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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!

Noah Schweber
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  • I think this is actually an open question: It's basically analogous to asking how many different equivalence classes (up to rational differences) of irrationals there are. – Justin Benfield Mar 13 '16 at 23:15
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    @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 $|\mathbb R|$. – Andrés E. Caicedo Mar 13 '16 at 23:26
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    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. – Asaf Karagila Mar 14 '16 at 04:48
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    @AsafKaragila "Attack ships on fire off the shoulder of Orion! Models in which every cardinal has cofinality $\omega$!" – Noah Schweber Mar 14 '16 at 05:03
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    "All those models will be lost in $V$... like reals in the $L$." – Asaf Karagila Mar 14 '16 at 05:07
  • I'd watch it. :) – Noah Schweber Mar 14 '16 at 05:10
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    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. – Asaf Karagila Mar 14 '16 at 05:16
  • It's plausible....... Solovay (1972) showed that if there is a measurable cardinal then there is a set-model of ZF that satisfies "$\mathbb R$ is a countable union of countable sets". But existence of a measurable cardinal is a fairly strong assumption, not provable in ZFC (unless 0=1 is provable in ZFC).... Many odd-sounding things are equiconsistent with ZF+$\neg AC$, such as a vector space with 2 Hamel (vector-space) bases that are cardinally incomparable. – DanielWainfleet Jan 07 '17 at 06:23
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    @AsafKaragila. Working title: The Man Who Knew $\omega_1$. – DanielWainfleet Jan 07 '17 at 06:26
  • @user254665: What result are you talking about? It's due to Feferman and Levy, from 1964, and requires absolutely no large cardinals. The results about a vector space with two bases is due to Lauchli, also from 1964 I think (but maybe 1962?), and it is originally a permutation model with atoms, while it can be transferred to ZF the transfer process is rarely one that lets you to fully control the resulting vector space. It could be better controlled by some methods (or perhaps meta-methods), but you still don't get the reals over the rationals, I believe. – Asaf Karagila Jan 07 '17 at 06:32

1 Answers1

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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.

Paul Larson
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