# 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 $g \in X$.

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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