# Spanning the reals with a small set - choicelessly

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!

• @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 $|\mathbb R|$. Commented Mar 13, 2016 at 23:26
• 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. Commented Mar 14, 2016 at 4:48
• @AsafKaragila "Attack ships on fire off the shoulder of Orion! Models in which every cardinal has cofinality $\omega$!" Commented Mar 14, 2016 at 5:03
• "All those models will be lost in $V$... like reals in the $L$." Commented Mar 14, 2016 at 5:07
• @AsafKaragila. Working title: The Man Who Knew $\omega_1$. Commented Jan 7, 2017 at 6:26

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.
• A late thought: does GST help answer my questions about "generic cardinalities of $\mathbb{R}$" (1, 2)? Commented Apr 17, 2021 at 21:05