In their 1943 paper On non-denumerable graphs, Erdos and Kakutani suggest as likely the following proposition.

(EK*) Suppose CH fails and $\lbrace M_n : n \in \omega \rbrace$ is a countable family of sets of rationally independent real numbers; then $\bigcup_{n \in \omega} M_n$ has inner measure zero.

Their Theorem 2 establishes the equivalence of CH with the assertion that $\mathbb{R} = \bigcup_{n \in \omega} M_n$ for some family $\lbrace M_n : n \in \omega \rbrace$ of sets of rationally independent numbers.

Has (EK*) ever been settled?

The question is prompted by the related question (and its discussion) monochromatic cycle-free colouring of the complete graph on R?


Addendum: P. Komjath has pointed out that the following argument is originally due to Erdos and Kunen. See page 136 here.

Claim 1: (Mycielski) Suppose $A$ is a compact subset of plane of positive area. Then there exist perfect sets $P, Q$ such that $Q$ has positive length and $P \times Q \subseteq A$.

Proof: (Martin) Let $M$ be a countable transitive model of ZFC containing a code of $A$. Let $N$ be a forcing extension of $M$ in which continuum is at least $\omega_3$ and Martin's axiom (MA) holds.

Claim 2: Assume MA + $2^{\omega} \geq \omega_3$. For every collection $\{K_{\alpha} : \alpha < \omega_2\}$ of compact sets of positive measure, there is a compact set $ K$ of positive measure such that $ K \subseteq K_{\alpha}$ for at least $ \omega_2$ many $ \alpha$'s.

Proof: WLOG, we may assume that each $K_{\alpha} \subseteq [0, 1]$ with $\mu(K_{\alpha}) > 0.5$. Let $U_{\alpha} = [0, 1] \backslash K_{\alpha}$. Consider the poset $P$ consisting of open subsets of $[0, 1]$ of measure strictly less than $0.5$ ordered by reverse inclusion; i.e., $U \leq V$ iff $V \subseteq U$. $P$ has ccc because the measure algebra is separable. By MA and $2^{\omega} \geq \omega_3$, it follows that every subset of $P$ of size $\omega_2$ has a centered subset of size $\omega_2$. Hence for some $S \in [\omega_2]^{\omega_2}$, any finite union of sets in $\{U_{\alpha}: \alpha \in S\}$ is in $P$. Claim 2 follows.

The claim implies that, in $N$, there is a compact set $Q$ of positive measure such that $W = \{x: Q \subseteq A_x = \{y: (x, y) \in A\}\}$ has size at least $\omega_2$. Since $A$ is closed, $W$ is a coanalytic set hence it is a union of $\omega_1$ many Borel sets. Thus $W$ must contain a prefect set $P$. So $P \times Q \subseteq A$.

Now the statement $(\exists P, Q)(P, Q \text{ are perfect}, Q \text{ has positive measure and }P \times Q \subseteq A)$ is $\Sigma^{1}_2(A)$ so by Shoenfield's absoluteness theorem it holds in $M$ and hence, by $\Pi_1^1$ absoluteness, also in $V$.

Claim 3: Assume $2^{\omega} \geq \omega_2$. Let $X_n$, for $n < \omega$, be pairwise disjoint $\mathbb{Q}$-linearly independent sets of reals. Then $X = \bigcup_n X_n$ has zero inner measure.

Proof: Suppose not and let $K$ be a compact set of positive length contained in $X$. Let $A = \{(x, y) : y \in K + x\}$. By Lemma 1, there are disjoint perfect sets $P, Q$ such that $P \times Q \subseteq A$. Hence for every $x \in P, y \in Q$, $y - x \in K$. Consider the complete bipartite graph $K_{P, Q}$. Let $c$ be an edge coloring of $K_{P, Q}$ such that $c(x, y) = n$ iff $y - x \in X_n$. Let $P' \in [P]^{\omega_1}, Q' \in [Q]^{\omega_2}$. Using Fodor's lemma get $x_0, x_1 \in P'$, $y_0, y_1 \in Q'$ such that $c(x_i, y_j) = n^{\star}$ for every $i, j < 2$. Let $a = y_1 - x_0$, $b = y_1 - x_1$, $c = y_0 - x_1$, $d = y_0 - x_0$. Then, $a, b, c, d \in X_{n^{\star}}, \{a, c\} \cap \{b, d\} = \phi$ but $a-b+c-d = 0$: A contradiction.

| cite | improve this answer | |
  • $\begingroup$ Fascinating answer. Why the change from $\aleph_3$ in Claim 2 to $\aleph_2$ in Claim 3? $\endgroup$ – Avshalom May 12 '15 at 10:15
  • $\begingroup$ Claim 3 is the Erods-Kakutani conjecture ($2^{\omega} \geq \omega_2$ is the negation of CH). Claim 2 is just an intermediate lemma in the proof of Claim 1. $\endgroup$ – Ashutosh May 12 '15 at 10:57
  • $\begingroup$ I follow. Claim 1/Lemma 1 wants $P$ and $Q$ disjoint. Is there a forcing-free proof of Claim 1? Many thanks for the lovely argument. $\endgroup$ – Avshalom May 12 '15 at 11:04
  • 1
    $\begingroup$ The reference for Mycielski's result is: J. Mycielski, Algebraic independence and measure, Fund. Math. 61 (1967), 165-169. I never read it but I don't think it uses forcing. $\endgroup$ – Ashutosh May 12 '15 at 11:19
  • 2
    $\begingroup$ "Kunen and I proved that if $c>\aleph_1$ then the union of $\aleph_0$ rationally independent sets always has inner measure 0." (then the proof, essentially the same as yours, is sketched) P. Erdos: Set theoretic, measure theoretic, combinatorial, and number theoretic problems concerning poit sets in Euclidean space, Real Analysis Exchange 4(1978-79), 113--138. $\endgroup$ – Péter Komjáth May 12 '15 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.