Is the following consistent?

There exists $X \subseteq [0, 1]$, such that $X$ does not have measure zero and for every $Y \subseteq X$, if $Y$ does not have measure zero, then there are $y_1 < y_2 < y_3 < y_4$ in $Y$ such that $y_2 - y_1 = y_4 - y_3$.

Edit: Under CH, there is no such $X$. This follows from a result of Erdos and Kakutani that says that CH is equivalent to the following statement:

The real line $\mathbb{R}$ can be partitioned into countably many rationally independent sets.

The category analogue has a positive answer.

Maybe I should add a comment to Pietro Majer's response:

It is a simple consequence of Lebesgue density theorem that every non null measurable $Y$ must contain equal distances. There are non trivial problems about avoiding patterns in positive measure sets (Erdos similarity problem) but this one is quite different.

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    $\begingroup$ You ask about consistency (rather than truth or provability). Do you know that the negation is consistent? $\endgroup$ – Joel David Hamkins Apr 9 '17 at 21:49
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    $\begingroup$ I edited the post to incorporate your query/suggestion. $\endgroup$ – Ashutosh Apr 9 '17 at 22:21

Any measurable $X$ with positive measure does it, e.g. $X:=[0,1]$. Indeed, given any measurable $Y\subset\mathbb{R}$ with positive measure, let $a\in\mathbb{R}$ be such that both $Y_-:=Y\cap (-\infty,a)$ and $Y_+:=Y\cap (a,+\infty)$ have positive measure. By Steinhaus theorem, $Y_--Y_-$ and $Y_+-Y_+$ are nbds of $0$, so in particular there are $y_1<y_2$ in $Y_-$ and $y_3<y_4$ in $Y_+$ as required.

  • $\begingroup$ (not sure if you want to allow non-measurable $Y$'s) $\endgroup$ – Pietro Majer Apr 9 '17 at 21:58
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    $\begingroup$ I do. I know that every positive measure set contains a similar copy of every finite set (Leb. density), The set theory tag was chosen to consider the problem for arbitrary set of reals. $\endgroup$ – Ashutosh Apr 9 '17 at 22:00

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