In their 1943 paper [On non-denumerable graphs](http://projecteuclid.org/download/pdf_1/euclid.bams/1183505209), 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?](https://mathoverflow.net/questions/35838/monochromatic-cycle-free-colouring-of-the-complete-graph-on-r)



Under $\neg CH$, have countable unions of rationally independent numbers inner measure zero?
Addendum: P. Komjath has pointed out that the following argument is originally due to Erdos and Kunen. See page 136 [here.](https://www.renyi.hu/~p_erdos/1978-40.pdf)


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.


