10
$\begingroup$

In this question Subset of the plane that intersects every line exactly twice someone ask for a reference of a paper where they proof the result : ''There exist a subset of the plane that intersects each line exactly twice'' (called $2$-point sets [I think]).

I was talking with an students and we were wondering if, in the absence of Choice, this result is still true.

To make my question more precise, I know that we only need Choice to well order the reals and, I believe, you can make the set to be non-measureble but I'm not sure if it is non-measurable itself (I believe that using CH you can make a $2$-point set of measure zero [but, to be honest, I haven't work out the details]).

So, being more concrete, how much choice is need to create such a set? Are there models of ZF without them?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ It seems that this is open whether or not such set can even be Borel, let alone $G_\delta$. In that case, I think you can probably construct one without appealing to the axiom of choice. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 3, 2015 at 23:14
  • 3
    $\begingroup$ Arnie Miller showed that you don't need a well ordering of reals in the sense that there are ZF models with 2-point sets where the set of reals cannot be well ordered: math.wisc.edu/~miller/res/two-pt.pdf He also showed that in $L$, there are conanalytic 2-point sets which is the best known upper bound - An analytic 2-point set is necessarily Borel. Mauldin has results connecting this to geometric measure theory: math.unt.edu/~mauldin/papers/no100.pdf $\endgroup$
    – Ashutosh
    Commented Dec 4, 2015 at 0:13
  • 1
    $\begingroup$ On measure and category: There is a 2-point set within $\{z \in \mathbb{R}^2 : |z| \in C\})$ where $C$ is any set of reals which, say, meets every interval on a perfect set . So there is always a meager null 2-point set. $\endgroup$
    – Ashutosh
    Commented Dec 4, 2015 at 0:56
  • $\begingroup$ Great question! This one is closely related: mathoverflow.net/questions/272527/…. $\endgroup$
    – Will Brian
    Commented Feb 21, 2019 at 13:45

1 Answer 1

Reset to default
3
$\begingroup$

I wasn't aware of A. Miller's paper http://www.math.wisc.edu/~miller/res/two-pt.pdf up until recently. In Miller's model, DC (dependent choice) is false (there is a Dedekind finite infinite set). In a paper with M. Beriashvili I show that ZF plus DC plus the existence of a 2-point set does not imply that there is a well-ordering of the reals, see https://ivv5hpp.uni-muenster.de/u/rds/mazurkiewicz_sets.pdf or https://ivv5hpp.uni-muenster.de/u/rds/mazurkiewicz.pdf .

Improve this answer
$\endgroup$
1
  • $\begingroup$ I think it's about time that you guys find the right general theorem for this (do contact me if you want to collaborate on this). It seems like a source of a lot of papers, and it seems to me that there is some bigger theorem in the background that you're just scratching instances of each and every time. $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 21, 2019 at 13:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .