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<k. 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 nth 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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2 thoughts on “Topological meaning of bounded sets

  1. Pingback: Adding up to a non-meager set | Dan Ma's Topology Blog

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

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