Divergence of the sum of the reciprocals of the primes

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:

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

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 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 $N_b$.

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

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 , 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

1. 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

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.

1. The space $X$ is paracompact.
2. 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

1. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976.
2. 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.
3. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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Making sense of the spaces with small diagonal

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 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

1. Arhangelskii, A., Bella, A., Few observations on topological spaces with small diagonal, Zb. Rad. Filoz. Fak. Nisu, 6, No. 2, 211-213, 1992.
2. Dow, A., Hart, P. Elementary chains and compact spaces with a small diagonal, Indagationes Mathematicae, 23, No. 3, 438-447, 2012.
3. Dow, A., Pavlov, O. More about spaces with a small diagonal, Fund. Math., 191, No. 1, 67-80, 2006.
4. Dow, A., Pavlov, O. Perfect preimages and small diagonal, Topology Proc., 31, No. 1, 89-95, 2007.
5. Gruenhage, G., Spaces having a small diagonal, Topology Appl., 22, 183-200, 2002.
6. Gruenhage, G., Generalized metrizable spaces, Recent Progress in General Topology III (K.P. Hart, J. van Mill, and P. Simon, eds.), Atlantis Press 2014.
7. Husek, M., Topological spaces without $\kappa$-accessible diagonal, Comment. Math. Univ. Carolin., 18, No. 4, 777-788, 1977.
8. 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.
9. Juhasz, I., Szentmikloss, Z. Convergent free sequences in compact spaces, Proc. Amer. Math. Soc., 116, No. 4, 1153-1160, 1992.
10. Zhou, H. X., On The Small Diagonals, Topology Appl., 13, 283-293, 1982.

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The Cichon’s Diagram

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

1. Bartoszynski, T., Judah H., Shelah S.,The Cichon Diagram, J. Symbolic Logic, 58(2), 401-423, 1993.
2. Bartoszynski, T., Judah H., Shelah S.,Set theory: On the structure of the real line, A K
Peters, Ltd.. Wellesley, MA, 1995.
3. 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.
4. Fremlin, D. H., Cichon’s diagram. In Seminaire d’Initiation ´a l’Analyse, 23, Universite Pierre et Marie Curie, Paris, 1984.
5. 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.
6. Laver, R., On the consistency of Borel’s conjecture, Acta Math., 137, 151-169, 1976.
7. Miller, A. W., Some Properties of Measure and Category, Trans. Amer. Math. Soc., 266 (1), 93-114, 1981.
8. 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.
9. 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.
10. Miller A. W., Fremlin D. H., On some properties of Hurewicz, Menger, and Rothberger, Fund. Math., 129, 17-33, 1988.
11. Pawlikowski, J., Reclaw I., Parametrized Cichon’s diagram and small sets, Fund. Math., 127, 225-239, 1987.
12. 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.
13. 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

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

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

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.

…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

1. 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.

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Daniel Ma topology

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$\copyright$ 2020 – Dan Ma

The ideal of bounded sets

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.

1. The set $F$ is bounded.
2. There exists a $\sigma$-compact set $X$ such that $F \subset X \subset \omega^\omega$.
3. 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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$\copyright$ 2020 – Dan Ma

Cardinals associated with an ideal

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.

1. $\varnothing \in \mathcal{I}$.
2. If $A \in \mathcal{I}$ and $B \subset A$, then $B \in \mathcal{I}$.
3. 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.

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

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

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

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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$\copyright$ 2020 – Dan Ma

Adding up to a non-meager set

The preceding post gives a topological characterization of bounded subsets of $\omega^\omega$. From it, we know what it means topologically for a set to be unbounded. In this post we prove a theorem that ties unbounded sets to Baire category.

A set is nowhere dense if its closure has empty interior. A set is a meager set if it is the union of countably many nowhere dense sets. By definition, the union of countably many meager sets is always a meager set. In order for meager sets to add up to a non-meager set (though taking union), the number of meager sets must be uncountable. What is this uncountable cardinal number? We give an indication of how big this number is. In this post we give a constructive proof to the following fact:

Theorem 1 …. Given an unbounded set $F \subset \omega^\omega$, there exist $\kappa=\lvert F \lvert$ many meager subsets of the real line whose union is not meager.

We will discuss the implications of this theorem after giving background information.

We use $\omega$ to denote the set of all non-negative integers $\{ 0,1,2,\cdots \}$. The set $\omega^\omega$ is the set of all functions from $\omega$ into $\omega$. It is called the Baire space when it is topologized with the product space topology. It is well known that the Baire space is homeomorphic to the space of irrational numbers $\mathbb{P}$ (see here).

The notion of boundedness or unboundedness used in Theorem 1 refers to the eventual domination order ($\le^*$) for functions in the product space. For $f,g \in \omega^\omega$, by $f \le^* g$, we mean $f(n) \le g(n)$ for all but finitely many $n$. A set $F \subset \omega^\omega$ is bounded if it has an upper bound with respect to the partial order $\le^*$, i.e. there is some $f \in \omega^\omega$ such that $g \le^* f$ for all $g \in F$. The set $F$ is unbounded if it is not bounded. To spell it out, $F$ is unbounded if for each $f \in \omega^\omega$, there exists $g \in F$ such that $g \not \le^* f$, i.e. $f(n) for infinitely many $n$.

All countable subsets of the Baire space are bounded (using a diagonal argument). Thus unbounded sets must be uncountable. It does not take extra set theory to obtain an unbounded set. The Baire space $\omega^\omega$ is unbounded. More interesting unbounded sets are those of a certain cardinality, say unbounded sets of cardinality $\omega_1$ or unbounded sets with cardinality less than continuum. Another interesting unbounded set is one that is of the least cardinality. In the literature, the least cardinality of an unbounded subset of $\omega^\omega$ is called $\mathfrak{b}$, the bounding number.

Another notion that is part of Theorem 1 is the topological notion of small sets – meager sets. This is a topological notion and is defined in topological spaces. For the purpose at hand, we consider this notion in the context of the real line. As mentioned at the beginning of the post, a set is nowhere dense set if its closure has empty interior (i.e. the closure contains no open subset). Let $A \subset \mathbb{R}$. The set $A$ is nowhere dense if no open set is a subset of the closure $\overline{A}$. An equivalent definition: the set $A$ is nowhere dense if for every nonempty open subset $U$ of the real line, there is a nonempty subset $V$ of $U$ such that $V$ contains no points of $A$. Such a set is “thin” since it is dense no where. In any open set, we can also find an open subset that has no points of the nowhere dense set in question. A subset $A$ of the real line is a meager set if it is the union of countably many nowhere dense sets. Another name of meager set is a set of first category. Any set that is not of first category is called a set of second category, or simply a non-meager set.

Corollaries

Subsets of the real line are either of first category (small sets) or of second category (large sets). Countably many meager sets cannot fill up the real line. This is a consequence of the Baire category theorem (see here). By definition, caountably many meager sets cannot fill up any non-meager subset of the real line. How many meager sets does it take to add up to a non-meager set?

Theorem 1 gives an answer to the above question. It can take as many meager sets as the size of an unbounded subset of the Baire space. If $\kappa$ is a cardinal number for which there exists an unbounded subset of $\omega^\omega$ whose cardinality is $\kappa$, then there exists a non-meager subset of the real line that is the union of $\kappa$ many meager sets. The bounding number $\mathfrak{b}$ is the least cardinality of an unbounded set. Thus there is always a non-meager subset of the real line that is the union of $\mathfrak{b}$ many meager sets.

Let $\kappa_A$ be the least cardinal number $\kappa$ such that there exist $\kappa$ many meager subsets of the real line whose union is not meager. Based on Theorem 1, the bounding number $\mathfrak{b}$ is an upper bound of $\kappa_A$. These two corollaries just discussed are:

• There always exists a non-meager subset of the real line that is the union of $\mathfrak{b}$ many meager sets.
• $\kappa_A \le \mathfrak{b}$.

The bounding number $\mathfrak{b}$ points to a non-meager set that is the union of $\mathfrak{b}$ many meager sets. However, the cardinal $\kappa_A$ is the least number of meager sets whose union is a non-meager set and this number is no more than the bounding number. The cardinal $\kappa_A$ is called the additivity number.

There are other corollaries to Theorem 1. Let $A(c)$ be the statement that the union of fewer than continuum many meager subsets of the real line is a meager set. For any cardinal number $\kappa$, let $A(\kappa)$ be the statement that the union of fewer than $\kappa$ many meager subsets of the real line is a meager set. We have the following corollaries.

• The statement $A(c)$ implies that there are no unbounded subsets of $\omega^\omega$ that have cardinalities less than continuum. In other words, $A(c)$ implies that the bounding number $\mathfrak{b}$ is continuum.
• Let $\kappa \le$ continuum. The statement $A(\kappa)$ implies that there are no unbounded subsets of $\omega^\omega$ that have cardinalities less than $\kappa$. In other words, $A(\kappa)$ implies that the bounding number $\mathfrak{b}$ is at least $\kappa$, i.e. $\mathfrak{b} \ge \kappa$.

Let $B(c)$ be the statement that the real line is not the union of less than continuum many meager sets. Clearly, the statement $A(c)$ implies the statement $B(c)$. Thus, it follows from Theorem 1 that $A(c) \Longrightarrow B(c) + \mathfrak{b}=2^{\aleph_0}$. This is a result proven in Miller [1]. Theorem 1.2 in [1] essentially states that $A(c)$ is equivalent to $B(c) + \mathfrak{b}=2^{\aleph_0}$. The proof of Theorem 1 given here is essentially the proof of one direction of Theorem 1.2 in [1]. Our proof has various omitted details added. As a result it should be easier to follow. We also realize that the proof of Theorem 1.2 in [1] proves more than that theorem. Therefore we put the main part of the constructive in a separate theorem. For example, Theorem 1 also proves that the additivity number $\kappa_A$ is no more than $\mathfrak{b}$. This is one implication in the Cichon’s diagram.

Proof of Theorem 1

Let $2=\{ 0,1 \}$. The set $2^\omega$ is the set of all functions from $\omega$ into $\{0, 1 \}$. When $2^\omega$ is endowed with the product space topology, it is called the Cantor space and is homemorphic to the middle-third Cantor set in the unit interval $[0,1]$. We use $\{ [s]: \exists \ n \in \omega \text{ such that } s \in 2^n \}$ as a base for the product topology where $[s]=\{ t \in 2^\omega: s \subset t \}$.

Let $F \subset \omega^\omega$ be an unbounded set. We assume that the unbounded set $F$ satisfies two properties.

• Each $g \in F$ is an increasing function, i.e. $g(i) for any $i.
• For each $g \in F$, if $j>g(n)$, then $g(j)>g(n+1)$.

One may wonder if the two properties are satisfied by any given unbounded set. Since $F$ is unbounded, we can increase the values of each function $g \in F$, the resulting set will still be an unbounded set. More specifically, for each $g \in F$, define $g^*\in \omega^\omega$ as follows:

• $g^*(0)=g(0)+1$,
• for each $n \ge 1$, $g^*(n)=g(n)+\text{max}\{ g^*(i): i.

The set $F^*=\{ g^*: g \in F \}$ is also an unbounded set. Therefore we use $F^*$ and rename it as $F$.

Fix $g \in F$. Define an increasing sequence of non-negative integers $n_0,n_1,n_2,\cdots$ as follows. Let $n_0$ be any integer greater than 1. For each integer $j \ge 1$, let $n_j=g(n_{j-1})$. Since $n_0>1$, we have $n_1=g(n_0)>g(1)$. It follows that for all integer $k \ge 1$, $n_k>g(k)$.

For each $g \in F$, we have an associated sequence $n_0,n_1,n_2,\cdots$ as described in the preceding paragraph. Now define $C(g)=\{ q \in 2^\omega: \forall \ k, q(n_k)=1 \}$. It is straightforward to verify that each $C(g)$ is a closed and nowhere dense subset of the Cantor space $2^\omega$. Let $X=\bigcup \{C(g): g \in F \}$. The set $X$ is a union of meager sets. We show that it is a non-meager subset of $2^\omega$. We prove the following claim.

Claim 1
For any countable family $\{C_n: n \in \omega \}$ where each $C_n$ is a nowhere dense subset of $2^\omega$, we have $X \not \subset \bigcup \{C_n: n \in \omega \}$.

According to Claim 1, the set $X$ cannot be contained in any arbitrary meager subset of $2^\omega$. Thus $X$ must be non-meager. To establish the claim, we define an increasing sequence of non-negative integers $m_0,m_1,m_2,\cdots$ with the property that for any $k \ge 1$, for any $i, and for any $s \in 2^{m_k}$, there exists $t \in 2^{m_{k+1}}$ such that $s \subset t$ and $[t] \cap C_i=\varnothing$.

The desired sequence is derived from the fact that the sets $C_n$ are nowhere dense. Choose any $m_0 to start. With $m_1$ determined, the only nowhere dense set to consider is $C_0$. For each $s \in 2^{m_1}$, choose some integer $y>m_1$ such that there exists $t \in 2^{y+1}$ such that $s \subset t$ and $[t] \cap C_0=\varnothing$. Let $m_2$ be an integer greater than all the possible $y$‘s that have been chosen. The integer $m_2$ can be chosen since there are only finitely many $s \in 2^{m_1}$.

Suppose $m_0<\cdots have been chosen. Then the only nowhere dense sets to consider are $C_0,\cdots,C_{k-1}$. Then for each $i \le k-1$, for each $s \in 2^{m_k}$, choose some integer $y>m_k$ such that there exists $t \in 2^{y+1}$ such that $s \subset t$ and $[t] \cap C_i=\varnothing$. As before let $m_{k+1}$ be an integer greater than all the possible $y$‘s that have been chosen. Again $m_{k+1}$ is possible since there are only finitely many $i \le k-1$ and only finitely many $s \in 2^{m_k}$.

Let $Z=\{ m_k: k \in \omega \}$. We make the following claim.

Claim 2
There exists $h \in F$ such that the associated sequence $n_0, n_1,n_2,\cdots$ satisfies the condition: $\lvert [n_k,n_{k+1}) \cap Z \lvert \ge 2$ for infinitely many $k$ where $[n_k,n_{k+1})$ is the set $\{ m \in \omega: n_k \le m < m_{k+1} \}$.

Suppose Claim 2 is not true. For each $g \in F$ and its associated sequence $n_0, n_1,n_2,\cdots$,

(*) there exists some integer $b$ such that for all $k>b$, $\lvert [n_k,n_{k+1}) \cap Z \lvert \le 1$.

Let $f \in \omega^\omega$ be defined by $f(k)=m_k$ for all $k$. Choose $\overline{f} \in \omega^\omega$ in the following manner. For each $k \in \omega$, define $d_k \in \omega^\omega$ by $d_k(n)=f(n+k)$ for all $n$. Then choose $\overline{f} \in \omega^\omega$ such that $d_k \le^* \overline{f}$ for all $k$.

Fix $g \in F$. Let $m_j$ be the least element of $[n_b, \infty) \cap Z$. Then for each $k>b$, we have $g(k) \le n_k \le m_{j+k}=f(j+k)=d_j(k)$. Note that the inequality $n_k \le m_{j+k}$ holds because of the assumption (*). It follows that $g \le^* d_j \le^* \overline{f}$. This says that $\overline{f}$ is an upper bound of $F$ contradicting that $F$ is an unbounded set. Thus Claim 2 must be true.

Let $h \in F$ be as described in Claim 2. We now prove another claim.

Claim 3
For each $n$, $C_n$ is a nowhere dense subset of $C(h)$.

Fix $C_n$. Let $p$ be an integer such that $[n_p,n_{p+1}) \cap Z$ has at least two points, say $m_k$ and $m_{k+1}$. We can choose $p$ large enough such that $n. Choose $s \in 2^{m_k}$. Since $n_p$ is arbitrary, $[s]$ is an arbitrary open set in $2^\omega$. Since $m_k$ is in between $n_p$ and $n_{p+1}$, $[s]$ contains a point of $C(h)$. Thus $[s] \cap C(h)$ is an arbitrary open set in $C(h)$. By the way $m_k$ and $m_{k+1}$ are chosen originally, there exists $t \in 2^{m_{k+1}}$ such that $s \subset t$ and $[t] \cap C_n=\varnothing$. Because $m_k$ and $m_{k+1}$ are in between $n_p$ and $n_{p+1}$, $[t] \cap C(h) \ne \varnothing$. This establishes the claim that $C_n$ is nowhere dense subset of $C(h)$.

Note that $C(h)$ is a closed subset of the Cantor space $2^\omega$ and hence is also compact. Thus $C(h)$ is a Baire space and cannot be the union of countably many nowhere dense sets. Thus $C(h) \not \subset \cup \{C_n: n \in \omega \}$. Otherwise, $C(h)$ would be the union of countably many nowhere dense sets. This means that $X=\bigcup \{C(g): g \in F \} \not \subset \cup \{C_n: n \in \omega \}$. This establishes Claim 1.

Considering the Cantor space $2^\omega$ as a subspace of the real line, each $C(g)$ is also a closed nowhere dense subset of the real line. The set $X=\bigcup \{C(g): g \in F \}$ is also not a meager subset of the real line. This establishes Theorem 1. $\square$

Reference

1. Miller A. W., Some properties of measure and category, Trans. Amer. Math. Soc., 266, 93-114, 1981.

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Topological meaning of bounded sets

In this post, we discuss a topological characterization of bounded sets in $\omega^\omega$. We also give an example of an unbounded meager set. We also briefly discuss the $\sigma$-ideal generated by $\sigma$-compact subsets of $\omega^\omega$ and $\sigma$-ideal of meager subsets of $\omega^\omega$.

Let $\omega$ be the first infinite ordinal. We consider $\omega$ to be the set of all non-negative integers $\{0, 1, 2, \cdots \}$. Let $\omega^\omega$ be the set of all functions from $\omega$ into $\omega$. When the set $\omega$ is considered a discrete space, the set $\omega^\omega$ is the product of countably many copies of $\omega$. As a product space, $\omega^\omega$ is homeomorphic to the set $\mathbb{P}$ of all irrational numbers (see here). The product space $\omega^\omega$ is called the Baire space in the literature.

Even though the two are topologically the same, working in the product space has its advantage. With the Baire space, we can define a partial order. For $f,g \in \omega^\omega$, define $f \le^* g$ if $f(n) \le g(n)$ for all but finitely many $n$. Let $F \subset \omega^\omega$. The set $F$ is a bounded set if there exists $f \in \omega^\omega$ which is an upper bound of $F$ according to the partial order $\le^*$. The set $F$ is an unbounded set if it is not bounded. We prove the following theorem.

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

1. The set $F$ is bounded.
2. There exists a $\sigma$-compact set $X$ such that $F \subset X \subset \omega^\omega$.
3. 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.

This theorem is Theorem 9.3 in p. 149 in [1], the chapter by Van Douwen in the Handbook of Set-Theoretic Topology. As mentioned above, the set $\mathbb{P}$ of irrational numbers is homeomorphic to the Base space $\omega^\omega$. For $\mathbb{P}$, the irrational numbers are points on a straight line. For the Baire space, the irrational numbers are functions in a product space. The second condition of Theorem 1 tells us what it means topologically for a subset of the Baire space to be bounded. The third condition tells us what a bounded set means if the set is placed on a straight line.

A subset $A$ of any topological space $Y$ is said to be nowhere dense set if for any non-empty open subset $U$ of $Y$, there exists a non-empty open subset $V$ of $U$ such that $V \cap A=\varnothing$. A subset $M$ of the topological space $Y$ is said to be a meager set if $M$ is the union of countably many nowhere dense sets. Meager sets are “small” sets. The discussion that follows shows that bounded sets are meager sets.

For $m \in \omega$ and $s \in \omega^m$, define $[s]=\{ t \in \omega^\omega: \exists \ n \in \omega \text{ such that } s \subset t \}$. Note that the set $\mathcal{B}$ of all $[s]$ over all $s \in \omega^m$ over all $m \in \omega$ is a base for the Baire space $\omega^\omega$. As discussed below in the proof of Theorem 1, any compact subset of $\omega^\omega$ is a subset of $A_g=\{ h \in \omega^\omega: \forall \ n, \ h(n) \le g(n) \}$ for some $g \in \omega^\omega$. Thus for any $[s]$, there exists some $[t]$ with $s \subset t$ such that $t$ is greater than some $g(n)$. Thus sets of the form $A_g$ and any compact subset of $\omega^\omega$ are nowhere dense sets. Then any $\sigma$-compact subset of $\omega^\omega$ is contained in the union of countably many sets of the form $A_g$ and is thus a meager set. The following theorem follows from Theorem 1.

Theorem 2
If $F$ is a bounded subset of $\omega^\omega$, then $F$ is a meager set.

A natural question: is the converse of Theorem 2 true? If true, boundedness would be a characterization of meager subsets of $\omega^\omega$. We give an example of an unbounded nowhere dense set, thus showing that the converse is not true.

Example of an Unbounded Meager Set

Let $\omega^{< \omega}$ be the union of all $\omega^n$ where $n \in \omega$. Each $\omega^n$ is the set of all functions $t:n=\{0,1,\cdots, n-1 \} \rightarrow \omega$. Recall that $\mathcal{B}$ is a base of $\omega^\omega$ that consists of sets of the form $[s]$ where $s \in \omega^{< \omega}$. Note that $[s]$ is the set of all $t \in \omega^\omega$ such that $s \subset t$. To find a meager set, we remove $[t]$ from each $[s] \in \mathcal{B}$. The points remaining in $\omega^\omega$ form a nowhere dense set. We remove $[t]$ in such a way that the resulting set is a dominating set, hence an unbounded set.

For $[t] \in \mathcal{B}$, we also notate $[t]$ by $[t]=[t(0),t(1),\cdots,t(n-1)]$ if $t \in \omega^n$. This would be the set of all $h \in \omega^\omega$ such that $h(i)=t(i)$ for all $i \le n-1$.

For each $t \in \omega^1=\omega^{ \{ 0 \} }$, let $A_t=[t(0),t(0)+1]$. Note that $A_t \subset \omega^2$ and $A_t \subset [t]$. We remove all such $A_t$ from $\omega^\omega$.

For each $t \in \omega^2=\omega^{ \{ 0,1 \} }$, we define $A_t$ where $A_t=[t(0),t(1),j]$ where $j= \text{max} \{ t(0),t(1) \}+1$. Note that $A_t \subset \omega^3$ and $A_t \subset [t]$. We remove all such $A_t$.

For each $t \in \omega^n=\omega^{ \{ 0,1,\cdots,n-1 \} }$, we define $A_t$ where $A_t=[t(0),t(1),\cdots, t(n-1),j]$ where $j= \text{max} \{ t(0),t(1),\cdots,t(n-1) \}+1$. Note that $A_t \subset \omega^{n+1}$ and $A_t \subset [t]$. We remove all such $A_t$.

Let $X=\omega^\omega \backslash \bigcup_{t \in \omega^{< \omega}} A_t$. The set $X$ is clearly a nowhere dense subset of $\omega^\omega$ since we remove an element of the base from each element of the base. We now show that $X$ is a dominating set. To this end, let $f \in \omega^\omega$. We define $g \in X$ such that $f \le^* g$. If $f \in X$, then we define $g=f$. Assume $f \notin X$. Choose the least $n$ such that $f \in A_t$ where $t \in \omega^n$. According to our notation $t=[t(0),t(1),\cdots,t(n-1)]$. Define $g \in \omega^\omega$ as follows.

$g(i) = \begin{cases} t(i) & \ \ \ \mbox{if } i \le n-2 \\ t(n-1)+2 & \ \ \ \mbox{if } i=n-1 \\ \text{max} \{g(0),g(1), \cdots, g(n-1) \}+99 & \ \ \ \mbox{if } i = n \\ \text{max} \{g(0),g(1), \cdots, g(n) \}+99 & \ \ \ \mbox{if } i = n+1 \\ \vdots & \ \ \ \ \ \ \ \ \ \ \vdots \\ \text{max} \{g(0),g(1), \cdots, g(k-1) \} +99& \ \ \ \mbox{if } i = k \text{ and } k \ge n \\ \vdots & \ \ \ \ \ \ \ \ \ \ \vdots \end{cases}$

Because $g(n-1) > t(n-1)$, the basic open set $[g(0),g(1),\cdots,g(n-1)]$ is not marked for removal. For $i \ge n$, because of the way $g(i)$ is defined, the basic open set $[g(0),g(1),\cdots,g(i)]$ is also not marked for removal. Thus $g \in X$.

Comment about the Example

The meager set that is a dominating set given above is not a Menger set (see here).

Theorem 2 and the example showing that the converse of Theorem 2 is not true speak to a situation involving two $\sigma$-ideals. One of the ideals is $\mathcal{K}$, which is the set of all subsets of $\omega^\omega$, each of which is contained in a $\sigma$-compact subset of $\omega^\omega$. The set $\mathcal{K}$ is a $\sigma$-ideal. Theorem 1 says that elements of $\mathcal{K}$ are precisely the bounded sets. Theorem 2 says that elements of $\mathcal{K}$ are meager sets.

The other ideal is $\mathcal{M}$, which is the set of all meager subsets of $\omega^\omega$. This is also a $\sigma$-ideal. Then $\mathcal{K} \subset \mathcal{M}$. The example shows that the two $\sigma$-ideals are not the same, in particular $\mathcal{M} \not \subset \mathcal{K}$. The ideal $\mathcal{K}$ is the $\sigma$-ideal generated by the $\sigma$-compact subsets of $\omega^\omega$. This $\sigma$-ideal is much smaller than the $\sigma$-ideal $\mathcal{M}$ of meager subsets of $\omega^\omega$.

Proof of Theorem 1

It is helpful to set up notations and have a background discussion before proving the theorem. For $g \in \omega^\omega$, define the following sets:

$A_g=\{ h \in \omega^\omega: \forall \ n, \ h(n) \le g(n) \}$

$B_g=\{ h \in \omega^\omega: h \le^* g \}$

The set $A_g$ is a compact set since it is the product of finite sets, i.e. $A_g=\prod_{n \in \omega} [0,g(n)]$. The set $B_g$ is a $\sigma$-compact set. To see this, $B_g=\bigcup_{n \in \omega} A_{h_n}$ for a sequence $h_n \in \omega^\omega$. The sequence $\{ h_n \}$ is obtained by considering, for each $k \in \omega$, all functions $t \in \omega^\omega$ where $t(i)=g(i)$ for all $i \ge k$ while $t(i)$ ranges over all non-negative integers for $i. There are only countably many functions $t$ for each $k$. Then enumerate all these functions in a sequence $h_0, h_1,h_2,\cdots$.

On the other hand, any compact subset of $\omega^\omega$ is a subset of $A_g$ for some $g \in \omega^\omega$. To see this, let $\pi_n$ be the projection from $\omega^\omega$ to the $n$th factor. Let $K \subset \omega^\omega$ be compact. Then for each $n$, $\pi_n(K)$ is compact in the discrete space $\omega$, hence finite. Since it is finite, for each $n$, $\pi_n(K) \subset [0,g(n)]$ for some $g(n) \in \omega$. Then $K \subset A_g$.

It follows that any $\sigma$-compact subset of $\omega^\omega$ is a subset of the union of countably many $A_g$, i.e. if $K \subset \omega^\omega$ is $\sigma$-compact, then $K \subset \bigcup_{n \in \omega} A_{g_n}$ for $g_0, g_1, g_2, \cdots \in \omega^\omega$.

$1 \rightarrow 2$
Suppose $F$ is bounded. Let $f \in \omega^\omega$ be an upper bound of $F$. It is clear that $F \subset B_f$, which is $\sigma$-compact.

$2 \rightarrow 3$
Let $F \subset \omega^\omega$. Suppose that $F \subset X \subset \omega^\omega$ where $X=\bigcup_{n \in \omega} X_n$ with each $X_n$ being a compact subset of $\omega^\omega$. For each $n$, let $Y_n=F \cap X_n$. Consider the sets $F$, $X_n$ and $Y_n$ as subsets of the real line. Since each $X_n$ is compact and $X_n \cap \mathbb{Q}=\varnothing$, each $Y_n$ is a closed subset of $F \cup \mathbb{Q}$. Thus $F$ is an $F_\sigma$ subset of $F \cup \mathbb{Q}$.

$3 \rightarrow 1$
Let $F \subset \omega^\omega$. Consider $F$ as a subset of $\mathbb{P}$. Suppose that $F=\bigcup_{n \in \omega} C_n$ where each $C_n$ is a closed subset of $F \cup \mathbb{Q}$. For each $n$, let $\overline{C_n}$ be the closure of $C_n$ in the real line. Because it is a closed subset of the real line, $\overline{C_n}$ is $\sigma$-compact. Since points of $\mathbb{Q}$ are not in the closure of $C_n$ in $F \cup \mathbb{Q}$, points of $\mathbb{Q}$ are not in the closure of $C_n$ in the real line. It follows that $\overline{C_n} \subset \mathbb{P}$. Now consider each $\overline{C_n}$ as a subset of $\omega^\omega$. According to the above discussion, each $\overline{C_n}$ is a subset of $\bigcup_{j \in \omega} A_{g_{n,j}}$ where $g_{n,0}, g_{n,1}, g_{n,2}, \cdots \in \omega^\omega$. Choose $f \in \omega^\omega$ such that $g_{n,j} \le^* f$ for all $n,j$ combinations. Then $f$ is an upper bound of $F$. This completes the proof of Theorem 1. $\square$

Reference

1. 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.

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The space of irrational numbers is not Menger

This post could very well be titled Killing Two Birds in One Stone. We present one proof that can show two results – the set $\mathbb{P}$ of all irrational numbers does not satisfy the Menger property and that $\mathbb{P}$ is homeomorphic to the product space $\omega^\omega$.

In this previous post, we introduce the notion of Menger spaces. A space $X$ is a Menger space (or has the Menger property) if for every sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$, there exists, for each $n$, a finite $\mathcal{V}_n \subset \mathcal{U}_n$ such that $\bigcup_{n \in \omega} \mathcal{V}_n$ is an open cover of $X$. In the definition if the open covers $\mathcal{U}_n$ are made identical, then the definition becomes that of a Lindelof space. Thus Menger implies Lindelof. For anyone encountering the notion of Menger spaces for the first time, one natural question is: are there Lindelof spaces that are not Menger? Another natural question: are there subsets of the real line that are not Menger?

The most handy example of “Lindelof but not Menger” is probably $\mathbb{P}$, the set of all irrational numbers. The way we show this fact in this previous post is by working in the product space $\omega^\omega$. This approach requires an understanding that $\mathbb{P}$ and the product space $\omega^\omega$ are identical topologically as well as working with the eventual domination order $\le^*$ in $\omega^\omega$. In this post, we give a direct proof that $\mathbb{P}$ is not Menger with the irrational numbers lying on a line. This proof will open up an opportunity to see that $\mathbb{P}$ is homeomorphic to $\omega^\omega$. The product space $\omega^\omega$ is also called the Baire space in the literature.

The sets $\mathbb{P}$ and $\omega^\omega$ are two different (but topologically equivalent) ways of looking at the irrational numbers. The first way is to look at the numbers on a line. With the rational numbers removed, the line will have countably many holes (but the holes are dense). The open sets in the line are open intervals or unions of open intervals. The second way is to view the irrational numbers as a product space – the product of countably many copies of $\omega$ with $\omega$ being the set of all non-negative numbers with the discrete topology. In many ways the product space version is easier to work with. For example, the eventual domination order $\le^*$ on $\omega^\omega$ help describe the classical covering properties such as Menger property and Hurewicz property.

A Direct Proof of Non-Menger

As mentioned above, the set $\mathbb{P}$ is the set of all irrational numbers. We let $\mathbb{Q}$ denote the set of all rational numbers.

To see that $\mathbb{P}$ does not satisfy the Menger property, we need to exhibit a sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $\mathbb{P}$ such that whenever we pick, for each $n$, a finite $\mathcal{V}_n \subset \mathcal{U}_n$, the set $\{ \mathcal{V}_n: n \in \omega \}$ cannot be a cover of $\mathbb{P}$, i.e. no matter how we pick the finite $\mathcal{V}_n$, there is always an irrational number $x$ that is missed by all $\mathcal{V}_n$. The open covers $\mathcal{U}_n$ are derived inductively, starting with $\mathcal{U}_0$.

To prepare for the inductive steps, we fix a scheme of choosing convergent sequences of rational numbers. For any interval $(a,b)$ where $a, b \in \mathbb{Q}$, we choose a decreasing sequence $\{ a_n \in (a,b) \cap \mathbb{Q}: n \in \omega \}$ such that $a_n \rightarrow a$ from the right with $a_0=b$ and an increasing sequence $\{ b_n \in (a,b) \cap \mathbb{Q}: n \in \omega \}$ such that $b_n \rightarrow b$ from the left with $b_0=a$. Enumerate $\mathbb{Q}$ in a countable sequence $\{ r_0, r_1, r_2, \cdots \}$ with $r_0=0$. Several points to keep in mind when choosing such sequences.

1. In each inductive step, either the sequences of the type $a_n$ or the sequences of the type $b_n$ are chosen but not both.
2. in the $j$-th stage of the inductive process, in picking the sequence $a_n$ or $b_n$ for the interval $(a,b)$, we make sure that the rational number $r_j$ is used in the sequence $a_n$ or $b_n$ if $r_j$ is in the interval $(a,b)$ and if $r_j$ is not previously chosen. So at the end, all rational numbers are used up as endpoints of intervals in all $\mathcal{U}_n$.
3. If sequence of the type $a_n$ is chosen out of the interval $(a,b)$, we derive the subintervals $\cdots (a_3,a_2),(a_2,a_1),(a_1,a_0)$ whose union is $(a,b)$. We want the sequence $a_n$ chosen in such a way that the length of each subinterval $(a_n,a_{n-1})$ is less than 1/2 of $b-a$.
4. If sequence of the type $b_n$ is chosen out of the interval $(a,b)$, we derive the subintervals $(b_0,b_1), (b_1,b_2),(b_2,b_3),\cdots$ whose union is $(a,b)$. We want the sequence $b_n$ chosen in such a way that the length of each subinterval $(b_n,b_{n+1})$ is less than 1/2 of $b-a$.

To start, let $\mathcal{U}_0=\{ (0,1),(1,2),(2,3),\cdots,(-1,0),(-2,-1),(-3,-2),\cdots \}$. In other words, $\mathcal{U}_0$ consists of all open intervals whose endpoints are the integers and whose lengths are 1. Label the elements of $\mathcal{U}_0$ as $\mathcal{U}_0=\{ O_0,O_1,O_2,\cdots \}$.

For each $O_i=(a,b) \in \mathcal{U}_0$, choose a sequence of rational numbers $b_n$ converging to $b$ from the left (according to the scheme described above). Make sure that the rational number $r_1$ is picked if $r_1 \in (a,b)$. From this, we obtain the intervals $(b_0,b_1), (b_1,b_2),(b_2,b_3),\cdots$ covering all the irrational numbers in $(a,b)$. Label these intervals as $O_{i,0},O_{i,1},O_{i,2},\cdots$. Then $\mathcal{U}_1$ consists of all such intervals obtained from each $O_i=(a,b) \in \mathcal{U}_0$.

Here’s how to define $\mathcal{U}_2$. For each $O_{i,j}=(a,b) \in \mathcal{U}_0$, choose a sequence of rational numbers $a_n$ converging to $a$ from the right (according to the scheme described above). Make sure that the rational number $r_2$ is picked if $r_2 \in (a,b)$ and if $r_2$ has not been chosen previously. From this, we obtain the intervals $\cdots (a_3,a_2),(a_2,a_1),(a_1,a_0)$ covering all the irrational numbers in $(a,b)$. Label these intervals as $O_{i,j,0},O_{i,j,1},O_{i,j,2},\cdots$. Then $\mathcal{U}_2$ consists of all such intervals obtained from each $O_{i,j}=(a,b) \in \mathcal{U}_1$.

From here on out, we continue the same induction steps to derive $\mathcal{U}_n$, making sure that when $n$ is odd, we choose sequence $b_n$ converging to the end point $b$ from the left and when $n$ is even, we choose sequence $a_n$ converging to the end point $a$ from the right for each $O_{i_0,i_2,\cdots,i_{n-1}}=(a,b) \in \mathcal{U}_{n-1}$, producing the intervals $O_{i_0,i_2,\cdots,i_{n-1},0}$, $O_{i_0,i_2,\cdots,i_{n-1},1}$, $O_{i_0,i_2,\cdots,i_{n-1},2}$, $O_{i_0,i_2,\cdots,i_{n-1},3}\cdots$.

Now the sequence $\{ \mathcal{U}_n: n \in \omega \}$ has been defined. Note that every rational number is used as the endpoint of some open interval in some $\mathcal{U}_n$. Each $\mathcal{U}_n$ is clearly an open cover of $\mathbb{P}$. Another useful observation is that

$\bigcap_{n \in \omega} U_n=\bigcap_{n \in \omega} \overline{U_n}=\{ x \}$

whenever $U_n \in \mathcal{U}_n$ for all $n$ and $U_{n+1} \subset U_n$ for all $n$. This is because each chosen interval has length less than 1/2 of the previous interval. Furthermore, the point $x$ in the intersection must be an irrational number since all rational numbers are used up as endpoints.

Let’s pick, for each $n$, a finite $\mathcal{V}_n \subset \mathcal{U}_n$. We now find an irrational number $x$ that is not in any interval of any $\mathcal{V}_n$.

Since $\mathcal{V}_0$ is finite, choose some $O_{m_0} \in \mathcal{U}_0$ such that $O_{m_0} \notin \mathcal{V}_0$. Since $\mathcal{V}_1$ is finite, choose some $O_{m_0,m_1} \in \mathcal{U}_1$ such that $O_{m_0,m_1} \notin \mathcal{V}_1$. In the same manner, choose $O_{m_0,m_1,m_2} \in \mathcal{U}_2$ such that $O_{m_0,m_1,m_2} \notin \mathcal{V}_2$.

Continuing the inductive process, we have open intervals $O_{m_0},O_{m_0,m_1},O_{m_0,m_1,m_2},\cdots$ from $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$, respectively. Furthermore, for each $n$, $O_{m_0,\cdots,m_n} \notin \mathcal{V}_n$. Then the intersection

$\bigcap_{n \in \omega} O_{m_0,\cdots,m_n}=\bigcap_{n \in \omega} \overline{O_{m_0,\cdots,m_n}}=\{ x \}$

must be an irrational number $x$. This is a real number that is not covered by $\mathcal{V}_n$ for all $n$. This shows that $\{ \mathcal{V}_n: n \in \omega \}$ cannot be an open cover of $\mathbb{P}$. This completes the proof that $\mathbb{P}$ does not satisfy the Menger property. $\square$

A Homeomorphism

In defining the sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $\mathbb{P}$, we have also define a one-to-one correspondence between $\mathbb{P}$ and the product space $\omega^\omega$.

First, each $x \in \mathbb{P}$ is uniquely identified with a sequence of non-negative integers $f_0,f_1,f_2,\cdots$.

$x \mapsto (f_0,f_1,f_2,\cdots)=f \in \omega^\omega$

Observe that each irrational number $x$ belongs to a unique element in each $\mathcal{U}_n$. Thus the sequence $f_0,f_1,f_2,\cdots$ is simply the subscripts of the open intervals from the open covers $\mathcal{U}_n$ such that $x \in O_{f_0,\cdots,f_n}$ for each $n$. With the way the intervals $O_{f_0,\cdots,f_n}$ are chosen, it must be the case that $O_{f_0} \supset O_{f_1} \supset O_{f_2} \cdots$. Furthermore the intervals collapse to the point $x$.

$\bigcap_{n \in \omega} O_{f_0,\cdots,f_n}=\bigcap_{n \in \omega} \overline{O_{f_0,\cdots,f_n}}=\{ x \}$……………(1)

The mapping goes in the reverse direction too. For each $f=(f_0,f_1,f_2,\cdots) \in \omega^\omega$, we can use $f$ to obtain the open intervals $O_{f_0,\cdots,f_n}$. These intervals collapse to a single point, which is the irrational number that is associated with $f=(f_0,f_1,f_2,\cdots)$.

Denote this mapping by $H: \mathbb{P} \rightarrow \omega^\omega$. For each $x \in \mathbb{P}$, $H(x)=f$ is derived in the manner described above. The function $H$ maps $\mathbb{P}$ onto $\omega^\omega$. It follows that $H$ is a one-to-one map and that both $H$ and the inverse $H^{-1}$ are continuous. To see this, we need to focus on the open sets in both the domain and the range.

For the correspondence $H(x)=(f_0,f_1,f_2,\cdots)=f$, we consider the two sets:

$\mathcal{B}_x=\{ O_{f_0}, O_{f_0,f_1}, O_{f_0,f_1,f_2}, \cdots, O_{f_0,\cdots,f_n}, \cdots \}$

$\mathcal{V}_f=\{ E(f,n): n \in \omega \}$

where $E(f,n)=\{ g \in \omega^\omega: f_j=g_j \ \forall \ j \le n \}$ for each $n$. Note that $\mathcal{B}_x$ is a local base at $x \in \mathbb{P}$. This is because the intervals $O_{f_0,\cdots,f_n}$ have lengths converging to zero and they collapse to the point $x$ in the manner described in (1) above. On the other hand, $\mathcal{V}_f$ is a local base at $H(x)=f \in \omega^\omega$. Furthermore, there is a clear correspondence between $\mathcal{B}_x$ and $\mathcal{V}_f$. Note that $H(O_{f_0,\cdots,f_n})=E(f,n)$. This means that both $H$ and the inverse $H^{-1}$ are continuous. Since $\{ \mathcal{B}_x: x \in \mathbb{P} \}$ is a base for $\mathbb{P}$, it is clear the map $H$ is one-to-one.

This previous post has a shorter (but similar) derivation of the homeomorphism between $\mathbb{P}$ and $\omega^\omega$.

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