Counterexamples to the conjecture of Menger

This post is a continuation of the preceding post on Menger and Hurewicz spaces. We aim to provide the details on Example 4 and Example 5 in that post. These are counterexamples to Karl Menger’s conjecture concerning Menger spaces.

In 1924, Karl Menger introduced a basis covering property in metric spaces. He conjectured that among metric spaces this basis covering property is equivalent to \sigma-compactness. In 1925 Witold Hurewicz introduced a selection principle that led to a definition called Menger space as defined in the preceding post and below, which he showed to be equivalent to Menger’s basis covering property. In 1928, Sierpinski showed that Continuum Hypothesis (CH) implies that Menger’s conjecture is false. The counterexample is the Lusin sets. This is Example 4 in the preceding post, which is further discussed here. Lusin sets can be constructed using CH. On the other hand, Lusin sets do not exist under Martin’s axiom and the negation of CH. Thus Lusin sets, as counterexample to Menger’s conjecture, are only consistent counterexamples.

In a 1988 paper, Miller and Fremlin [3] gave the first ZFC counterexample to Menger’s conjecture. This is the subject of Example 5 in the preceding post. We plan to give a more detailed account of this example here.

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. If in the definition the \mathcal{V}_n can always be made to consist of only one element, then the space is said to be a Rothberger space. It is clear from definition that \sigma-compact \rightarrow Menger \rightarrow Lindelof. On the other hand, Rothberger \rightarrow Menger. The following diagram is from the preceding post.

Figure 1

      \displaystyle \begin{aligned} &  \sigma \text{-} \bold C \bold o \bold m \bold p \bold a \bold c \bold t  \\&\ \ \ \ \ \downarrow \\& \bold H \bold u \bold r \bold e \bold w \bold i \bold c \bold z  \\&\ \ \ \ \ \downarrow \\& \bold M \bold e \bold n \bold g \bold e \bold r  \ \ \ \ \ \leftarrow \ \ \ \bold R \bold o \bold t \bold h \bold b \bold e \bold r \bold g \bold e \bold r \\&\ \ \ \ \ \downarrow \\& \bold L \bold i \bold n \bold d \bold e \bold l \bold o \bold f  \end{aligned}

Example 4 – Lusin Sets

In this example, we show that any Lusin set is a Menger space that is not \sigma-compact. This fact is established through a series of claims listed below. The proofs are found at the end of the post.

Let X \subset \mathbb{R} be uncountable. The set X is said to be a Lusin set (alternative spelling: Luzin) if the intersection of X with any first category subset of the real line is countable, i.e. for every first category subset A of \mathbb{R}, X \cap A is countable. In other words, an uncountable subset X of the real line is a Lusin set if every first category subset of the real line can contain at most countably many points of X. Thus any Lusin set (if it exists) must necessarily be a subset of the real line that is of second category. Sets of first category are also called meager sets. For more information about Lusin sets, see [2].

Let X \subset \mathbb{R}. Let B \subset \mathbb{R}. The set X is said to be concentrated on the set B if for any open set U with B \subset U, X \backslash U is countable. In other words, X is concentrated on the set B if any open set containing the B contains all but countably many points of X.

Claim 1
Let X \subset \mathbb{R}. If X is a Lusin set, then X is concentrated on any countable dense subset of X.

Claim 2
Let X \subset \mathbb{R}. Suppose that D is a countable dense subset of X. Then if X is concentrated on the set D, then X is a Rothberger space and hence a Menger space.

Claim 3
Let X \subset \mathbb{R}. If X is an uncountable \sigma-compact, then X contains a Cantor set.

Note. By Cantor set here, we mean any subset of the real line that is homeomorphic to the middle third Cantor in the unit interval or the product space \{0, 1 \}^\omega.

Claim 4
Let X \subset \mathbb{R}. If X is a Lusin set, then X is a Menger space that is not \sigma-compact.

Lusin sets can be constructed under CH while they cannot exist under Martin’s axiom and the negation of CH. Thus any Lusin set is a consistent counterexample to Menger’s conjecture.

Example 5 (Miller and Fremlin)

The example makes use of two cardinals \mathfrak{b} and \mathfrak{d} that are associated with subsets of \omega^\omega, which is the set of all functions from \omega into \omega. Define an order \le^* on \omega^\omega as follows. For any f, g \in \omega^\omega, we say f \le^* g if f(n) \le g(n) for all but finitely many n. A subset A of \omega^\omega is a bounded set if there exists some f \in \omega^\omega such that for all g \in A, g \le^* f, i.e. f is an upper bound of A according to \le^*. The set A is said to be unbounded if it is not bounded. The set A is a dominating set if for each f \in \omega^\omega, there exists g \in A such that f \le^* g.

The cardinal \mathfrak{b} is called the bounding number and is the least cardinality of an unbounded subset of \omega^\omega. The cardinal \mathfrak{d} is called the dominating number and is the least cardinality of a dominating subset of \omega^\omega. It is always the case that \omega_1 \le \mathfrak{b} \le \mathfrak{d} \le \mathfrak{c} where \mathfrak{c} is the cardinality of the continuum. For more information on bounded and dominating sets and the associated cardinals, see the preceding post or [4].

Furthermore, the cardinal \text{non}(\text{Menger}) is the least cardinality of a non-Menger subset of the real line. It is shown in the preceding post that \mathfrak{d}=\text{non}(\text{Menger}). As a result, any subset of the real line with cardinality less than \mathfrak{d} must be a Menger space. It is consistent with ZFC that \mathfrak{b} < \mathfrak{d}. When this happens, any subset of the real line with cardinality \omega_1 is a Menger space that is not \sigma-compact. But this by itself is only a consistent example and not a ZFC example. We continue to fine tune until we get a ZFC example.

Another notion that is required is that of a scale. Let X \subset \omega^\omega. The set X is said to be a scale if it is a dominating set and is well ordered by \le^*. The following claims will produce the desired example. In proving the following claims, we make of the fact that the set of irrational numbers, as a subset of the real line, is homeomorphic to the product space \omega^\omega. Thus in dealing with a set of irrational numbers, we sometimes think of it as a subset of \omega^\omega.

Claim 5
Let X \subset \omega^\omega. If X is a dominating set of cardinality \omega_1, then there exists Y \subset X such that Y is a scale.

Claim 6
Let Y \subset \omega^\omega be compact. Then Y is a bounded set.

Claim 7
Let X \subset \omega^\omega. If X is a scale of cardinality \omega_1, then X is concentrated on \mathbb{Q} where \mathbb{Q} is the set of all rational numbers.

Claim 8
If X is an uncountable \sigma-compact subset of the real line, then X contains a Cantor set C such that C contains no rational number.

Note. By Cantor set here, we mean any subset of the real line that is homeomorphic to the middle third Cantor in the unit interval or the product space \{0, 1 \}^\omega.

Claim 9
There exists a subset X of the real line with \lvert X \lvert=\omega_1 such that X is a Menger space that is not \sigma-compact.

\text{ }

\text{ }

\text{ }

The Proof Section

Proof of Claim 1
Let X be a Lusin set. Let D be any countable dense subset of X. Let U be any open subset of the real line such that D \subset U. Let W=U \cup (\mathbb{R} \backslash \overline{X}). Note that W is a dense open subset of the real line. Then \mathbb{R} \backslash W is a closed nowhere dense subset of the real line. Since X is a Lusin set, \mathbb{R} \backslash W can contain at most countably many points of X. Note that \mathbb{R} \backslash \overline{X} contains no points of X. The set \mathbb{R} \backslash U can contain at most countably many points of X. It follows that X \backslash U is countable. This shows that X is concentrated on any countable dense subset of X. \square

Proof of Claim 2
Let X \subset \mathbb{R}. Suppose that D is a countable subset of X such that D is a dense subset of X. Suppose X is concentrated on the set D. We show that X is a Rothberger space. To this end, let \{ \mathcal{U}_n: n \in \omega \} be a sequence of open covers of X. Enumerate the set D=\{ d_0, d_2, d_4, \cdots \} (indexed by the even integers). For each even n, choose V_n \in \mathcal{U}_n such that d_n \in V_n. Each V_n is an open subset of X. Find W_n, open in the real line, such that V_n=W_n \cap X. Let W=\bigcup_{\text{even } n} W_n. Note that D \subset W. Since X is concentrated on D, there are at most countably many points of X not in W. Say, the set of these points is labeled as \{d_1, d_3, d_5, \cdots \} (indexed by the odd integers). Then for each odd n, choose V_n \in \mathcal{U}_n such that d_n \in V_n. Overall, we can choose, for each n, one set V_n \in \mathcal{U}_n such that \{V_0,V_1,V_2,V_3,\cdots \} is a cover of X. This shows that X is a Rothberger space, hence a Menger space. \square

Proof of Claim 3
Let X be an uncountable \sigma-compact subset of the real line. Let X=\bigcup_{n \in \omega} Y_n where each Y_n is a compact subset of the real line. Then for at least one n, Y_n is uncountable. Let Y_n be one such. It is well known that any uncountable closed subset of the real line contains a Cantor set. For an algorithm of how to construct a Cantor set, see here. \square

Proof of Claim 4
This is where we put the preceding three claims together to prove that any Lusin set is a desired counterexample. Let X be a Lusin set. By Claim 1, X is concentrated on any countable dense subset of X. Since there is some countable subset D \subset X such that D is dense in X, by Claim 2, X is a Menger space.

If X is \sigma-compact, then by Claim 3 X would contain a Cantor set C. The set C is a nowhere dense subset of the real line. Because X is a Lusin set, X can have at most countably many points in C (a contradiction). Thus X cannot be \sigma-compact. \square

Proof of Claim 5
Let X=\{ x_\alpha: \alpha<\omega_1 \} be a dominating set of size \omega_1. We now derive Y=\{ y_\alpha: \alpha<\omega_1 \} \subset X such that Y is also a dominating set and such that for any \alpha<\beta, we have y_\alpha \le^* y_\beta. We derive by induction. First let y_0=x_0. Suppose that for \beta<\omega_1, Y_\beta=\{ y_\alpha \in X: \alpha<\beta \} has been chosen such that for any \gamma<\tau<\beta, we have y_\gamma \le^* y_\tau. The set Y_\beta \cup \{ x_\beta \} is a countable subset of \omega^\omega. As a countable set, it is bounded. There exists f \in \omega^\omega such that for each g \in Y_\beta \cup \{ x_\beta \}, we have g \le^* f. Since X is dominating, there exists x_\mu \in X such that f \le^* x_\mu. Let y_\beta=x_\mu. Now Y=\{ y_\alpha: \alpha<\omega_1 \} has been inductively derived. It is clear that Y is a scale. \square

Proof of Claim 6
Let Y \subset \omega^\omega be compact. Let B_n be the projection of Y into the nth coordinate \omega. The set B_n must be a finite set since \omega is discrete. Note that Y \subset \prod_{n \in \omega} B_n. Furthermore, \prod_{n \in \omega} B_n is a bounded set. \square

Proof of Claim 7
Let X=\{ x_\alpha: \alpha<\omega_1 \} be a scale of size \omega_1. Let U be an open subset of the real line such that \mathbb{Q} \subset U. For each positive integer n, consider the set K_n=[-n,n] \backslash U. The set K_n is a compact subset of the real line consisting entirely irrational numbers. Thus we can consider each K_n as a subset of \omega^\omega. By Claim 6, any compact subset of \omega^\omega is bounded. Thus for each n, there exists f_n \in \omega^\omega such that f_n is an upper bound of K_n, i.e. for each h \in K_n, we have h \le^* f_n. For each n, let g_n \in X such that f_n \le^* g_n (using the fact that X is dominating).

For each n, let X_n=X \cap K_n. Note that X \backslash U=\bigcup_{n \ge 1} X_n. Furthermore, for each h \in X_n, we have h \le^* f_n \le^* g_n. Since X is a scale of size \omega_1, there are only countably many points in X that are \le^* g_n. Thus each X_n is countable. As a result, X \backslash U is countable. This proves that the scale X is concentrated on the set \mathbb{Q} of rationals. \square

Proof of Claim 8
This is similar to Claim 3, except that this time we wish to obtain a Cantor set containing no rational numbers. Let enumerate the rational numbers in a sequence \{ r_0, r_1, r_2, \cdots \}. In the algorithm indicated here, at the nth step of the inductive process, we can make sure that the chosen intervals miss r_n. The selected intervals would then collapse in a Cantor set consisting entirely of irrational numbers. \square

Proof of Claim 9
This is the step where we put everything together to obtain ZFC example of Menger but not \sigma-compact. Recall the dominating number \mathfrak{d}. We consider two cases. Case 1. \omega_2 \le \mathfrak{d}. Case 2. \mathfrak{d}=\omega_1. The first case is that \mathfrak{d} is at least \omega_2, the second uncountable cardinal. The second case is that \mathfrak{d} is less than \omega_2, hence is \omega_1, the first uncountable cardinal. In either case, we can obtain a subset of the real line that is Menger but not \sigma-compact.

Case 1. \omega_2 \le \mathfrak{d}
Then any subset of the real line with cardinality \omega_1 must be Menger (see Theorem 10 in the preceding post). Such a set cannot be \sigma-compact. If it is, it would contain a Cantor set which has cardinality continuum (see Claim 3). In this scenario, \omega_1 is strictly less than continuum since \omega_1 < \omega_2 \le \mathfrak{d} \le \mathfrak{c}.

Case 2. \mathfrak{d}=\omega_1
Let Y \subset \omega^\omega be a dominating set of cardinality \omega_1. This set is not a Menger space. If a subset of \omega^\omega is a Menger space, it must be a non-dominating set (see Theorem 4 in the preceding post). By Claim 5, we can assume that Y is a scale. Let X=Y \cup \mathbb{Q}. By Claim 7, X is concentrated on \mathbb{Q}. By Claim 2, X is a Menger space.

We claim that X is not \sigma-compact. Suppose X=\bigcup_{n \in \omega} H_n where each H_n is compact. Then H_m is uncountable for some m. By Claim 8, there is a Cantor set C such that C \subset H_n and that C contains no rational numbers. Let U=\mathbb{R} \backslash C. Here U is an open set containing the rational numbers but uncountably many points of X are outside of U. This means that X is not concentrated on \mathbb{Q}, going against Claim 7. Thus X is not \sigma-compact. \square

Remarks on Example 5

At heart, the derivation of Example 5 is deeply set-theoretic. It relies heavily on set theory. Just that when the set theory goes one way, there is a counterexample and when the set theory goes the opposite way, there is also a counterexample. Taken altogether, it appears no extra set theory is needed. However, when each case is taken by itself, extra set theory is required. Case 2 says there is a scale of cardinality \omega_1. This is possible when \mathfrak{b}=\mathfrak{d}=\omega_1. This equality is consistent with ZFC. It is also consistent that \mathfrak{b}=\mathfrak{d}=\kappa where \kappa is a regular cardinal and \kappa>\omega_1. It is also consistent that \omega_1=\mathfrak{b}<\mathfrak{d} or \omega_1<\mathfrak{b}<\mathfrak{d}. In general, a scale exists when \mathfrak{b}=\mathfrak{d} (the scale is then of cardinality \mathfrak{b}). So Case 2 is possible because of set theory. For more information about the bounding number \mathfrak{b} and the dominating number \mathfrak{d}, see [4].

Case 1, being the opposite of Case 2, is also deeply set-theoretically sensitive. However, the two cases taken together have the effect that set theory beyond ZFC is not used. This argument is a dichotomy. The dichotomic proof covers all bases. The manipulation of set theory to produce a ZFC result is clever. It is indeed a very nifty proof! Though the proof is not constructive, the example shows that Menger’s conjecture is false in ZFC. There are better counterexamples, e.g. [1]. However, the example of Miller and Fremlin shows that Menger’s conjecture is indeed false in ZFC.

Reference

  1. Bartoszyński T., Tsaban B., Hereditary topological diagonalizations and the Menger–Hurewicz Conjectures, Proc. Amer. Math. Soc., 134 (2), 605-615, 2005.
  2. Miller, A. W., Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 201-233, 1984.
  3. Miller A. W., Fremlin D. H., On some properties of Hurewicz, Menger, and Rothberger, Fund. Math., 129, 17-33, 1988.
  4. 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.

\text{ }

\text{ }

\text{ }

Dan Ma topology

Daniel Ma topology

Dan Ma math

Daniel Ma mathematics

\copyright 2020 – Dan Ma

3 thoughts on “Counterexamples to the conjecture of Menger

  1. Pingback: Introducing Menger and Hurewicz spaces | Dan Ma's Topology Blog

  2. Pingback: Menger spaces are Lindelof-like | Dan Ma's Topology Blog

  3. Pingback: Hurewicz spaces are sigma-compact-like | 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