19
$\begingroup$

Working in ZF (so, no choice): is it possible that there is a set of reals $X$ such that

  • $\vert X\vert<\mathbb{R}$, but

  • $X$ generates $\mathbb{R}$ as a subgroup under addition?

This seems weird, but I can't even show that we can't generate $\mathbb{R}$ with a Dedekind-finite set!

$\endgroup$
10
  • 3
    $\begingroup$ @JustinBenfield The problems are not analogous (or basically analogous): $\mathbb R$ always embeds into the collection of equivalence classes of the Vitali equivalence relation. What is not provable without choice is that this collection is not larger than $|\mathbb R|$. $\endgroup$ Commented Mar 13, 2016 at 23:26
  • 2
    $\begingroup$ Maybe because "I've seen things you people wouldn't believe", but spanning the reals with a small set doesn't sound that weird. No idea about the answer, though. Nice question. $\endgroup$
    – Asaf Karagila
    Commented Mar 14, 2016 at 4:48
  • 5
    $\begingroup$ @AsafKaragila "Attack ships on fire off the shoulder of Orion! Models in which every cardinal has cofinality $\omega$!" $\endgroup$ Commented Mar 14, 2016 at 5:03
  • 6
    $\begingroup$ "All those models will be lost in $V$... like reals in the $L$." $\endgroup$
    – Asaf Karagila
    Commented Mar 14, 2016 at 5:07
  • 3
    $\begingroup$ @AsafKaragila. Working title: The Man Who Knew $\omega_1$. $\endgroup$ Commented Jan 7, 2017 at 6:26

1 Answer 1

7
$\begingroup$

This is in the Geometric Set Theory book with Jindra, in particular on pages 190-191 of the version here: https://people.clas.ufl.edu/zapletal/files/balanced14.pdf

The partial order there forces over a Solovay model. The conditions are disjoint pairs of set of reals $(a,b)$, where $a$ is finite and $b$ is countable, with the order of coordinatewise inclusion.

The GST machinery shows that the partial order doesn't add reals. The union of the finite parts of a generic filter will then be a set of reals of cardinality less than the continuum such that, by genericity, every real is a sum of two of them. In fact, the set will be Dedekind-finite.

$\endgroup$
3
  • $\begingroup$ Wow, very nice, thanks! $\endgroup$ Commented Apr 14, 2021 at 20:20
  • $\begingroup$ Very nice indeed. Maybe it's a good idea to understand GST from the lens of standard constructions like symmetric extensions (which would make these an iteration of symmetric extensions)? $\endgroup$
    – Asaf Karagila
    Commented Apr 14, 2021 at 21:06
  • $\begingroup$ A late thought: does GST help answer my questions about "generic cardinalities of $\mathbb{R}$" (1, 2)? $\endgroup$ Commented Apr 17, 2021 at 21:05

You must log in to answer this question.

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