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!

  • 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 $|\mathbb R|$. $\endgroup$ – Andrés E. Caicedo Mar 13 '16 at 23:26
  • 3
    $\begingroup$ @AsafKaragila "Attack ships on fire off the shoulder of Orion! Models in which every cardinal has cofinality $\omega$!" $\endgroup$ – Noah Schweber Mar 14 '16 at 5:03
  • 3
    $\begingroup$ "All those models will be lost in $V$... like reals in the $L$." $\endgroup$ – Asaf Karagila Mar 14 '16 at 5:07
  • 1
    $\begingroup$ They're making a sequel nowadays. You're closer to Hollywood. Maybe you can find the writers and have them change the script a little bit. $\endgroup$ – Asaf Karagila Mar 14 '16 at 5:16
  • 1
    $\begingroup$ @AsafKaragila. Working title: The Man Who Knew $\omega_1$. $\endgroup$ – DanielWainfleet Jan 7 '17 at 6:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.