7
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ I edited the post to incorporate your query/suggestion. $\endgroup$ – Ashutosh Apr 9 '17 at 22:21
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ (not sure if you want to allow non-measurable $Y$'s) $\endgroup$ – Pietro Majer Apr 9 '17 at 21:58
  • 1
    $\begingroup$ 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

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.