# 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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