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)<g(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)<g(j) for any i<j.
  • 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<n \} + n+1.

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<k, 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<m_1 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<m_{k-1}<m_k 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<k. 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.

\text{ }

\text{ }

\text{ }

Dan Ma topology

Daniel Ma topology

Dan Ma math

Daniel Ma mathematics

\copyright 2020 – Dan Ma

1 thought on “Adding up to a non-meager set

  1. Pingback: The ideal of meager sets | Dan Ma's Topology Blog

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s