-H$ contains uncountably many points of $A$ . Then these points, except for countably many points, must belong to $H^*=\bigcap \limits_{i=1}^\infty U_i$ . The above set inclusion shows that these points must belong to $H$ too, a contradiction. Thus $Y-H$ can only contain countably many points of $A$ , equivalently the $G_\delta$ -set $H$ contains all but countably many points of $A \cap Y$ . $\blacksquare$

The following proposition follows from Proposition 3 and is a useful fact that will help us see that the product of an uncountable Luzin set and $\mathbb{P}$ is not normal.

Proposition 4
Let $Y$ be an uncountable Luzin set such that $\mathbb{Q} \subset Y$ . Then $Y-\mathbb{Q}$ cannot be an $F_\sigma$ -set in the Euclidean space $Y$ , equivalently $\mathbb{Q}$ cannot be a $G_\delta$ -set in the space $Y$ .

Proof of Proposition 4
By Proposition 3, any dense $G_\delta$ -subset of $Y$ must be co-countable. $\blacksquare$

The following proposition is another useful observation about Luzin sets. Let $A \subset \mathbb{R}$ . Let $D \subset \mathbb{R}$ be a countable dense subset of the real line. The set $A$ is said to be concentrated about $D$ if for every open subset $O$ of the real line such that $D \subset O$ , $O$ contains all but countably many points of $A$ . The following proposition can be readily checked based on the definition of Luzin sets.

Proposition 5
For any $A \subset \mathbb{R}$ , $A$ is a Luzin set if and only if $A$ is concentrated about every countable dense subset of the real line.

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Lindelof Subspace of The Michael Line

Let $A$ be an uncountable Luzin set. We can assume that $A$ is dense in the real line. If not, just add a countble subset of $\mathbb{P}$ that is dense in the real line. Let $L=A \cup \mathbb{Q}$ . It is clear that adding countably many points to a Luzin set still results in a Luzin set. Thus $L$ is also a Luzin set. Now consider $L$ as a subspace of the Michael line $\mathbb{M}$ . Then points of $L-\mathbb{Q}$ are discrete and points in $\mathbb{Q}$ have Euclidean open neighborhoods. By Proposition 5, the set $L$ is concentrated about every countable dense subset of the real line. In particular, it is concentrated about $\mathbb{Q}$ . Thus as a subspace of the Michael line, $L$ is a Lindelof space, since every open set containing $\mathbb{Q}$ contains all but countably many points of $L$ .

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The Non-Normal Product $L \times \mathbb{P}$

We highlight the following two facts about the Luzin set $L=A \cup \mathbb{Q}$ as discussed in the preceding section.

$L-\mathbb{Q}$ is not an $F_\sigma$ -set in $L$ (as Euclidean space).
$A=L-\mathbb{Q}$ is dense in the real line.

The first bullet point follows from Proposition 4. The second bullet point is clear since we assume the Luzin set $A$ we start with is dense. Recall that when thinking of $L$ as a subspace of the Michael line, $L-\mathbb{Q}$ are isolated and $\mathbb{Q}$ retains the usual real line open sets. Because of the above two bullet points, $L \times \mathbb{P}$ is not normal. The proof that $L \times \mathbb{P}$ is not normal is the corollary of the proof that $\mathbb{M} \times \mathbb{P}$ is not normal. Note that in the proof for showing $\mathbb{M} \times \mathbb{P}$ is not normal, the two crucial points about the proof are that the isolated points of the Michael line cannot be an $F_\sigma$ -set and are dense in the real line (found in “Michael Line Basics”).

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Michael Space

The example $L \times \mathbb{P}$ that we construct here was hinted in footnote 4 in [6]. In a later publication, E. Michael constructed an uncountable Lindelof subspace of the Michael line (see Lemma 3.1 in [5]). That construction should produce a similar set as the Luzin sets since the approach in [5] is a mirror image of the Luzin set construction. The approach in the Luzin set construction in Theorem 1 is to pick points not in the union of countably many closed nowhere dense sets, while the approach in [5] was to pick points in dense $G_\delta$ -sets in a transfinite induction process.

A Michael space is a Lindelof space whose product with $\mathbb{P}$ is not normal. The example shown here shows that under CH, there exists a Michael space. However, the question of whether there exists a Michael space in ZFC is still unsolved. This is called the Michael problem. A recent mention of this unsolved problem is [3] (page 160). A Michael space can also be constructed using Martin’s axiom (see [1]).

A space is said to be a productively Lindelof space if its product with every Lindelof space is Lindelof. Is $\mathbb{P}$ a productively Lindelof space? As we see here, under CH the answer is no. Another way of looking at the Michael problem: is it possible to show that $\mathbb{P}$ is not productively Lindelof in ZFC alone?

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Reference

Alster, K., The product of a Lindelof space with the space of irrationals under Martin’s Axiom, Proc. Amer. Math. Soc., 110 (1990) 543-547.
Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
Miller, A. W., Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 201-233, 1984.
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Dan Ma's Topology Blog

Expository Articles In General Topology

Monthly Archives: November 2012

Bing’s Example G

Topological Spaces with Caliber Omega 1

Cartesian Products of Two Paracompact Spaces – Continued

Cartesian Products of Two Paracompact Spaces

The Michael Line and the Continuum Hypothesis