Kuratowski Theorem

Posted on February 6, 2023 by Dan Ma

We give a proof of Kuratowski theorem. This is the Kuratowki Theorem involving the projection map from a product space $X \times Y$ onto one of the factors. All spaces in consideration are at least Hausdorff.

A map $f:L \rightarrow M$ from the space $L$ into the space $M$ is said to be a closed map if $f(C)$ is closed in $M$ for every closed subset $C$ of $L$ . Here’s the statement of the theorem.

Kuratowski Theorem
The following conditions are equivalent.

The space $Y$ is compact.
For every space $X$ , the projection $\pi:X \times Y \rightarrow X$ is a closed map.
For every normal space $X$ , the projection $\pi:X \times Y \rightarrow X$ is a closed map.

Before going to the proof, note that the theorem is not true when none of the factors is compact. Consider the product space $\mathbb{R} \times \mathbb{R}$ where $\mathbb{R}$ is the real number line. The graph of the curve $\frac{1}{x}$ with $x$ ranges over the non-zero real numbers is a closed subset of $\mathbb{R} \times \mathbb{R}$ . The projection of graph of $\frac{1}{x}$ into the x-axis is the set $(- \infty,0) \cup (0,\infty)$ , which is not closed in $\mathbb{R}$ .

The Proof

The direction $2 \rightarrow 3$ is clear. We prove $1 \rightarrow 2$ and $3 \rightarrow 1$ .

$1 \rightarrow 2$
Let $Y$ be compact. Let $C$ be a closed subset of $X \times Y$ . We show that $X \backslash \pi(C)$ is open in $X$ . To this end, let $x \in X$ and $x \notin \pi(C)$ . Note that $\{x \} \times Y \subset (X \times Y) \backslash C$ . By the Tube Lemma (see here), there exists an open set $O \subset X$ with $x \in O$ such that $\{x \} \times Y \subset O \times Y \subset (X \times Y) \backslash C$ . It follows that $O \cap \pi(C)=\varnothing$ . This means that $X \backslash \pi(C)$ is open and $\pi(C)$ is closed in $X$ .

$3 \rightarrow 1$
Suppose condition 3 holds. We show that $Y$ is compact. Suppose not and let $\mathcal{C}$ be a collection of closed subsets of $Y$ with the finite intersection property such that $\bigcap \mathcal{C}=\varnothing$ (see here for the characterization of compactness using the finite intersection property). Choose a point $p \notin Y$ . Let $X=\{ p \} \cup Y$ . Define a topology on $X$ using the following base:

$\displaystyle \mathcal{B}=\{ \{ y \}: y \in Y \} \cup \mathcal{B}_0$

where sets in $\mathcal{B}_0$ are of the form $\{ p \} \cup (\bigcap_{C \in T} C) \cup A$ where $T \subset \mathcal{C}$ is finite and $A \subset Y$ . In words, we can describe the topology on $X$ in this way. The set $Y$ is a discrete subspace of $X$ and each basic open set containing the point $p$ is the union of three sets – the singleton set $\{ p \}$ . the intersection of finitely many sets in $\mathcal{C}$ and a subset of $Y$ . It follows that the space $X$ is Hausdorff. To show this, we only need to separate $p$ and any point in $Y$ by disjoint open sets. Let $y \in Y$ . Note that $\bigcap \mathcal{C}=\varnothing$ . Thus, $y \notin C$ for some $C \in \mathcal{C}$ . As a result, $\{ y \}$ and $\{p \} \cup C$ are disjoint open sets separating $y$ and $p$ , respectively. It also follows that the space $X$ is normal. To see this, let $H$ and $K$ be disjoint closed subsets of $X$ such that $p \in H$ . Then $X \backslash K$ and $K$ are open sets separating $H$ and $K$ , respectively.

Let $E=\overline{D}$ where $D=\{ (x,x): x \in Y \} \subset X \times Y$ . By assumption, $\pi(E)$ is closed in $X$ . Note that $Y \subset \pi(E)$ . Since $\overline{Y}=X$ , we have $X=\pi(E)$ . This means that $p \in \pi(E)$ . Furthermore, there exists $t \in Y$ such that $(p,t) \in E$ . We claim that $t \in C$ for all $C \in \mathcal{C}$ . This contradicts that the intersection of $\mathcal{C}$ is empty. Let $C \in \mathcal{C}$ . Let $U \subset Y$ be open such that $t \in U$ . Consider the open set $O=(\{p \} \cup C) \times U$ , which contains the point $(p,t)$ . Since $(p,t) \in E$ , and since $(p,t) \in O$ , the open set $O$ contains points of $D$ . Thus $(x,x) \in O$ for some $x \in Y$ . This means that $x \in C$ and $x \in U$ . What we have just shown is that for every open set $U$ containing $t$ , $U \cap C \ne \varnothing$ for all $C \in \mathcal{C}$ . It follows that $t \in C$ for all $C \in \mathcal{C}$ , producing that contradiction indicated earlier. Thus, every collection of closed subsets of $Y$ with the finite intersection property must have non-empty intersection. This implies that $Y$ is compact. With this observation, the proof of Kuratowski Theorem is complete. $\square$

Perfect Mapping

The function $f:L \rightarrow M$ is a perfect map if $f$ is continuous and a closed map such that each point inverse is compact. When each point inverse under the mapping $f$ is compact, the map is said to have compact fibers. A corollary of Kuratowski Theorem is that the projection map in Kuratowski Theorem is also a perfect map.

Theorem 1
Let $X$ be a space and $Y$ be a compact space. Then the projection $\pi:X \times Y \rightarrow X$ is a perfect map.

The projection map $\pi:X \times Y \rightarrow X$ is always continuous. According to Kuratowski Theorem, it is a closed map. For each $x \in X$ , $\pi^{-1}(x)=\{ x \} \times Y$ , which is compact. Thus, each point inverse of the projection map is compact. $\square$

$\text{ }$

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$\copyright$ 2023 – Dan Ma

A Pixley-Roy Meta-Lindelof Space

Posted on November 1, 2022 by Dan Ma

The discussion in the preceding post pointed to a natural question of whether we have a theorem of CCC + meta-Lindelof $\rightarrow$ Lindelof or there is a counterexample. In this post, we present a counterexample. The example is a Pixley-Roy hyperspace $\mathcal{F}[X]$ , which is always metacompact, hence meta-Lindelof. We then make sure the ground $X$ is chosen appropriately to ensure that the Pixley-Roy space has the CCC and is not Lindelof.

The following diagram is shown in the preceding post.

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

In the diagram, if there is an up arrow from meta-Lindelof to Lindelof, then all 4 notions in the diagram are equivalent. But we know Lindelofness and paracompactness are not equivalent. Thus, there must be meta-compact spaces that are not Lindelof. Such examples can be found in the preceding post. On the other hand, among separable spaces, meta-Lindelof implies Lindelof. Thus, among separable spaces, the diagram has a closed loop, implying all 4 notions are equivalent. Naturally, we would like to weaken the separability to the countable chain condition (CCC). Would the diagram be a closed loop for CCC spaces? The answer is no. The counterexample is a Pixley-Roy space as discussed below.

Pixley-Roy Spaces

All spaces are at least Hausdorff. Let $X$ be a space. Let $\mathcal{F}[X]$ be the set of all non-empty finite subsets of $X$ . For any $F \in \mathcal{F}[X]$ and any open $U \subset X$ with $F \subset U$ , define $[F,U]$ as follows:

$[F,U]=\{ D \in \mathcal{F}[X]: F \subset D \subset U \}$

The set of all possible $[F,U]$ is a base for a topology on $\mathcal{F}[X]$ . This topology is called the Pixley-Roy topology (or Pixley-Roy hyperspace topology). The set $\mathcal{F}[X]$ with this topology is called a Pixley-Roy space or Pixley-Roy hyperspace. The space $X$ that defines the hyperspace is called the ground space.

We are interested in a Pixley-Roy space $\mathcal{F}[X]$ that has the CCC, is meta-Lindelof and not Lindelof. The space $\mathcal{F}[X]$ is always metacompact, hence meta-Lindelof whenever the ground space $X$ is Hausdorff. To obtain the desired example, we only need to use a gound space $X$ that is uncountable (to ensure that the hyperspace is not Lindelof) and has a countable network (to ensure that the hyperspace has the CCC). This preceding post discusses the needed facts as well as other basic facts of $\mathcal{F}[X]$ . In the remainder of this post, we show that if the ground space $X$ has a countable network, then the Pixley-Roy space has the CCC.

Spaces with Countable Networks

A collection $\mathcal{N}$ of subsets of the space $X$ is said to be a network if for every $x \in X$ , and for every open $U \subset X$ with $x \in U$ , there exists $N \in \mathcal{N}$ such that $x \in N \subset U$ . This sounds like the definition of a base for a topology. Note that the sets in the network $\mathcal{N}$ do not have to be open. Thus, a network is not necessarily a base. Though not as strong as having a countable base, a space having a countable network is a strong property. For example, a space with a countable network is both hereditarily separable and hereditarily Lindelof. The Lindelof property can fail to be preserved when taking product. The property of having a countable network is preserved by taking countable product.See here for a discussion of spaces with countable network.

We now show that if the ground space $X$ has a countable network, then the Pixley-Roy hyperspace $\mathcal{F}[X]$ has the countable chain condition (CCC), which means that every pairwise disjoint collection of open sets must be countable. Let $\mathcal{N}$ be a countable network for the ground space $X$ . Let $\{ [F_\alpha,U_\alpha]: \alpha \in \omega_1 \}$ be an uncountable collection of open sets in $\mathcal{F}[X]$ . We show that there must exist 2 sets from this collection having non-empty intersection.

To make the argument clear, we can assume that the network $\mathcal{N}$ is closed under finite intersection. If not, we can just make sure that $\mathcal{N}$ include all finite intersections its elements. For each $\alpha$ , there exists $B_\alpha \in \mathcal{N}$ such that $F_\alpha \subset B_\alpha \subset U_\alpha$ . Since $\mathcal{N}$ is countable, there must exists some $B \in \mathcal{N}$ such that $B=B_\alpha$ for uncountably many $\alpha \in \omega_1$ . Consider two such, say $\beta$ and $\gamma$ . Then we have $F_\beta \subset B \subset U_\beta$ and $F_\gamma \subset B \subset U_\gamma$ . Note that the finite set $F_\beta \cup F_\gamma$ belongs to both $[F_\beta,U_\beta]$ and $[F_\gamma,U_\gamma]$ . This completes the proof that the Pixley-Roy space $\mathcal{F}[X]$ satisfies the CCC if the ground space has a countable network.

Counterexamples

Based on the above discussion, for any uncountable Hausdorff space $X$ with a countable network, the Pixley-Roy space $\mathcal{F}[X]$ is a CCC meta-Lindelof space that is not Lindelof. In fact, we can simply one such space that has a countable base (any base is a network). As a concrete example, we can use the real line as the ground space $X$ . Thus, the Pixley-Roy space $\mathcal{F}[\mathbb{R}]$ is a CCC meta-Lindelof space that is not Lindelof.

$\text{ }$

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$\copyright$ 2022 – Dan Ma

Posted 11/1/2022
Updated 11/4/2022

Examples of Meta-Lindelof Spaces

Posted on October 29, 2022 by Dan Ma

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.

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).

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

Divergence of the sum of the reciprocals of the primes

Posted on May 29, 2022 by Dan Ma

The sum of the reciprocals of the prime numbers diverges, i.e., the sum $\sum \limits_{p} \frac{1}{p}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\cdots$ , where $p$ ranges over all the primes, diverges. Facts about prime numbers are always interesting, especially a fundamental fact such as this one. With the divergence of the sum of the reciprocals, a corollary is that there are infinitely many primes. The first proof of this fundamental fact about prime numbers was given by Euler in 1737. The proof discussed below is due to Paul Erdős. It is an elegant and nifty proof that earned a place in the book called Proofs from THE BOOK [1]. Erdős’ proof is by way of a contradiction. It is also a simple proof. After assuming the convergence of the sum, the rest of the proof only utilizes basic information about numbers (integers in particular). There is a Wikipedia entry on the divergence of the sum of the reciprocal of primes. We also briefly mention a conjecture of Erdős and the Green-Tao Theorem.

Erdős’ Proof

Let $p_1,p_2,p_3,\cdots$ be the listing of all the prime numbers. Suppose that $\sum \limits_{p_i} \frac{1}{p_i}$ converges. There is some positive integer $m$ such that $\sum_{i \ge m+1} \frac{1}{p_i}<\frac{1}{2}$ . In the remainder of the proof, the prime numbers $p_1,p_2, \cdots,p_m$ are called small primes and $p_{m+1},p_{m+2},\cdots$ are called big primes. For any positive integer $N$ , we have the following inequality:

…………….

$\displaystyle \sum \limits_{i \ge m+1} \frac{N}{p_i}<\frac{N}{2}$

For each $N$ indicated above, let $N_s$ be the number of all positive integers $n \le N$ such that the prime divisors of $n$ are all small primes. Let $N_b$ be the number of all positive integers $n \le N$ such that at least one prime divisor of $n$ is a big prime. By definition, $N_s+N_b=N$ for all $N$ . We show that $N_s+N_b<N$ for some $N$ , a contradiction. To obtain the contradiction, upper bounds are established for both $N_s$ and $N_b$ .

The symbol $\left\lfloor y \right\rfloor$ refers to the floor function, which is greatest integer less than or equal to $y$ . Consider $\left\lfloor \frac{N}{p_i} \right\rfloor$ , which is identical to the number of positive integers $n \le N$ that are multiples of $p_i$ . We now have the following upper bound for $N_b$ .

…………….

$\displaystyle N_b \le \sum \limits_{i \ge m+1} \left\lfloor \frac{N}{p_i} \right\rfloor \le \sum \limits_{i \ge m+1} \frac{N}{p_i} <\frac{N}{2}$

A positive integer $n$ is square-free if its prime decomposition contains no repeated prime factors. For example, $15=3 \times 5$ is square-free while $18=2 \times 3^2$ is not square-free. Note that any positive integer $n$ can be expressed as the product of two parts, one of which is square-free and the other is a square, i.e., $n=a \times b^2$ with $a$ being square-free. If a positive integer is itself square-free, the square part is 1. Otherwise, we can always rearrange the prime decomposition to obtain $n=a \times b^2$ .

We are now ready to examine $N_s$ . For each $n \le N$ that has only small prime divisors, write $n=a_n \times b_n^2$ where $a_n$ is square-free part. Each $a_n$ is a product of small primes. There are only $m$ small primes. Thus, there can be at most $2^m$ many different square-free parts $a_n$ . On the other hand, since $b_n^2 \le n$ , we have $b_n \le \sqrt{n} \le \sqrt{N}$ . The following is an upper bound on $N_s$ .

(3) ……………. $\displaystyle N_s \le 2^m \times \sqrt{N}$

The inequality (3) holds for all positive integers $N$ . We would like to find an $N$ such that $2^m \times \sqrt{N} \le \frac{N}{2}$ or $2^{m+1} \le \sqrt{N}$ . A good choice is $N=2^{2 m+2}$ . With this particular $N$ , inequality (3) becomes:

(4) ……………. $\displaystyle N_s \le 2^m \times \sqrt{N} \le \frac{N}{2}$

Summing (2) and (4) produces $N_b+N_s <N$ , which contradicts the fact that $N_s+N_b=N$ . Thus, we can conclude that the sum of the reciprocals of the primes must diverge.

Erdős conjecture on arithmetic progressions

A brief mention of the connection with a conjecture posed by Erdős. Erdős posed many mathematical problems. One problem related to the topic at hand is the conjecture on arithmetic progressions. It states that if the sum of the reciprocals of a sequence of positive integers diverges, then the sequence contains arbitrarily long arithmetic progressions. The sequence of primes satisfies the hypothesis of the conjecture. Are there arbitrarily long arithmetic progressions of primes? For example, 199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879, 2089 is a 10-term arithmetic progression of primes with difference 210. For each positive integer $k$ , are there arithmetic progression of primes of length $k$ ? This problem about primes would be a worthy conjecture on its own. This is actually an old problem that dates back to 1770. In the 20th century, incremental progresses included “there are infinitely many triples of primes in arithmetic progression” (1939) and “there are infinitely many four-term progressions consisting of three primes and a number that is either a prime or semiprime” (1981), according to Wolfram Math World. This conjecture was settled in 2004 by Ben Green and Terrence Tao. The Green_Tao Theorem states that the prime numbers contain arbitrary long arithmetic progressions.

Reference

Aigner, M., Gunter, M.Z., Proofs from THE BOOK, third edition, Springer-Verlag, Berlin, 2004.

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Revisiting example 106 from Steen and Seebach

Posted on December 25, 2021 by Dan Ma

The example 106 from Counterexamples in Topology by Steen and Seebach [3] is the space $\omega_1 \times I^I$ where the first factor $\omega_1$ is the space of countable ordinals with the usual order topology and the second factor $I^I$ is the product of continuum many copies of the unit interval $I=[0,1]$ .

This space was previously discussed in this site. One of the key results from that discussion is that $\omega_1 \times I^I$ is not normal, a result not shown in Steen and Seebach. The proof that was given in this site (see here) is based on an article published in 1976 [1], long before the publication date of the first edition of Steen and Seebach in 1970. It turns out that the non-normality of $\omega_1 \times I^I$ was given as an exercise in Steen and Seebach in the problem section at the end of the book (problem 127 in page 211, Dover edition). Problem 127: Show that $[0, \Omega) \times I^I$ is not normal. This indicates that the result not shown in Steen and Seebach was because it was given as a problem and not because the tool for solving it was not yet available. The fact that it is given as an exercise also means that there is a more basic proof of the non-normality of $\omega_1 \times I^I$ . So, once this is realized, I set out to find a simpler proof or at least one that does not rely on the result from [1]. Interestingly, this proof brings out a broader discussion that is worthwhile and goes beyond the example at hand. The goal here is to examine the more basic proof and the broader discussion.

A Classic Example

Before talking about the promised proof, we consider the product of $\omega_1$ and its immediate successor.

As noted at the beginning, the space $\omega_1$ is the set of all countable ordinals with the order topology. The ordinal $\omega_1+1$ is the immediate successor of $\omega_1$ . It can be regarded as the result of adding one more point to $\omega_1$ . The extra point is $\omega_1$ , i.e., $\omega_1 +1=\omega_1 \cup \{ \omega_1 \}$ with $\omega_1$ greater than all points $\beta < \omega_1$ . The ordinal $\omega_1+1$ with the order topology is a compact space. Using interval notation, $\omega_1=[0, \omega_1)$ and $\omega_1+1=[0, \omega_1]$ . As ordinals, $\omega_1$ is the first uncountable ordinal and $\omega_1+1$ is the first uncountable successor ordinal. For more information, see here.

The product $[0, \omega_1) \times [0, \omega_1]$ is a classic example of a product of a normal space (the first factor) and a compact space (the second factor) that is not normal. This example and others like it show that normality is easily broken upon taking product even if one of the factors is as nice as a compact space. The non-normality of $[0, \omega_1) \times [0, \omega_1]$ is discussed here. In that proof, two disjoint closed sets $H$ and $K$ are given such that they cannot be separated by disjoint open sets. The $H$ and $K$ are:

$H=\{ (\alpha, \alpha): \alpha < \omega_1 \}$

$K=\{(\alpha, \omega_1): \alpha < \omega_1 \}$

The Basic Proof

To show that $\omega_1 \times I^I$ is not normal, we show that one of its closed subspaces is not normal. That closed subspace is $[0, \omega_1) \times [0, \omega_1]$ . To this end, we show that $[0, \omega_1]$ can be embedded in the product space $I^I$ . With a non-normal closed subspace, it follows that $\omega_1 \times I^I$ is not normal. The remainder of the proof is to give the embedding.

We show that $[0, \omega_1]$ can be embedded as a closed subspace of $I^{\omega_1}$ , the product of $\omega_1$ many copies of $I$ . This means that $[0, \omega_1]$ is also a closed subspace of $I^I$ .

For each $\beta < \omega_1$ , define $T_\beta: \omega_1 \rightarrow I$ as follows:

$T_\beta(\gamma) = \begin{cases} 1 & \ \ \ \mbox{if } \gamma < \beta \\ 0 & \ \ \ \mbox{if } \gamma \ge \beta \end{cases}$

Furthermore, define $T: \omega_1 \rightarrow I$ by letting $T(\beta)=1$ for all $\beta < \omega_1$ . Consider the correspondence $\beta \rightarrow T_\beta$ with $\beta < \omega_1$ and $\omega_1 \rightarrow T$ . The mapping is clearly one-to-one from $[0, \omega_1]$ onto $\{ T_\beta: \beta < \omega_1 \} \cup \{ T \}$ . Upon closer inspection, the mapping in each direction is continuous (this is a good exercise to walk through). Thus, the mapping is a homeomorphism. It follows that $[0, \omega_1]$ can be considered a subspace of $I^{\omega_1}$ . Since $[0, \omega_1]$ is compact, it must be a closed subspace. With the cardinality of $\omega_1$ being less than or equal to continuum, it follows that $[0, \omega_1]$ can be embedded as a closed subspace of $I^I$ .

Stone-Cech Compactification

The first broader discussion is that of Stone-Cech compactification. More specifically, $\beta \omega_1=\omega_1+1$ , i.e., the Stone-Cech compactification of the first uncountable ordinal is its immediate successor.

To see that $\beta \omega_1=\omega_1+1$ , note that every continuous function defined on $[0,\omega_1)$ is bounded and is eventually constant (see result B here). As a result, every continuous function defined on $[0,\omega_1)$ can be extended to a continuous function defined on $[0, \omega_1]$ . For any continuous function $f: \omega_1 \rightarrow \mathbb{R}$ , we can simply define $f(\omega_1)$ to be the eventual constant value. A subspace $W$ of a space $Y$ is $C^*$ -embedded in $Y$ if every bounded continuous real-valued function on $W$ can be extended to $Y$ . According to theorem 19.12 in [4], if $Y$ is a compactification of $X$ and if $X$ is $C^*$ -embedded in $Y$ , then $Y$ is the Stone-Cech compactification of $X$ . Thus $[0,\omega_1)$ is $C^*$ -embedded in $[0,\omega_1]$ and $[0,\omega_1]$ is the Stone-Cech compactification of $[0,\omega_1)$ . In this instance, the Stone-Cech compactification agrees with the one-point compactification. Consider the following class theorem about normality in product space. The theorem is Corollary 3.4 in the chapter on products of normal spaces in the handbook of set-theoretic topology [2].

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

The space $X$ is paracompact.
The product space $X \times \beta X$ is normal.

Based on the discussion presented above, the non-normality of $\omega_1 \times I^I$ is due to the non-normality of $[0, \omega_1) \times [0, \omega_1]$ . Based on this theorem, the non-normality of $[0, \omega_1) \times [0, \omega_1]$ is due to the non-paracompactness of $[0, \omega_1)$ . See result G here for a proof that $[0, \omega_1)$ is not paracompact.

The discussion up to this point points to two ways to prove that $\omega_1 \times I^I$ is not normal. One way is the basic proof indicated above. The other way is to use Theorem 1, along with the homeomorphic embedding from $[0, \omega_1]$ into $I^I$ , the fact that $\beta \omega_1=\omega_1+1$ and the fact that $[0, \omega_1)$ is not paracompact. Both are valuable. The first way is basic and is a constructive proof. Because it is more hands-on, it is a better proof to learn from. The second way provides a broader perspective that is informative but requires quoting a couple of fairly deep results. Perhaps it is best used as a second proof for perspective.

Countable Tightness

The essence of the basic proof above goes like this: if the space $Y$ contains a copy of $\omega_1+1=[0, \omega_1]$ , then the product space $[0, \omega_1) \times Y$ is not normal. The contrapositive statement would be the following:

Corollary
Let $Y$ be a space. If the product space $\omega_1 \times Y$ is normal, then $Y$ cannot contain a copy of $\omega_1+1$ .

In the space of $\omega_1+1=[0, \omega_1]$ , note the following about the last point: $\omega_1 \in \overline{[0, \omega_1)}$ but $\omega_1 \notin \overline{C}$ for any countable $C \subset [0, \omega_1)$ , i.e., the last point is the limit point of the set of all the points preceding it but is not in the closure of any countable set. This means that the space $\omega_1+1=[0, \omega_1]$ does not have countable tightness (or is not countably tight). See here for definition. The property of countable tightness is hereditary. If $Y$ contains a copy of $\omega_1+1$ , then $Y$ is not countably tight (or is uncountably tight). This brings us to the following theorem.

Theorem 2
Let $Y$ be an infinite compact space. Then $\omega_1 \times Y$ is normal if and only if $Y$ has countable tightness.

Whenever we consider the normality of a product with the first factor being $\omega_1$ and the second factor being a compact space, the real story is the tightness of that compact space. If the tightness is countable, the product is normal. Otherwise, the product is not normal. The theorem is another reason that $\omega_1 \times I^I$ is not normal. Instead of embedding $[0, \omega_1]$ into $I^I$ , we can actually show that $I^I$ does not have countable tightness. This is the approach that was taken in this previous post.

Theorem 2 is the result from 1976 alluded to earlier [1]. A proof of Theorem 2 is found in this previous post. For results concerning normality in a product space with a compact factor (the other factor does not have to be $\omega_1$ ), see the chapter on products of normal spaces in the handbook of set-theoretic topology [2].

Reference

Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976.
Przymusinski T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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Making sense of the spaces with small diagonal

Posted on January 26, 2021 by Dan Ma

This is a small attempt toward making sense of the spaces with small diagonal. What makes the property of small diagonal interesting is the longstanding open problem that is discussed below. We use examples to help make sense of the definition of small diagonal. Then we discuss briefly the open problem regarding small diagonal. We discuss the following three examples of compact spaces:

$\omega_1 + 1$ with the ordered topology
The one-point compactification of a discrete space of cardinality $\omega_1$
The double arrow space

By a space we mean a topological space that is Tychonoff, i.e. Hausdorff and completely regular (defined here). Let $X$ be a space. The subset $\Delta=\{ (x,x): x \in X \}$ of the square $X^2=X \times X$ is called the diagonal of the space $X$ .

We want to focus on two diagonal properties. The space $X$ is said to have a $G_\delta$ -diagonal if $\Delta$ is a $G_\delta$ -set in $X^2$ , i.e. $\Delta$ is the intersection of countably many open subsets of $X^2$ . The space $X$ is said to have a small diagonal if for each uncountable subset $A$ of $X^2 \backslash \Delta$ , there is an open subset $O$ of $X^2$ such that $\Delta \subset O$ and $O$ misses uncountably many points of $A$ .

How do these two diagonal properties relate? Any space that has a $G_\delta$ -diagonal also has a small diagonal. This fact can be worked out quite easily based on the definitions. The opposite direction is a totally different matter. In fact, the question of whether having a small diagonal implies having a $G_\delta$ -diagonal is related to a longstanding open question.

Let’s focus on compact spaces. A classic metrization theorem for compact spaces states that any compact space with a $G_\delta$ -diagonal is metrizable (see here). A natural question is: if a compact space has a small diagonal, must it be metrizable? Indeed this is a well known open problem, a longstanding problem that has not completely resolved completely. Before discussion the open problem, we look at the three examples indicated above.

The three examples are all non-metrizable compact spaces and thus do not have a $G_\delta$ -diagonal. We show that they also not have a small diagonal. To appreciate the definition of small diagonal, it is helpful to look spaces that do not have a small diagonal.

Example 1

Consider $X=\omega_1+1$ with the order topology. Using interval notion, $X=[0,\omega_1]$ . This is the space of all countable ordinals $\alpha<\omega_1$ plus the point $\omega_1$ at the end. This is a compact space that is not metrizable. For example, any compact metrizable space would be separable. The last point $\omega_1$ cannot be in the closure of any countable subset. According to the classic theorem mentioned above, $X=[0,\omega_1]$ cannot have a $G_\delta$ -diagonal. We show that it does have a small diagonal too.

Let $A=\{ (\gamma, \gamma+1): \gamma< \omega_1 \}$ . Note that any open set of the form $(\alpha, \omega_1] \times (\alpha, \omega_1]$ contains the point $(\omega_1, \omega_1)$ and contains all but countably many points of $A$ . As a result, any open set containing the diagonal $\Delta$ contains all but countably many points of $A$ . This violates the definition of having a small diagonal. Thus the space $X=[0,\omega_1]$ does not have a small diagonal.

More on the Definition

Example 1 suggests a different angle in looking at the definition. The set $A=\{ (\gamma, \gamma+1): \gamma< \omega_1 \}$ is a $\omega_1$ -length sequence convergent to the diagonal, meaning that any open set containing the diagonal contains a tail of the sequence (in this case containing all but countably many elements in the sequence). The convergent sequence view point is the way Husek defined small diagonal [7].

In [7], $X$ is said to have an $\omega_1$ -accessible diagonal if there is an $\omega_1$ -length sequence $\{ (x_\alpha, y_\alpha) \in X^2 \backslash \Delta: \alpha < \omega_1 \}$ that converges to the diagonal $\Delta$ , meaning that any open set containing $\Delta$ contains all but countably many terms in the sequence. This is exactly what is occurring in the space $X=[0,\omega_1]$ . The set $A=\{ (\gamma, \gamma+1): \gamma< \omega_1 \}$ is precisely a convergent sequence converging to the diagonal. The space $X=[0,\omega_1]$ has an $\omega_1$ -accessible diagonal.

Seen in this light, spaces with a small diagonal are the the spaces that do not have an $\omega_1$ -accessible diagonal (or spaces that have an $\omega_1$ -inaccessible diagonal). Though the definition of Husek is more descriptive, the term small diagonal, suggested by E. van Douwen, has become more popular. The small diagonal defined above is more positive sounding. For example, we say the space $X$ has a small diagonal. In Husek [7], the spaces of interest would be the spaces without an $\omega_1$ -accessible diagonal. This definition is a negative one (defined by the lack of certain thing) and takes more syllables to express. Personally speaking, we prefer the term small diagonal though we understand that $\omega_1$ -accessible diagonal is more descriptive.

A slightly different (but equivalent) way of stating the definition of a space with a small diagonal: a space $X$ has a small diagonal if for any uncountable $A \subset X^2 \backslash \Delta$ , there is an uncountable $B \subset A$ such that $\Delta \cap \overline{B}=\varnothing$ . In the literature, the definition of a space with a small diagonal is usually either this one or the one given at the beginning of this post.

Example 2

Now consider the space $X=\omega_1 \cup \{ \infty \}$ where points in $\omega_1$ are isolated and open neighborhoods of the point $\infty$ are of the form $\{ \infty \} \cup (\omega_1 \backslash F)$ with $F$ being any finite subset of $\omega_1$ . This is usually called the one-point compactification of a discrete space (in this case of size $\omega_1$ ). This space is compact. It is also non-metrizable since it has an uncountable discrete subset. Thus it cannot have a $G_\delta$ -diagonal. The weight of $X$ here is $\omega_1$ . Any compact space whose weight is $\omega_1$ cannot have a small diagonal. This is Fact 1 below.

Some Basic Results

Fact 1
Let $X$ any compact space with $w(X)=\omega_1$ . Then $X$ does not have a small diagonal.

Proof of Fact 1
Let $\mathcal{B}=\{ B_\alpha: \alpha < \omega_1 \}$ be a base for $X$ . Let $\mathcal{B}_1=\{ U \times V: U, V \in \mathcal{B} \}$ , which is a base for $X^2$ . Define $\mathcal{U}=\{ \cup F: F \subset \mathcal{B}_1 \text{ and } \lvert F \lvert < \omega \text{ and } \Delta \subset \cup F \}$ . Since $X$ is compact, the diagonal $\Delta$ is compact. As a result, $\mathcal{U} \ne \varnothing$ . Furthermore, $\lvert \mathcal{U} \lvert=\omega_1$ .

We claim that $\mathcal{U}$ is a base for the diagonal $\Delta$ . To show this, let $W$ be an open subset of $X^2$ such that $\Delta \subset W$ . We can assume that $W$ is the union of elements of the base $\mathcal{B}_1$ . Let $W=\cup \mathcal{A}$ for some $\mathcal{A} \subset \mathcal{B}_1$ . Since $\Delta$ is compact, $\Delta \subset \cup \mathcal{A}_0$ for some finite $\mathcal{A}_0 \subset \mathcal{A}$ . Note that $\cup \mathcal{A}_0 \in \mathcal{U}$ and that $\Delta \subset \cup \mathcal{A}_0 \subset W$ .

Enumerate $\mathcal{U}$ as $\mathcal{U}=\{U_\alpha: \alpha<\omega_1 \}$ . Since $X$ is compact and has uncountable weight, $X$ is not metrizable. Hence the diagonal $\Delta$ is not a $G_\delta$ -set. As a result, for each $\alpha<\omega_1$ , $(\bigcap_{\beta<\alpha} U_\beta) \backslash \Delta \ne \varnothing$ . Furthermore, for any countable $\{ y_0, y_1, y_2, \cdots \} \subset X^2 \backslash \Delta$ , $[(\bigcap_{\beta<\alpha} U_\beta) \cap (\bigcap_{n \in \omega} X^2 \backslash \{ y_n \}) ]\backslash \Delta \ne \varnothing$ .

Pick $x_0 \in U_0$ . For any $\alpha<\omega_1$ with $\alpha > 0$ , choose $x_\alpha \in (\bigcap_{\beta<\alpha} U_\beta) \backslash (\Delta \cup \{x_\delta: \delta < \alpha \})$ .

Then the sequence $\{ x_\alpha: \alpha < \omega_1 \}$ converges to the diagonal $\Delta$ . To see this, fix $U_\gamma \in \mathcal{U}$ . From the way the sequence is chosen, $x_\alpha \in U_\gamma$ for all $\alpha > \gamma$ . This concludes the proof that $X$ does not have a small diagonal. $\square$

Fact 2
Let $X$ any compact space with $w(X) \le \omega_1$ . Then if $X$ has a small diagonal, then $w(X)=\omega$ and thus $X$ is metrizable.

Fact 2 is an easy corollary of Fact 1. Fact 2 says that any compact space with “small” weight (no more than $\omega_1$ ) is metrizable if it has a small diagonal. Both Example 1 and Example 2 are compact non-metrizable spaces with small weight. Therefore they do not have small diagonal. The following basic fact is also useful.

Fact 3
Let $X$ any compact space. Then if $X$ has a small diagonal, then $t(X)=\omega$ , i.e. $X$ is countably tight.

Proof of Fact 3
Suppose $X$ is uncountably tight. We show that it does not have a small diagonal. A sequence of points $\{ x_\alpha \in X: \alpha< \tau \}$ is a free sequence of length $\tau$ if for each $\alpha< \tau$ , $\overline{\{ x_\gamma: \gamma < \alpha \}} \cap \overline{\{ x_\gamma: \gamma \ge \alpha \}}=\varnothing$ . For any compact space with tightness at least $\omega_1$ , there exists a free sequence of length $\omega_1$ (see Lemma 2 here). Thus in the space $X$ in question, there exists a free sequence $\{ x_\alpha \in X: \alpha< \omega_1 \}$ .

The main result in [9] says that if a compact space contains a free sequence of length $\omega_1$ , then it contains a free sequence of the same length that is convergent. We now assume that the above free sequence $\{ x_\alpha \in X: \alpha< \omega_1 \}$ is also convergent, i.e. it converges to some point $x \in X$ . Thus every open set containing $x$ contains all but countably many $x_\alpha$ . Consider the sequence $\{ (x_\alpha, x_{\alpha+1}) \in X^2: \alpha< \omega_1 \}$ . Observe that every open set in $X^2$ containing the point $(x,x)$ contains all but countably pairs $(x_\alpha, x_{\alpha+1})$ . This implies that every open set containing the diagonal $\Delta$ contains all but countably many points in the sequence. This shows that the compact $X$ does not have a small diagonal. $\square$

According to Fact 3, the compact ordinal $\omega_1+1=[0, \omega_1]$ does not have a small diagonal since it is uncountably tight at the last point $\omega_1$ .

Example 3

We now consider the double arrow space. Let $D=[0,1] \times \{0, 1 \}$ . The space $D$ consists of two copies of the unit interval, an upper one and a lower one. See the first two diagrams in this previous post. For $0 \le a <1$ , a basic open set containing the point $(a, 1)$ in the upper interval is of the form $\biggl( [a,b) \times \{ 1 \} \biggr) \cup \biggl( (a,b) \times \{0 \} \biggr)$ . For $0<a \le 1$ , a basic open set containing the point $(a, 0)$ in the lower interval is of the form $\biggl( (c,a) \times \{ 1 \} \biggr) \cup \biggl( (c,a] \times \{0 \} \biggr)$ . The rightmost point in the upper interval $(1,1)$ and the leftmost point in the lower interval $(0,0)$ are made isolated points.

The double arrow space $D$ is compact, perfectly normal and not metrizable (discussed here). Thus $D$ does not have a $G_\delta$ -diagonal. We do not have a direct way of showing that it does not have a small diagonal. We rely on a result from [5], which says that every compact metrizably fibered space with a small diagonal is metrizable. In light of this result, we only need to show that the double arrow space $D$ is metrizably fibered. A space $Y$ is metrizably fibered if there is a continuous map from $Y$ onto some metrizable space $M$ such that each point inverse is metrizable.

Starting with the double arrow space $D$ , let $f:D \rightarrow [0,1]$ be defined by $f((a,0))=f((a,1))=a$ for each $0 \le a \le 1$ . This is a two-to-one continuous map from the double arrow space $D$ onto the unit interval. The map $f$ is essentially a quotient map, the result of identifying $\{ (a,0), (a, 1) \}$ as one point $a$ . By the result in [5], the double arrow space $D$ cannot have a small diagonal.

The Open Problem

As mentioned earlier, what makes the property of small diagonal is a related longstanding open problem. The statement “every compact space with a $G_\delta$ -diagonal is metrizable” is true in ZFC. What about the following statement?

(*) Every compact space with a small diagonal is metrizable.

This question whether this statement is true was raised in Husek [7]. As of the writing of this article, the problem is still unsolved. There are partial results, some consistent results and some ZFC results.

Assuming CH, the answer to Hesek’s question is positive. Husek [7] showed that under CH, every compact space $X$ with a small diagonal such that the tightness of $X$ is countable is metrizable. Fact 3 from above states that every compact space with a small diagonal has countble tightness. Combining the two results, it follows that under CH any compact space with a small diagonal is metrizable. Fact 3 followed from a result by Juhasz and Szentmikloss [9]. Dow and Pavlov [3] showed that under PFA, every compact space with a small diagonal is metrizable.

There are partial answers in ZFC. We mention three results. The first one is Fact 2 discussed above, which is that compact spaces with small diagonal are metrizable if there is a weight restriction (weight no more than $\omega_1$ ). If there exists a non-metrizable compact space with a small diagonal, its weight would have to be greater than $\omega_1$ .

Gruenhage [5] showed that every compact metrizably fibered space with a small diagonal is metrizable (this result is discussed in Example 3 above). Dow and Hart [2] showed that every compact space with a small diagonal that is weight $\omega_1$ fibered is metrizable. The notion of weight $\omega_1$ fibered is a generalization of metrizably fibered. A space $X$ is weight $\omega_1$ fibered if there is a continuous surjection $f: X \rightarrow Y$ such that $Y$ and each point inverse $f^{-1}(y)$ have weight at most $\omega_1$ .

The statement “every countably compact space with a $G_\delta$ -diagonal is metrizable” is a true statement in ZFC (see here). How about the statement “every countably compact space with a small diagonal is metrizable”, the statement (*) above with compact replaced by countably compact? Gruenhage [5] showed that this statement is consistent with and independent of ZFC. To prove or disprove this statement extra set theory assumptions beyond ZFC are required. This provides an interesting contrast between compact and countably compact with respect to the open problem. By broadening Husek’s question from compact to countably compact, the statement cannot be settled in ZFC. Dow and Pavlov [3] provided two consistent examples of “countably compact non-metrizable with small diagonal” that are improvements upon examples from Gruenhage.

Now consider the statement “Lindelof space with a small diagonal must have a $G_\delta$ -diagonal.” Dow and Pavlov [3] provided a consistent negative answer, an example of a Lindelof space with a small diagonal that does not have a $G_\delta$ -diagonal (under negation of CH).

Another broad natural question is: what do compact spaces with small diagonal look like? The main problem is, of course, trying to see if these spaces are metrizable. There have been attempts to explore this general question of what these spaces look like. According to Juhasz and Szentmikloss [9], these spaces have countable tightness (Fact 2 above). However, it is not known if compact spaces with small diagonal have points of countable character. Dow and Hart [2] uncovered a surprising connection that if there is a subset of the real line that is a Luzin set, every compact space with a small diagonal does have points of countable character. Dow and Hart in the same paper also showed that in every compact space with a small diagonal, CCC subspaces have countable $\pi$ -weight.

This is a brief walk through of the open problem based on the statement (*) indicated above. To find out more, consult with the references listed below. Beyond the main open problem, there are many angles to be explored.

Reference

Arhangelskii, A., Bella, A., Few observations on topological spaces with small diagonal, Zb. Rad. Filoz. Fak. Nisu, 6, No. 2, 211-213, 1992.
Dow, A., Hart, P. Elementary chains and compact spaces with a small diagonal, Indagationes Mathematicae, 23, No. 3, 438-447, 2012.
Dow, A., Pavlov, O. More about spaces with a small diagonal, Fund. Math., 191, No. 1, 67-80, 2006.
Dow, A., Pavlov, O. Perfect preimages and small diagonal, Topology Proc., 31, No. 1, 89-95, 2007.
Gruenhage, G., Spaces having a small diagonal, Topology Appl., 22, 183-200, 2002.
Gruenhage, G., Generalized metrizable spaces, Recent Progress in General Topology III (K.P. Hart, J. van Mill, and P. Simon, eds.), Atlantis Press 2014.
Husek, M., Topological spaces without $\kappa$ -accessible diagonal, Comment. Math. Univ. Carolin., 18, No. 4, 777-788, 1977.
Juhasz, I., Cardinals Functions II, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 63-109, 1984.
Juhasz, I., Szentmikloss, Z. Convergent free sequences in compact spaces, Proc. Amer. Math. Soc., 116, No. 4, 1153-1160, 1992.
Zhou, H. X., On The Small Diagonals, Topology Appl., 13, 283-293, 1982.

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The Cichon’s Diagram

Posted on March 22, 2020 by Dan Ma

The Cichon’s Diagram is a diagram that shows the relationships among ten small cardinals – four cardinals associated with the $\sigma$ -ideal of sets of Lebesgue measure zero, four cardinals associated with the $\sigma$ -ideal of sets of meager sets, the bounding number $\mathfrak{b}$ , and the dominating number $\mathfrak{d}$ . What makes this interesting is that elements of analysis, topology and set theory flow into the same spot. Here’s the diagram.

Figure 1 – The Cichon’s Diagram

In this diagram, $\alpha \rightarrow \beta$ means $\alpha \le \beta$ . The preceding three posts (the first post, the second post and the third post) give the necessary definitions and background to understand the diagram. In addition to the above diagram, the following relationships also hold.

Figure 2 – The Cichon’s Diagram – Additional Relationships

The Cardinal Characteristics of a $\sigma$ -Ideal

For any $\sigma$ -ideal $\mathcal{I}$ on a set $X$ , there are four associated cardinals – $\text{add}(\mathcal{I})$ , $\text{non}(\mathcal{I})$ , $\text{cov}(\mathcal{I})$ and $\text{cof}(\mathcal{I})$ . The first one is the additivity number, which is the least number of elements of $\mathcal{I}$ whose union is not an element of $\mathcal{I}$ . The second cardinal is called the uniformity number, which is the least cardinality of a subset of $X$ that is not an element of $\mathcal{I}$ . The third cardinal is called the covering number, which is the least cardinality of a subfamily of $\mathcal{I}$ that is also a covering of $X$ . The fourth cardinal is called the cofinality number, which is the least cardinality of a subfamily of $\mathcal{I}$ that is cofinal in $\mathcal{I}$ . For more information, see the first post. The four cardinals are related in a way that is depicted in the following diagram. Again, $\alpha \Rightarrow \beta$ means $\alpha \le \beta$ .

Figure 3 – Cardinal Characteristics of a $\sigma$ -Ideal

…Cichon… $\displaystyle \begin{array}{ccccccccccccc} \text{ } & \text{ } & \text{ } &\text{ } & \bold n \bold o \bold n ( \mathcal{I} ) &\text{ } &\Longrightarrow &\text{ } &\bold c \bold o \bold f (\mathcal{I} )&\text{ } &\Longrightarrow & \text{ } & \lvert \mathcal{I} \lvert\\ \text{ } & \text{ } & \text{ } &\text{ } & \text{ } &\text{ } &\text{ } &\text{ } &\text{ }&\text{ } &\text{ } & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } &\text{ } & \Uparrow &\text{ } &\text{ } &\text{ } &\Uparrow&\text{ } &\text{ } & \text{ }& \text{ } \\ \text{ }& \text{ } & \text{ } &\text{ } & \text{ } &\text{ } &\text{ } &\text{ } &\text{ }&\text{ } &\text{ } & \text{ }& \text{ } \\ \aleph_1 & \text{ } & \Longrightarrow & \text{ } &\bold a \bold d \bold d ( \mathcal{I} ) & \text{ } &\Longrightarrow & \text{ } &\bold c \bold o \bold v ( \mathcal{I} ) &\text{ } &\text{ } &\text{ } & \text{ } \end{array}$

Figure 3 explains the basic orientation of the Cichon’s Diagram. Filling it with three $\sigma$ -ideals produces the Cichon’s Diagram.

The Three $\sigma$ -Ideals in the Cichon’s Diagram

Let $\mathcal{K}$ be the $\sigma$ -ideal of bounded subsets of $\omega^\omega$ . It is known that $\mathfrak{b}=\text{add}(\mathcal{K})=\text{non}(\mathcal{K})$ (this is called the bounding number) and $\mathfrak{d}=\text{cov}(\mathcal{K})=\text{cof}(\mathcal{K})$ (this is called the dominating number). The ideal $\mathcal{K}$ is discussed in this previous post. Let $\mathcal{M}$ be the $\sigma$ -ideal of meager subsets of the real line $\mathbb{R}$ (this is discussed in this previous post). Let $\mathcal{L}$ be the $\sigma$ -ideal of Lebesgue measure zero subsets of the real line.

Thus the Cichon’s Diagram (Figure 1 above) houses information about three $\sigma$ -ideals. The two numbers for the $\sigma$ -ideal $\mathcal{K}$ are situated in the middle of the diagram ( $\mathfrak{b}$ and $\mathfrak{d}$ ). The four numbers for the $\sigma$ -ideal $\mathcal{M}$ are situated in the center portion of the diagram. The four numbers for the $\sigma$ -ideal $\mathcal{L}$ are located on the left side and the right side. The Cichon’s Diagram (Figure 1) is flanked by $\aleph_1$ on the lower left and by continnum $2^{\aleph_0}$ on the upper right.

More on the Cichon’s Diagram

One interesting aspect of the Cichon’s Diagram: it is a small diagram with small cardinals where elements of analysis (measure) and topology (category) come together. The following diagram shows the path that includes both the bounding number and the dominating number.

Figure 4 – The Cichon’s Diagram – The Main Path

The path circled in the above diagram involves all three $\sigma$ -ideals. It is also one of the longest increasing paths in the diagram.

$\aleph_1 \le \text{add}(\mathcal{L}) \le \text{add}(\mathcal{M}) \le \mathfrak{b} \le \mathfrak{d} \le \text{cof}(\mathcal{M}) \le \text{cof}(\mathcal{L}) \le 2^{\aleph_0}$

There are fifteen arrows in Figure 1. The proofs of these arrows (or inequalities) require varying degrees of effort. Three are basic information – $\aleph_1 \le \text{add}(\mathcal{L})$ , $\mathfrak{b} \le \mathfrak{d}$ and $\text{cof}(\mathcal{L}) \le 2^{\aleph_0}$ . Because $\mathcal{L}$ is a $\sigma$ -ideal, its additivity number must be uncountable. By definition, $\mathfrak{b} \le \mathfrak{d}$ . The $\sigma$ -ideal $\mathcal{L}$ has a cofinal subfamily consisting of Borel sets. Thus $\text{cof}(\mathcal{L}) \le 2^{\aleph_0}$ .

Four of the arrows follow from the relative magnitude of the four cardinals of a $\sigma$ -ideal as shown in Figure 3 – $\text{add}(\mathcal{L}) \le \text{cov}(\mathcal{L})$ , $\text{non}(\mathcal{L}) \le \text{cof}(\mathcal{L})$ , $\text{add}(\mathcal{M}) \le \text{cov}(\mathcal{M})$ and $\text{non}(\mathcal{M}) \le \text{cof}(\mathcal{M})$ .

Three of the arrows are proved in this previous post – $\mathfrak{b} \le \text{non}(\mathcal{M})$ , $\mathfrak{d} \le \text{cov}(\mathcal{M})$ and $\text{add}(\mathcal{M}) \le \mathfrak{b}$ . The last inequality follows from this fact: if $F \subset \omega^\omega$ is an unbounded set, then there exist $\lvert F \lvert$ many meager subsets of the real line whose union is a non-meager set, essentially a result in Miller [8].

The proofs of the remaining five arrows can be found in [3] – $\mathfrak{d} \le \text{cof}(\mathcal{M})$ , $\text{add}(\mathcal{L}) \le \text{add}(\mathcal{M})$ , $\text{cov}(\mathcal{L}) \le \text{non}(\mathcal{M})$ , $\text{cof}(\mathcal{M}) \le \text{cof}(\mathcal{L})$ and $\text{cov}(\mathcal{M}) \le \text{non}(\mathcal{L})$ . The proofs of two additional relationships displayed in Figure 2 can also be found in [3].

The fifteen arrows in the Cichon’s Diagram represent the only inequalities among the ten cardinals (not counting $\aleph_1$ and $2^{\aleph_0}$ ) that are provable in ZFC [1] and [5]. As illustration, we give an example of non-ZFC provable relation in the next section.

An Example of an Inequality Not Provable in ZFC

In the following diagram, the cardinals $\mathfrak{b}$ and $\text{cov}(\mathcal{M})$ are encircled. These two numbers are not connected by arrows.

Figure 5 – The Cichon’s Diagram – An Example of Non-ZFC Provable

We sketch out a proof that no inequalities can be established between $\mathfrak{b}$ and $\text{cov}(\mathcal{M})$ . First Martin’s Axiom (MA) implies that $\mathfrak{b} \le \text{cov}(\mathcal{M})$ . Topologically, the statement MA ( $\kappa$ ) means that any compact Hausdorff space $X$ that satisfies the countable chain condition cannot be the union of $\kappa$ or fewer many nowhere dense sets. The Martin’s Axiom (MA) is the statement that MA ( $\kappa$ ) holds for all $\kappa$ less than $2^{\aleph_0}$ . It follows that MA implies that $\text{cov}(\mathcal{M})$ cannot be less than $2^{\aleph_0}$ and thus $\text{cov}(\mathcal{M})=2^{\aleph_0}$ . It is always the case that the bounding number $\mathfrak{b}$ is $\le 2^{\aleph_0}$ .

On the other hand, in Laver’s model [6] for the Borel conjecture, $\mathfrak{b} > \text{cov}(\mathcal{M})$ . In Laver’s model, every subset of the real line that is of strong measure zero is countable. Since any set with the Rothberger property is of strong measure zero, every subset of the real line that has the Rothberger property is countable in Laver’s model. Let $\text{non}(\text{Rothberger})$ be the least cardinality of a subset of the real line that does not have the Rothberger property. Thus in Laver’s model, $\text{non}(\text{Rothberger})=\aleph_1$ . It is well known that $\text{non}(\text{Rothberger})=\text{cov}(\mathcal{M})$ ; see Theorem 5 in [10]. Thus in Laver’s model, $\text{cov}(\mathcal{M})=\aleph_1$ .

In Laver’s model, $\mathfrak{b} > \aleph_1$ . Note that $\mathfrak{b}= \aleph_1$ implies that there is an uncountable subset of the real line that is concentrated about $\mathbb{Q}$ , the set of all rational numbers; see Theorem 10.2 in [12]. Any concentrated set is of strong measure zero; see Theorem 3.1 in [9]. Thus it must be the case that $\mathfrak{b} > \aleph_1=\text{cov}(\mathcal{M})$ in Laver’s model.

Remarks

The Cichon’s Diagram is a remarkable diagram. It blends elements of analysis and topology into a small diagram. The fifteen arrows shown in the diagram are obviously far from the end of the story. The Cichon’s Diagram had been around for a long time. Much had been written about it. The article [13] posted some questions about the diagram. See [1], [2], [4] and [11] for further information on the cardinals in the diagram.

Reference

Bartoszynski, T., Judah H., Shelah S.,The Cichon Diagram, J. Symbolic Logic, 58(2), 401-423, 1993.
Bartoszynski, T., Judah H., Shelah S.,Set theory: On the structure of the real line, A K
Peters, Ltd.. Wellesley, MA, 1995.
Blass, A., Combinatorial Cardinal Characteristics of the Continuum, Handbook of Set Theory (M. Foreman, A. Kanamori, eds), Springer Science+Business Media B. V., Netherlands, 395-489, 2010.
Fremlin, D. H., Cichon’s diagram. In Seminaire d’Initiation ´a l’Analyse, 23, Universite Pierre et Marie Curie, Paris, 1984.
Garcia, H., da Silva S. G., Identifying Small with Bounded: Unboundedness, Domination, Ideals and Their Cardinal Invariants, South American Journal of Logic, 2 (2), 425-436, 2016.
Laver, R., On the consistency of Borel’s conjecture, Acta Math., 137, 151-169, 1976.
Miller, A. W., Some Properties of Measure and Category, Trans. Amer. Math. Soc., 266 (1), 93-114, 1981.
Miller, A. W., A Characterization of the Least Cardinal for Which the Baire Category Theorem Fails, Proc. Amer. Math. Soc., 86 (3), 498-502, 1982.
Miller, A. W., Special Subsets of the Real Line, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 201-233, 1984.
Miller A. W., Fremlin D. H., On some properties of Hurewicz, Menger, and Rothberger, Fund. Math., 129, 17-33, 1988.
Pawlikowski, J., Reclaw I., Parametrized Cichon’s diagram and small sets, Fund. Math., 127, 225-239, 1987.
Van Douwen, E. K., The Integers and Topology, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 111-167, 1984.
Vaughn, J. E., Small uncountable cardinals and topology, Open Problems in Topology (J. van Mill and G.M. Reed, eds), Elsevier Science Publishers B.V. (North-Holland), 1990.

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The ideal of meager sets

Posted on March 22, 2020 by Dan Ma

This is the third in a series of four posts leading to a diagram called The Cichon’s Diagram. This post focuses on the $\sigma$ -ideal of meager subsets of the real line. The links to the previous posts: the first post and the second post.

The next post is: the Cichon’s Diagram.

The notion of meager sets can be defined on any topological space. Let $Y$ be a space. A subset $A$ of $Y$ is a nowhere dense set if $\overline{A}$ , the closure of $A$ in the space $Y$ , contains no open sets. Equivalently, $A$ is a nowhere dense set if for each non-empty open subset $U$ of $Y$ , there exists a non-empty open subset $V$ of $U$ such that $V \cap A=\varnothing$ . We can always find a part of any open set that misses a nowhere dense set. Thus nowhere dense sets are considered “thin” sets. A subset of $Y$ is said to be a meager set if it is the union of countably many nowhere dense sets. A meager set is also called a set of first category. A non-meager set is then called a set of second category.

Though the notion of meager sets can be considered in any space, we would like to focus on the real line $\mathbb{R}$ or the space of all irrational numbers $\mathbb{P}$ . Note that $\mathbb{P}$ is homeomorphic to $\omega^\omega$ (see here). Instead of working with $\mathbb{P}$ , we work with $\omega^\omega$ , which is the product space of countably many copies of the countable discrete space $\omega$ .

$\sigma$ -Ideal of Meager Sets

The notion of meager sets is a topological notion of small sets. The real line and the space of irrationals $\omega^\omega$ are “big” sets. This means that they are not the union of countably many meager sets (this fact is a consequence of the Baire category theorem). Let $\mathcal{M}$ be the set of all subsets of the real line that are meager sets. It is straightforward to verify that $\mathcal{M}$ is a $\sigma$ -ideal on the real line $\mathbb{R}$ . Because of the Baire category theorem, $\mathcal{M}$ is a proper ideal, i.e. $\mathbb{R} \notin \mathcal{M}$ . Naturally, we would like to consider the four cardinals associated with this ideal – $\text{add}(\mathcal{M})$ (the additivity number), $\text{cov}(\mathcal{M})$ (the covering number), $\text{non}(\mathcal{M})$ (the uniformity number) and $\text{cof} (\mathcal{M})$ (the cofinality number). These four numbers are displayed in the following diagram.

Figure 1 – Cardinal Characteristics of the $\sigma$ -Ideal of Meager Sets

In the above diagram, an arrow means $\le$ . So $\alpha \Rightarrow \beta$ means $\alpha \le \beta$ . The inequalities displayed in this diagram always hold for any $\sigma$ -ideal. The only inequality that requires explanation is $\text{cof} (\mathcal{M}) \le 2^{\aleph_0}$ . Any meager set is a subset of an $F_\sigma$ -set. To see this, let $A=\bigcup_{n \in \omega} X_n$ where each $X_n$ is a nowhere dense subset of the real line. Then $A \subset \bigcup_{n \in \omega} \overline{X_n}$ . Each $\overline{X_n}$ is also nowhere dense. Thus the set of all $F_\sigma$ nowhere dense sets is cofinal in $\mathcal{M}$ . This cofinal set has cardinality continuum. Of course, if continuum hypothesis holds ( $\aleph_1=2^{\aleph_0}$ ), then all four cardinals are identical and are $\aleph_1$ .

$\sigma$ -Ideal of Bounded Sets

In some respects, it is more advantageous to consider the $\sigma$ -ideal of meager subsets of $\mathbb{P}$ , the set of all irrational numbers, or equivalently $\omega^\omega$ . Thus we consider the $\sigma$ -ideal of meager subsets of $\omega^\omega$ . We also use $\mathcal{M}$ denote this $\sigma$ -ideal. Note that the calculation of the four cardinals $\text{add}(\mathcal{M})$ , $\text{cov}(\mathcal{M})$ , $\text{non}(\mathcal{M})$ and $\text{cof} (\mathcal{M})$ yields the same values regardless of whether $\mathcal{M}$ is the $\sigma$ -ideal of meager subsets of the real line or of $\omega^\omega$ . In the remainder of this post, $\mathcal{M}$ is the $\sigma$ -ideal of meager subsets of $\omega^\omega$ , the space of the irrational numbers.

Let $\mathcal{S}$ be the collection of all $\sigma$ -compact subsets of $\omega^\omega$ . In this previous post, the following $\sigma$ -ideal is discussed.

$\mathcal{K}=\{ A \subset \omega^\omega: \exists \ B \in \mathcal{S} \text{ such that } A \subset B \}$

It is straightforward to verify that $\mathcal{K}$ is indeed a $\sigma$ -ideal on $\omega^\omega$ . This is what we know about this $\sigma$ -ideal from this previous post.

$A \in \mathcal{K}$ if and only if $A$ is a bounded subset of $\omega^\omega$ .
$\mathfrak{b}=\text{add}(\mathcal{K})=\text{non}(\mathcal{K})$ .
$\mathfrak{d}=\text{cov}(\mathcal{K})=\text{cof}(\mathcal{K})$ .

So the sets in $\mathcal{K}$ are simply the bounded sets. For this $\sigma$ -ideal, the additivity number and the uniformity numbers are $\mathfrak{b}$ , the bounding number. The covering number and the cofinality number are the dominating number $\mathfrak{d}$ . As we will see below, these facts provide insight on the $\sigma$ -ideal $\mathcal{M}$ .

The following lemma connects the $\sigma$ -ideal $\mathcal{K}$ with the $\sigma$ -ideal $\mathcal{M}$ .

Lemma 1
Let $A$ be a compact subset of $\omega^\omega$ . Then $A$ is a closed and nowhere dense subset of $\omega^\omega$ . Hence any $\sigma$ -compact subset of $\omega^\omega$ is a meager subset of $\omega^\omega$ .

Proof of Lemma 1
Since $A$ is compact, for each $n$ , the projection of $A$ into the $n$ th factor of $\omega^\omega$ is compact and thus finite. Let $[0, g(n)]=\{ j \in \omega: 0 \le j \le g(n) \}$ be a finite set that contains the $n$ th projection of $A$ . Thus $A \subset \prod_{n \in \omega} [0, g(n)]$ . It is straightforward to verify that $\prod_{n \in \omega} [0, g(n)]$ is nowhere dense in $\omega^\omega$ . Thus $A$ is a closed nowhere dense subset of $\omega^\omega$ . It follows that any $\sigma$ -compact subset of $\omega^\omega$ is a meager subset of $\omega^\omega$ . $\square$

Theorem 2
As a result of Lemma 1, we have $\mathcal{K} \subset \mathcal{M}$ . However, $\mathcal{M} \not \subset \mathcal{K}$ . Thus the two $\sigma$ -ideals are not the same.

Any example that proves Theorem 2 would be an unbounded meager set. One such example is constructed in this previous post.

More on $\sigma$ -Ideal of Meager Sets

For $\mathcal{K}$ , the $\sigma$ -ideal generated by $\sigma$ -compact subsets of $\omega^\omega$ , and for $\mathcal{M}$ , the $\sigma$ -ideal of meager sets in $\omega^\omega$ , we are interested in the four associated cardinals add, non, cov and cof in each $\sigma$ -ideal. For $\mathcal{K}$ , the four cardinals are just two, $\mathfrak{b}$ and $\mathfrak{d}$ . We would like to relate these six cardinals, plus $\aleph_1$ and $2^{\aleph_0}$ . They are represented in the following diagram.

Figure 5 – Partial Cichon’s Diagram

…Cichon… $\displaystyle \begin{array}{ccccccccccccc} \text{ } & \text{ } & \text{ } &\text{ } & \bold n \bold o \bold n ( \mathcal{M} ) &\text{ } &\Longrightarrow &\text{ } &\bold c \bold o \bold f (\mathcal{M} )&\text{ } &\Longrightarrow & \text{ } & 2^{\aleph_0}\\ \text{ } & \text{ } & \text{ } &\text{ } & \text{ } &\text{ } &\text{ } &\text{ } &\text{ }&\text{ } &\text{ } & \text{ } & \text{ }\\ \text{ }& \text{ } & \text{ } &\text{ } & \Uparrow &\text{ } &\text{ } &\text{ } &\Uparrow &\text{ } &\text{ } & \text{ } & \text{ }\\ \text{ } & \text{ } & \text{ } &\text{ } & \text{ } &\text{ } &\text{ } &\text{ } &\text{ }&\text{ } &\text{ } & \text{ }& \text{ } \\ \text{ } & \text{ } & \text{ } &\text{ } & \mathfrak{b} &\text{ } &\Longrightarrow &\text{ } &\mathfrak{d}&\text{ } &\text{ } & \text{ }& \text{ } \\ \text{ }& \text{ } & \text{ } &\text{ } & \text{ } &\text{ } &\text{ } &\text{ } &\text{ }&\text{ } &\text{ } & \text{ }& \text{ } \\ \text{ } & \text{ } & \text{ } &\text{ } & \Uparrow &\text{ } &\text{ } &\text{ } &\Uparrow &\text{ } &\text{ } & \text{ } & \text{ }\\ \text{ } & \text{ } & \text{ } &\text{ } & \text{ } &\text{ } &\text{ } &\text{ } &\text{ }&\text{ } &\text{ } & \text{ }& \text{ } \\ \aleph_1 & \text{ } & \Longrightarrow & \text{ } &\bold a \bold d \bold d ( \mathcal{M} ) & \text{ } &\Longrightarrow & \text{ } &\bold c \bold o \bold v ( \mathcal{M} ) &\text{ } &\text{ } &\text{ } & \text{ } \end{array}$

As in the other diagrams, arrows mean $\le$ . So $\alpha \Longrightarrow \beta$ means $\alpha \le \beta$ . Furthermore, we have two additional relations.

Additional Relationships

…Cichon… $\text{add}(\mathcal{M})=\text{min}(\mathfrak{b},\ \text{cov}(\mathcal{M}))$
…Cichon… $\text{cof}(\mathcal{M})=\text{max}(\mathfrak{d},\ \text{non}(\mathcal{M}))$

As shown, Figure 5 is not complete. It only has information on the $\sigma$ -ideal $\mathcal{M}$ on meager sets. The usual Cichon’s diagram would also include the four associated cardinals for $\mathcal{L}$ , the $\sigma$ -ideal of Lebesgue measure zero sets. In this post we focus on $\mathcal{M}$ . The full Cichon’s diagram will be covered in a subsequent post. insert

In Figure 5, the cardinals go from smaller to the larger from left to right and bottom to top. It starts with $\aleph_1$ on the lower left and moves toward the continuum on the upper right. Because of Theorem 3, the cardinals associated with the $\sigma$ -ideal $\mathcal{K}$ are represented by $\mathfrak{b}$ and $\mathfrak{d}$ in the diagram. We next examine the inequalities between the cardinals associated with $\mathcal{M}$ and $\mathfrak{b}$ and $\mathfrak{d}$ .

There are four inequalities to account for. First, $\mathfrak{b} \le \text{non}(\mathcal{M})$ and $\text{cov}(\mathcal{M}) \le \mathfrak{d}$ . The first inequality follows from the fact that $\mathfrak{b} = \text{non}(\mathcal{K})$ and that $\mathcal{K} \subset \mathcal{M}$ . The second inequality follows from the fact that $\mathfrak{d} = \text{cov}(\mathcal{K})$ and that $\mathcal{K} \subset \mathcal{M}$ .

The inequality $\text{add}(\mathcal{K}) \le \mathfrak{b}$ follows from the fact that if $F \subset \omega^\omega$ is an unbounded set, then there exist $\lvert F \lvert$ many meager sets whose union is a non-meager set. This fact is established in this previous post (see Theorem 1 in that post).

For the inequality $\mathfrak{d} \le \text{cof}(\mathcal{M})$ , see Corollary 5.4 of [1]. For the additional inequalities, see Theorem 5.6 in [1].

The next post is on the full Cichon’s Diagram.

Reference

Blass, A., Combinatorial Cardinal Characteristics of the Continuum, Handbook of Set Theory (M. Foreman, A. Kanamori, eds), Springer Science+Business Media B. V., Netherlands, 395-489, 2010.

$\text{ }$

Dan Ma topology

Daniel Ma topology

Dan Ma math

Daniel Ma mathematics

The ideal of bounded sets

Posted on March 22, 2020 by Dan Ma

This is the second in a series of posts leading to a diagram called The Cichon’s Diagram. In this post, we examine an ideal that will provide insight on the ideal of meager sets, which is part of the Cichon’s Diagram. For the definitions of ideal and $\sigma$ -ideal, see the first post.

The next two posts are: the third post and the fourth post – the Cichon’s Diagram.

Let $\omega$ be the set of all non-negative integers, i.e. $\omega=\{ 0,1,2,\cdots \}$ . Let $X=\omega^\omega$ , the set of all functions from $\omega$ into $\omega$ . We can also think of $X$ as a topological space since it is a product space of countably many copies of the discrete space $\omega$ . As a product space, $X=\omega^\omega$ is homeomorphic to $\mathbb{P}$ , the space of all irrational numbers with the usual real line topology (see here).

Recall that for $f,g \in \omega^\omega$ , $f \le^* g$ means that $f(n) \le g(n)$ for all but finitely many $n$ . This is a partial order that is called the eventual domination order. A subset $F$ of $\omega^\omega$ is a bounded set if there is a $g \in \omega^\omega$ such that $g$ is an upper bound of $F$ with respect to the partial order $\le^*$ , i.e. for each $f \in F$ , we have $f \le^* g$ . The set $F$ is an unbounded set of it is not bounded. The set $F$ is a dominating set if for each $g \in \omega^\omega$ , there exists $f \in F$ such that $g \le^* f$ , i.e. the set $F$ is cofinal in $\omega^\omega$ with respect to the eventual domination order $\le^*$ .

We are interested in the least cardinality of an unbounded set and the least cardinality of a dominating set. The former is denoted by $\mathfrak{b}$ and is called the bounding number while the latter is denoted by $\mathfrak{d}$ and is called the dominating number.

An Interim Ideal

We define two ideals on $X=\omega^\omega$ . Let $\mathcal{S}$ be the collection of all $\sigma$ -compact subsets of $\omega^\omega$ .

$\mathcal{K}_\sigma=\{ A \subset \omega^\omega: \exists \ B \in \mathcal{S} \text{ such that } A \subset B \}$

$\mathcal{K}_b=\{ A \subset \omega^\omega: A \text{ is a bounded set} \}$

The first one $\mathcal{K}_\sigma$ is the set of all subsets of $\omega^\omega$ , each of which is contained in a $\sigma$ -compact set. The second one $\mathcal{K}_b$ is simply the set of all bounded subsets. It is straightforward to verify that $\mathcal{K}_\sigma$ is a $\sigma$ -ideal on $\omega^\omega$ . Note that any countable set $\{ f_0, f_1,f_2,\cdots \} \subset \omega^\omega$ is a bounded set (via a diagonal argument). Thus the union of countably many bounded sets $A_0,A_1,A_2,\cdots$ with $A_n$ having an upper bound $f_n$ must be a bounded set. The $f_n$ have an upper bound $f$ , which is an upper bound of the union of the sets $A_n$ . Thus $\mathcal{K}_b$ is a $\sigma$ -ideal on $\omega^\omega$ .

Furthermore, since $\omega^\omega$ is not $\sigma$ -compact, $\mathcal{K}_\sigma$ is a proper ideal. Likewise $\omega^\omega$ is an unbounded set, $\mathcal{K}_b$ is a proper ideal. The ideal $\mathcal{K}_\sigma$ is called the $\sigma$ -ideal generated by $\sigma$ -compact subsets of $\omega^\omega$ . The ideal $\mathcal{K}_b$ is the $\sigma$ -ideal of bounded subset of $\omega^\omega$ . However, these two ideals are one and the same.

Theorem 1
Let $F \subset \omega^\omega$ . Then the following conditions are equivalent.

The set $F$ is bounded.
There exists a $\sigma$ -compact set $X$ such that $F \subset X \subset \omega^\omega$ .
With $F$ as a subset of the real line, the set $F$ is an $F_\sigma$ -subset of $F \cup \mathbb{Q}$ where $\mathbb{Q}$ is the set of all rational numbers.

Theorem 1 is the Theorem 1 found in
. The sets satisfying Condition 1 of this theorem are precisely the elements of the $\sigma$ -ideal $\mathcal{K}_b$ . The sets satisfying Condition 2 of this theorem are precisely the elements of the $\sigma$ -ideal $\mathcal{K}_\sigma$ . According to this theorem, the two $\sigma$ -ideals are the same. Each is a different characterization of the same $\sigma$ -ideal. As a result, we drop the subscript and call this $\sigma$ -ideal $\mathcal{K}$ .

Four Cardinals

With the $\sigma$ -ideal $\mathcal{K}$ from the preceding section, we would like to examine the four associated cardinals $\text{add}(\mathcal{K})$ (the additivity number), $\text{non}(\mathcal{K})$ (the uniformity number), $\text{cov}(\mathcal{K})$ (the covering number) and $\text{cof}(\mathcal{K})$ (the cofinality number). For the definitions of these numbers, see the first post.

Figure 1 – Cardinal Characteristics of the $\sigma$ -Ideal Generated by $\sigma$ -Compact Sets

…Cichon… $\displaystyle \begin{array}{ccccccccccccc} \text{ } & \text{ } & \text{ } &\text{ } & \bold n \bold o \bold n ( \mathcal{K} ) &\text{ } &\Longrightarrow &\text{ } &\bold c \bold o \bold f (\mathcal{K} )&\text{ } &\Longrightarrow & \text{ } & 2^{\aleph_0}\\ \text{ } & \text{ } & \text{ } &\text{ } & \text{ } &\text{ } &\text{ } &\text{ } &\text{ }&\text{ } &\text{ } & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } &\text{ } & \Uparrow &\text{ } &\text{ } &\text{ } &\Uparrow&\text{ } &\text{ } & \text{ }& \text{ } \\ \text{ }& \text{ } & \text{ } &\text{ } & \text{ } &\text{ } &\text{ } &\text{ } &\text{ }&\text{ } &\text{ } & \text{ }& \text{ } \\ \aleph_1 & \text{ } & \Longrightarrow & \text{ } &\bold a \bold d \bold d ( \mathcal{K} ) & \text{ } &\Longrightarrow & \text{ } &\bold c \bold o \bold v ( \mathcal{K} ) &\text{ } &\text{ } &\text{ } & \text{ } \end{array}$

In the diagram, $\alpha \Rightarrow \beta$ means that $\alpha \le \beta$ . The additivity number $\text{add}(\mathcal{K})$ is lowered bounded by $\aleph_1$ on the lower right in the diagram since the ideal $\mathcal{K}$ is a $\sigma$ -ideal. The middle of the diagram shows the relationships that hold for any $\sigma$ -ideal. To see that $\text{cof}(\mathcal{K}) \le 2^{\aleph_0}$ , define $B_f=\{ h \in \omega^\omega: h \le^* f \}$ for each $f \in \omega^\omega$ . The set of all $B_f$ is cofinal in $\mathcal{K}$ . The inequality holds since there are $2^{\aleph_0}$ many sets $B_f$ .

We can further refine Figure 1. The following theorem shows how.

Theorem 2
The values of the four cardinals associated with the $\sigma$ -ideal $\mathcal{K}$ are the bounding numbers $\mathfrak{b}$ and the dominating number $\mathfrak{d}$ . Specifically, we have the following equalities.

$\mathfrak{b}=\text{add}(\mathcal{K})=\text{non}(\mathcal{K})$

$\mathfrak{d}=\text{cov}(\mathcal{K})=\text{cof}(\mathcal{K})$

Proof of Theorem 2
Based on the discussion in the first post, $\text{add}(\mathcal{K}) \le \text{non}(\mathcal{K})$ and $\text{cov}(\mathcal{K}) \le \text{cof}(\mathcal{K})$ always hold. We establish the equalities by showing the following.

$\mathfrak{b} \le \text{add}(\mathcal{K}) \le \text{non}(\mathcal{K}) = \mathfrak{b}$

$\mathfrak{d} \le \text{cov}(\mathcal{K}) \le \text{cof}(\mathcal{K}) \le \mathfrak{d}$

Viewing $\mathcal{K}$ as a $\sigma$ -ideal of bounded sets, $\text{non}(\mathcal{K})$ is the least cardinality of an unbounded set. Thus $\mathfrak{b}=\text{non}(\mathcal{K})$ .

To see $\mathfrak{b} \le \text{add}(\mathcal{K})$ , let $\mathcal{A} \subset \mathcal{K}$ such that $\lvert \mathcal{A} \lvert=\text{add}(\mathcal{K})$ and $Y=\bigcup \mathcal{A} \notin \mathcal{K}$ . Note that each $A \in \mathcal{A}$ is a bounded set with an upper bound $f(A) \in \omega^\omega$ . We claim that $F=\{ f(A): A \in \mathcal{A} \}$ is unbounded. This is because $Y=\bigcup \mathcal{A}$ is unbounded. Since there exists an unbounded set $F$ with cardinality $\text{add}(\mathcal{K})$ , it follows that $\mathfrak{b} \le \text{add}(\mathcal{K})$ .

To see $\text{cof}(\mathcal{K}) \le \mathfrak{d}$ , let $F \subset \omega^\omega$ be a dominating set such that $\lvert F \lvert=\mathfrak{d}$ . Note that for each $f \in \omega^\omega$ , the set $B_f=\{ h \in \omega^\omega: h \le^* f \}$ is a bounded set and thus $B_f \in \mathcal{K}$ . It can be verified that $\mathcal{B}=\{ B_f: f \in F \}$ is cofinal in $\mathcal{K}$ . Since there is a cofinal set $\mathcal{B}$ with cardinality $\mathfrak{d}$ , it follows that $\text{cof}(\mathcal{K}) \le \mathfrak{d}$ .

To see $\mathfrak{d} \le \text{cov}(\mathcal{K})$ , let $\mathcal{W} \subset \mathcal{K}$ such that $\lvert \mathcal{W} \lvert=\text{cov}(\mathcal{K})$ and $\bigcup \mathcal{W}=\omega^\omega$ . For each $A \in \mathcal{W}$ , let $f(A)$ be an upper bound of $A$ . It can be verified that the set $F=\{ f(A): A \in \mathcal{W} \}$ is a dominating set. Since we have a dominating set $F$ with cardinality $\text{cov}(\mathcal{K})$ , we have $\mathfrak{d} \le \text{cov}(\mathcal{K})$ . This completes the proof of Theorem 2. $\square$

With additional information from Theorem 2, Figure 1 can be revised as follows:

Figure 2 – Revised Figure 1

…Cichon… $\displaystyle \begin{array}{ccccccccccccc} \text{ } & \text{ } & \text{ } &\text{ } & \mathfrak{b}=\bold n \bold o \bold n ( \mathcal{K} ) &\text{ } &\Longrightarrow &\text{ } &\mathfrak{d}=\bold c \bold o \bold f (\mathcal{K} )&\text{ } &\Longrightarrow & \text{ } & 2^{\aleph_0}\\ \text{ } & \text{ } & \text{ } &\text{ } & \text{ } &\text{ } &\text{ } &\text{ } &\text{ }&\text{ } &\text{ } & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } &\text{ } & \parallel &\text{ } &\text{ } &\text{ } &\parallel&\text{ } &\text{ } & \text{ }& \text{ } \\ \text{ }& \text{ } & \text{ } &\text{ } & \text{ } &\text{ } &\text{ } &\text{ } &\text{ }&\text{ } &\text{ } & \text{ }& \text{ } \\ \aleph_1 & \text{ } & \Longrightarrow & \text{ } &\mathfrak{b}=\bold a \bold d \bold d ( \mathcal{K} ) & \text{ } &\Longrightarrow & \text{ } &\mathfrak{d}=\bold c \bold o \bold v ( \mathcal{K} ) &\text{ } &\text{ } &\text{ } & \text{ } \end{array}$

Note that there are only four cardinals in this diagram – $\aleph_1$ , $\mathfrak{b}$ , $\mathfrak{d}$ and $2^{\aleph_0}$ . Of course, if continuum hypothesis holds, there would only one number in the diagram, namely $\aleph_1$ .

The next post is on the $\sigma$ -ideal $\mathcal{M}$ of meager sets.

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Cardinals associated with an ideal

Posted on March 22, 2020 by Dan Ma

This is the first in a series of posts leading to a diagram called the Cichon’s diagram. The diagram displays relationships among twelve small cardinals, eight of which are defined by using $\sigma$ -ideals. The purpose of this post is to set up the scene.

The next three posts are: the second post, the third post and the fourth post – the Cichon’s Diagram.

Ideals and $\sigma$ -Ideals

Let $X$ be a set. Let $\mathcal{I}$ be a collection of subsets of $X$ . We say that $\mathcal{I}$ is an ideal on $X$ if the following three conditions hold.

$\varnothing \in \mathcal{I}$ .
If $A \in \mathcal{I}$ and $B \subset A$ , then $B \in \mathcal{I}$ .
If $A, B \in \mathcal{I}$ , then $A \cup B \in \mathcal{I}$ .

If $X \notin \mathcal{I}$ , then $\mathcal{I}$ is said to be a proper ideal. Note that if $X \in \mathcal{I}$ , then $\mathcal{I}$ would simply be the power set of $X$ . Thus we would only want to focus on proper ideals. Thus by ideal we mean proper ideal.

We say that $\mathcal{I}$ is a $\sigma$ -ideal on $X$ if it is an ideal with the additional property that it is closed under taking countable unions, i.e. if for each $n \in \omega$ , $A_n \in \mathcal{I}$ , then $\bigcup_{n \in \omega} A_n \in \mathcal{I}$ . For the discussion that follows, we even require that all singleton subsets of $X$ are members of any $\sigma$ -ideal $\mathcal{I}$ .

Elements of an ideal or a $\sigma$ -ideal are considered “small sets” or “negligible sets”. The definition of $\sigma$ -ideal does indeed reflect how small sets should behave. The empty set is naturally a small set. Any subset of a small set should be a small set. The union of countably many small sets should also be a small set as is any countable set (the union of countably many singleton sets).

Let $\mathcal{B}$ be a collection of subsets of the set $X$ . We assume that $\mathcal{B}$ is closed under taking countable unions. It is easy to verify that the set

$\mathcal{I}=\{ A \subset X: \exists \ B \in \mathcal{B} \text{ such that } A \subset B \}$

is a $\sigma$ -ideal on $X$ . This $\sigma$ -ideal $\mathcal{I}$ is said to be generated by the set $\mathcal{B}$ . The set $\mathcal{B}$ is called a base for the $\sigma$ -ideal $\mathcal{I}$ . A subbase for a $\sigma$ -ideal is simply a collection of subsets of $X$ . Then a base would be generated by taking countable unions of sets in the subbase.

Four Cardinals

We now discuss the cardinal characteristics associated with a $\sigma$ -ideal. As before, let $X$ be a set and $\mathcal{I}$ be a $\sigma$ -ideal on $X$ . As discussed above, we require that all singleton sets are in $\mathcal{I}$ . We define the following four cardinals.

Additivity Number

$\text{add}(\mathcal{I})=\text{min} \{ \lvert \mathcal{A} \lvert: \mathcal{A} \subset \mathcal{I} \text{ and } \bigcup \mathcal{A} \notin \mathcal{I} \}$

Covering Number
$\text{cov}(\mathcal{I})=\text{min} \{ \lvert \mathcal{A} \lvert: \mathcal{A} \subset \mathcal{I} \text{ and } \bigcup \mathcal{A} = X \}$

Uniformity Number
$\text{non}(\mathcal{I})=\text{min} \{ \lvert A \lvert: A \subset X \text{ and } A \notin \mathcal{I} \}$

Cofinality Number
$\text{cof} (\mathcal{I})=\text{min} \{ \lvert \mathcal{B} \lvert: \mathcal{B} \subset \mathcal{I} \text{ and } \mathcal{B} \text{ is cofinal in } \mathcal{I} \}$

A subset $\mathcal{B}$ of $\mathcal{I}$ is said to be cofinal in $\mathcal{I}$ if for each $A \in \mathcal{I}$ , there exists $B \in \mathcal{B}$ such that $A \subset B$ , i.e. $\mathcal{B}$ is cofinal in the partial order $\subset$ . Such a $\mathcal{B}$ is a base for $\mathcal{I}$ .

The numbers $\text{add}(\mathcal{I})$ , $\text{cov}(\mathcal{I})$ and $\text{cof} (\mathcal{I})$ are the minimum cardinalities of certain subfamilies of the $\sigma$ -ideal $\mathcal{I}$ which fail to be small, i.e. not in $\mathcal{I}$ . The additivity number $\text{add}(\mathcal{I})$ is the least cardinality of a subfamily of $\mathcal{I}$ whose union is not in $\mathcal{I}$ . The covering number $\text{cov}(\mathcal{I})$ is the minimum cardinality of a subfamily of $\mathcal{I}$ whose union is the entire set $X$ . The covering number is the minimum cardinality of a covering of $X$ with elements of $\mathcal{I}$ . The cofinality number $\text{cof} (\mathcal{I})$ is the least cardinality of a subfamily of $\mathcal{I}$ that is a cofinal in $\mathcal{I}$ . Equivalently, the cofinality number is the least cardinality of a base that generates the $\sigma$ -ideal. The uniformity number $\text{non}(\mathcal{I})$ is the least cardinality of a subset of $X$ that is not an element of $\mathcal{I}$ .

The elements of the $\sigma$ -ideal $\mathcal{I}$ are “small” sets. The additivity number $\text{add}(\mathcal{I})$ is the smallest number of small sets whose union is not small. The covering number $\text{cov}(\mathcal{I})$ is then the smallest number of small sets that fill up the entire set $X$ . The uniform number is the least cardinality of a non-small set.

Of these four cardinals, the smallest one is $\text{add}(\mathcal{I})$ and the largest one is $\text{cof} (\mathcal{I})$ . Because $\mathcal{I}$ is a $\sigma$ -ideal, all four cardinals must be uncountable, hence $\ge \aleph_1$ . Obviously $\mathcal{I}$ is confinal in $\mathcal{I}$ . Thus $\text{cof} (\mathcal{I}) \le \lvert \mathcal{I} \lvert$ . The following inequalities also hold.

$\aleph_1 \le \text{add}(\mathcal{I}) \le \text{cov}(\mathcal{I}) \le \text{cof} (\mathcal{I}) \le \lvert \mathcal{I} \lvert$

$\aleph_1 \le \text{add}(\mathcal{I}) \le \text{non}(\mathcal{I}) \le \text{cof} (\mathcal{I}) \le \lvert \mathcal{I} \lvert$

$\displaystyle \aleph_1 \le \text{add}(\mathcal{I}) \le \text{min} \{ \text{cov}(\mathcal{I}), \text{non}(\mathcal{I}) \} \le \text{max} \{ \text{cov}(\mathcal{I}), \text{non}(\mathcal{I}) \} \le \text{cof} (\mathcal{I}) \le \lvert \mathcal{I} \lvert$

Displaying the Four Cardinals in a Diagram

The inequalities shown in the preceding section can be displayed in a diagram such as the following.

Figure 1 – Cardinal Characteristics of a $\sigma$ -Ideal

In the above diagram, $a \Rightarrow b$ means $a \le b$ . The smallest cardinal $\text{add}(\mathcal{I})$ is lower bounded by $\aleph_1$ on the lower left since $\mathcal{I}$ is a $\sigma$ -ideal. The largest cardinal $\text{cof}(\mathcal{I})$ is upper bounded by the cardinality of $\mathcal{I}$ on the upper right since $\mathcal{I}$ is cofinal in $\mathcal{I}$ . The diagram tells us that the additivity number is less than or equal to the minimum of the uniformity number and the covering number. On the other hand, the cofinality number is greater than or equal to the maximum of the uniformity number and the covering number.

Examples

In the subsequent posts, we would like to focus on two $\sigma$ -ideals, hence eight associated cardinals. To define these two ideals, let $X=\mathbb{R}$ , the real line. Let $\mathcal{M}$ be set of all meager subsets of the real line and let $\mathcal{L}$ be the set of all subsets of the real line that are of Lebesgue measure zero.

$\mathcal{M}=\{ A \subset \mathbb{R}: A \text{ is a Meager set} \}$

$\mathcal{L}=\{ A \subset \mathbb{R}: A \text{ is of Lebesgue measure zero} \}$

In the real line, a set is nowhere dense if its closure contains no open set. A meager set is the union of countably many nowhere dense sets. It is straightforward to verify that $\mathcal{M}$ is a $\sigma$ -ideal on the real line $\mathbb{R}$ . Because of the Baire category theorem, $\mathbb{R} \notin \mathcal{M}$ . Thus it is a proper ideal. Similarly, it is straightforward to verify that $\mathcal{L}$ is a $\sigma$ -ideal as well as a proper ideal.

Before we examine the ideal $\mathcal{M}$ , we consider the $\sigma$ -ideal of bounded subsets of $\omega^\omega$ in the next post.

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A Pixley-Roy Meta-Lindelof Space

Examples of Meta-Lindelof Spaces

Divergence of the sum of the reciprocals of the primes

Revisiting example 106 from Steen and Seebach

Making sense of the spaces with small diagonal

The Cichon’s Diagram

The ideal of meager sets

The ideal of bounded sets

Cardinals associated with an ideal