Examples of Meta-Lindelof Spaces

Paracompact spaces and Lindelof spaces are familiar notions of covering properties. The covering properties resulting from replacing the para with meta in the case of paracompact and adding meta to Lindelof are also spaces that had been extensively studied. This is an introduction of meta-Lindelof spaces by way of examples.

Definitions

All spaces under discussion are Hausdorff and regular. A space $X$ is paracompact if every open cover of $X$ has a locally finite open refinement. A space $X$ is metacompact if every open cover of $X$ has a point-finite open refinement. A space $X$ is Lindelof if every open cover of $X$ has a countable subcover, i.e., every open cover of $X$ has a countable subcollection that is also a cover of $X$ . A space $X$ is meta-Lindelof if every open cover of $X$ has a point-countable open refinement, i.e., every open cover $\mathcal{U}$ of $X$ has a subcollection $\mathcal{V}$ such that $\mathcal{V}$ is a refinement of $\mathcal{U}$ and that $\mathcal{V}$ is a point-countable collection. From the definitions, we have the implications shown in the following diagram.

$\displaystyle \begin{aligned} & \\& \bold P \bold a \bold r \bold a \bold c \bold o \bold m \bold p \bold a \bold c \bold t \ \ \ \ \ \leftarrow \ \ \ \ \ \bold L \bold i \bold n \bold d \bold e \bold l \bold o \bold f \\&\ \ \ \ \ \ \ \ \ \downarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow\\& \bold M \bold e \bold t \bold a \bold c \bold o \bold m \bold p \bold a \bold c \bold t \ \ \ \ \rightarrow \ \ \ \bold M \bold e \bold t \bold a \bold - \bold L \bold i \bold n \bold d \bold e \bold l \bold o \bold f \end{aligned}$

It is well known that any regular Lindelof space is paracompact (the left arrow at the top of the diagram). The two down arrows and the right arrow at the bottom in the diagram follow from the definitions. If there is an up arrow from meta-Lindelof to Lindelof, the diagram would be a closed loop, which would mean all 4 notions are equivalent. At minimum, there are paramcompact spaces that are not Lindelof. For example, any non-separable metric space is paracompact and not Lindelof. Thus, in general it is impossible for the diagram to be a closed loop. In other words, there must be meta-Lindelof spaces that are not Lindelof. However, with additional assumptions, an up arrow from meta-Lindelof to Lindelof is possible. Theorem 1 below shows that the diagram is a closed loop for separable spaces. It is well known that paracompact spaces with the countable chain condition (CCC) are Lindelof (see here). Perhaps an up arrow from meta-Lindelof to Lindelof is possible for spaces with CCC. We discuss some partial results at the end of this article.

Spaces that are not meta-Lindelof

We present two spaces that are not meta-Lindelof (Example 1 and Example 2). The following theorem will make the examples clear.

Theorem 1
Let $X$ be a separable space. If $X$ is meta-Lindelof, then $X$ is Lindelof.

Proof
Let $\mathcal{U}$ be an open cover of $X$ . Since $X$ is meta-Lindelof, there is a point-countable open refinement $\mathcal{V}$ of $\mathcal{U}$ . Let $A$ be a countable dense subset of $X$ . Let $\mathcal{W}$ be defined by $\mathcal{W}= \{ V \in \mathcal{V}: V \cap A \not = \varnothing \}$ . Since $\mathcal{V}$ is point-countable, each point of $A$ can belong to only countably many $V$ . It follows that $\mathcal{W}$ is countable. Furthermore, $\mathcal{W}=\mathcal{V}$ . The set inclusion $\mathcal{W} \subset \mathcal{V}$ is clear. The set inclusion $\mathcal{V} \subset \mathcal{W}$ follows from the fact that $A$ is a dense subset. To complete the proof, for each $V \in \mathcal{V}$ , choose $U(V) \in \mathcal{U}$ such that $V \subset U(V)$ . Then $\{ U(V): V \in \mathcal{V} \}$ is a countable subcover of $\mathcal{U}$ . This completes the proof that $X$ is Lindelof. $\square$

In the above proof, the countable dense set forces the point-countable open refinement to be countable, thus leading to a countable subcover of the original open cover. We now look at examples.

Example 1
The idea for this example is that any separable space that is known to be non-Lindelof must not be meta-Lindelof due to the above theorem. A handy example is the Sorgenfrey plane $S \times S$ . Recall that the Sorgenfrey line $S$ is the real number line topologized by the base consisting of the half open intervals of the form $[a, b)$ . It is well known that $S$ is hereditarily Lindelof, separable and not metrizable. The Sorgenfrey line $S$ is a classic example of a Lindelof space whose square is not even normal, thus not Lindelof (see here). By Theorem 1, the Sorgenfrey plane is not meta-Lindelof.

Example 2
This is another example that is separable and non-Lindelof, hence not meta-Lindelof. The space is the Mrowka space (or Psi-space), which is defined here. To define the space, let $\mathcal{A}$ be uncountable almost disjoint family of subsets of $\omega$ , that is, for any $A, B \in \mathcal{A}$ with $A \ne B$ , $A \cap B$ is finite. The underlying set is $X=\omega \cup \mathcal{A}$ . The points in $\omega$ are isolated. An open neighborhood of $A \in \mathcal{A}$ is of the form $\{ A \} \cup (A \backslash F)$ where $F \subset \omega$ is a finite set. The space $X$ is not Lindelof since $\mathcal{A}$ is an uncountable closed and discrete subset. It is separable since the set of integers, $\omega$ , is a dense subset. By Theorem 1, the space $X$ must not be meta-Lindelof. This space is a classic example of a space that has a $G_\delta$ -diagonal but is not submetrizable.

Example 3
Example 1 is not normal. Example 2 is not normal if the almost disjoint family $\mathcal{A}$ has cardinality continuum (due to Jones’ Lemma). For Example 2 to be normal, additional set theory axiom is required. Example 3 is a normal example of a space that is not meta-Lindelof.

Consider $X=\omega_1$ , the space of the countable ordinals with the ordered topology. This space is not Lindelof since the open cover consisting of $[0,\alpha]$ , where $\alpha \in \omega_1$ , is an open cover that has no countable subcover. The space $X$ is also not paracompact since some open covers cannot be locally finite. The same idea can show that certain open covers cannot be point-countable open cover. This is due to the pressing down lemma (see here). To see this, for each limit ordinal $\alpha \in \omega_1$ , let $O_\alpha=(f(\alpha), \alpha]$ be an open set containing $\alpha$ . Then the function $f$ is a so called pressing down function. By the pressing down lemma, there exist some $\delta$ such that the set $\{ \alpha: f(\alpha)=\delta \}$ is a stationary subset of $X=\omega_1$ . This implies that the point $\delta$ would belong to uncountably many open sets $O_\alpha$ . Thus, any open cover of $X=\omega_1$ containing the sets $O_\alpha$ cannot be point-countable.

To wrap up the example, let $\mathcal{V}$ be any open refinement of the open cover consisting of the open sets $[0, \alpha]$ . We choose $O_\alpha$ described above such that each $O_\alpha$ is contained in some $V_\alpha \in \mathcal{V}$ . This means that $\mathcal{V}$ cannot be point-countable. Thus, one open cover of $X$ has no point-countable open refinement, showing that the space of all countable ordinals, $\omega_1$ , cannot be meta-Lindelof.

Example 4
This example makes use of Example 3. Consider $X$ be the product space of $\omega_1$ many copies of the real line $\mathbb{R}$ . Each $x \in X$ is a function $x: \omega_1 \rightarrow \mathbb{R}$ . Let $x_\alpha$ denote $x(\alpha)$ for each $\alpha \in \omega_1$ . Let $Y$ be the set of all points $x$ in $X$ where $x_\alpha$ is non-zero for at most countably many $\alpha \in \omega_1$ . This space is called the $\Sigma$ -product of lines. The topology of $Y$ is the inherited product topology from the product space $X$ . The $\Sigma$ -product of separable metric spaces is collectionwise normal (see here). The $\Sigma$ -product of uncountably many spaces, each of which has at least 2 points, always contains a closed copy of the space $\omega_1$ (see here).

Observe that closed subspaces of a meta-Lindelof space are always meta-Lindelof. Thus the $\Sigma$ -product $Y$ defined here cannot be a meta-Lindelof space. This example is revisited in Example 6 below.

Meta-Lindelof but not Lindelof

Example 5
This example is the Michael line $\mathbb{M}$ , which is the real line topologized by making the irrational numbers isolated and letting the rational numbers retain the usual Euclidean open sets (a basic introduction is found here). The usual Euclidean open sets of the real line are also open in the Michael line. Thus, $\mathbb{M}$ is a submetrizable space.

To see that it is meta-Lindelof, let $\mathcal{U}$ be an open cover of the Michael line $\mathbb{M}$ . Choose $\mathcal{N}=\{ U_0,U_1,U_2, \cdots \} \subset \mathcal{U}$ such that $\mathcal{N}$ covers the rational numbers. Let $\mathcal{V}=\{ \{ y \}: y \in \mathbb{R} \backslash \cup \mathcal{N} \} \cup \mathcal{N}$ . Thus $\mathcal{V}$ is a point-countable open refinement of $\mathcal{U}$ . This shows that the Michael line $\mathbb{M}$ is meta-Lindelof. This fact can also be seen from the above diagram, which shows that paracompact $\rightarrow$ metacompact $\rightarrow$ meta-Lindelof. Note that the Michael line is a paracompact space that is not Lindelof.

The Michael line as an example of a meta-Lindelof space is quite informative. Note that in any Lindelof space, all closed and discrete subsets must be countable (such a space is said to have countable extent). However, a meta-Lindelof space can have uncountable closed and discrete subset. There is a closed and discrete subset of the Michael line of cardinality continuum. To see this, note that there is a Cantor set in the real line consisting entirely of irrational numbers. For example, we can construct a Cantor set within the interval $A=[\sqrt{2},\sqrt{8}]$ . Let $r_0,r_1,r_2,r_4,\cdots$ enumerate all rational numbers within this interval. In the first step, we remove a middle part of the interval $A$ such that the two remaining closed intervals have irrational endpoints and will miss $r_0$ . In the $n$ th step of the construction, we remove the middle part of each of the remaining intervals such that all remaining intervals have irrational endpoints and will miss the rational number $r_n$ . Let $C$ be the resulting Cantor set, which consists only of irrational numbers since it misses all rational numbers in $A$ . The set $C$ , as a subset of the Michael line, is closed and discrete.

The above paragraph shows one direction in which the two notions of “Lindelof” diverge. Lindelof spaces have countable extent. On the other hand, meta-Lindelof space can have uncountable closed and discrete subsets.

Example 5 actually presents a template for producing meta-Lindelof spaces. In Example 5, start with the real line with the usual topology. Identify a “Lindelof” part (the rationals) and make the remainder discrete (the irrationals). The template is further described in the following theorem.

Theorem 2
Let $Y$ be a space and let $W$ be a Lindelof subspace of $Y$ . Define a new space $Z$ such that the underlying set is $Y$ and the new topology is defined as follows. Let points $y \in Y \backslash W$ be isolated and let points $x \in W$ retain the original open sets in $Y$ .

Then the space $Z$ is a meta-Lindelof space.
If the original space $Y$ is a non-Lindelof space, then $Z$ is also not Lindelof.

To see $Y$ with the new topology is meta-Lindelof, follow the same proof in showing the Michael line is meta-Lindelof. To prove part 2, let $\mathcal{U}$ be an open cover of $Y$ (in the original topology) such that no countable subcollection of $\mathcal{U}$ is a cover of $Y$ . Let $\mathcal{C}$ be a countable subcollection of $\mathcal{U}$ such that $\mathcal{C}$ covers $W$ . There must be uncountably many points of $Y$ not covered by $\mathcal{C}$ . Then $\mathcal{Q}=\{ \{ y \}: y \in Z \backslash \cup \mathcal{C} \} \cup \mathcal{C} \}$ is an open cover of $Z$ that has no countable subcover.

Example 6
This example starts with the space $Y$ in Example 4. To define $Y$ , let $X=\Pi_{\alpha \in \omega_1} \mathbb{R}$ , the product space of $\omega_1$ many copies of the real line $\mathbb{R}$ . Let $Y$ and $W$ be the subspaces of $X$ defined as follows:

$\displaystyle Y=\{ y \in X: \lvert \{ \alpha \in \omega_1: x(\alpha) \ne 0 \} \lvert \le \omega \}$

$\displaystyle W=\{ y \in X: \lvert \{ \alpha \in \omega_1: x(\alpha) \ne 0 \} \lvert < \omega \}$

The space $Y$ is called the $\Sigma$ -product of lines and is collectionwise normal. Note that the $\Sigma$ -product of separable metric spaces is collectionwise normal (see here). The space $W$ is called the $\sigma$ -product of lines, which consists of all points in the product space $X$ with non-zero values in at most finitely many coordinates. Note that the $\sigma$ -product of separable metric spaces is a Lindelof space (see here).

The $\Sigma$ -product $Y$ is not Lindelof because it contains a closed copy of $\omega_1$ (see here). Thus, $Y$ is a non-Lindelof space with a Lindelof subspace $W$ , which is exactly what is required in Theorem 2. Define the space $Z$ as described in Theorem 2. The resulting $Z$ is meta-Lindelof but not Lindelof.

Note that the Lindelof $W$ is a dense subset of $\Sigma$ -product $Y$ . Thus, though $Y$ is not Lindelof, it has a dense Lindelof subspace. In Example 4, we see that $Y$ is not meta-Lindelof since it contains a non-meta-Lindelof space as a closed subspace. However, re-topologizing it by making the points not in $W$ isolated and making the points in $W$ retain the open sets in the $\Sigma$ -product topology produces a meta-Lindelof space.

Concluding Remarks

A few observations can be made from the six examples discussed above.

Among the separable spaces, meta-Lindelofness and Lindelofness are the same (Theorem 1). This gives a handy way to show that any separable space that is not Lindelof is not meta-Lindelof.
It is possible that the product of Lindelof spaces is not meta-Lindelof (Example 1).
A question can be asked for Example 2, which is a Moore space. Any Lindelof Morre space is second countable. Example 2 is not meta-Lindelof. Must a meta-Lindelof Moore space be second countable?
The first uncountable ordinal $\omega_1$ with the order topology is not paracompact since it is not possible to cover the limit ordinals with a locally finite collection of open sets. The same idea shows that $\omega_1$ does not even satisfy the weaker property of meta-Lindelof (Example 3).
One take away from Example 4 is that meta-Lindelof spaces resemble Lindelof spaces in one respect, that is, meta-Lindelofness is hereditary with respect to closed subspaces. Because the $\Sigma$ -product of the real lines contains a closed copy of $\omega_1$ , it is not meta-Lindelof.
The Michael line (Example 5) demonstrates one critical difference between meta-Lindelof and Lindelof. Any Lindelof space has countable extent. It is different for meta-Lindelof spaces. In the Michael line, there is an uncountable closed and discrete subset. One way to see this is to construct a Cantor set in the real line consisting of only irrational numbers.
The Michael line (Example 5) gives the hint of a recipe for producing meta-Lindelof spaces. The recipe is described in Theorem 2. Example 6 is a demonstration of the recipe. The $\Sigma$ -product discussed inExample 4 is non-Lindelof with a dense Lindelof subspace $W$ . By letting $W$ retain the original $\Sigma$ -product open sets and making the complement of $W$ discrete, we produce another meta-Lindelof space that is not Lindelof.

The rest of the remark focuses on the diagram given earlier (repeated belo

Theorem 1 shows that for separable spaces, meta-Lindelof $\rightarrow$ Lindelof. As a result, the diagram becomes a closed loop, meaning all 4 notions in the diagram are equivalent for separable spaces. As pointed out earlier, CCC may be a good candidate for replacing separable since among CCC spaces, paracompactness equates Lindelofness. It turns out that for CCC spaces, meta-Lindelof does not imply Lindelof. A handy example is the Pixley-Roy space $\mathcal{F}[\mathbb{R}]$ , which is non-Lindelof, metacompact (hence meta-Lindelof) and satisfies the CCC (see the follow up discussion in the next post). We found the following partial result in p. 971 of the Handbook of Set-Theoretic Topology (Chapter 22 on Borel Measures by Gardner and Pfeffer).

Theorem 3 (MA + not CH)
Each locally compact meta-Lindelof space satisfying the CCC is Lindelof.

Theorem 3 is Corollary 4.9 in the article in the Handbook. It implies that for locally compact CCC space, the above diagram is a closed loop but only under Martin’s axiom and the negation of CH.

The Handbook was published in 1984. We wonder if there is any update since then. For any reader who has updated information regarding the consistency result in Theorem 3, please kindly comment in the space below.

For more information on meta-Lindelof and other covering properties, see Chapter 9 in the Handbook of Set-Theoretic Topology (the chapter by D. Burke on covering properties).

$\text{ }$

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Daniel Ma meta-Lindelof spaces

Posted: 10/29/2022
Updated: 10/30/2022
Updated: 11/1/2022

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Examples of Meta-Lindelof Spaces

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