# How much choice is needed to show that $|A\times \omega|=|A|$?

A recent question here (which I believe was deleted, unfortunately) came down to showing the title question, that $$|A\times\omega|=|A|$$ for all infinite sets $$A$$. In that situation, the questioner could assume AC so the question became easy, but it got me thinking about the situation without choice. I'm familiar with Tarski's theorem that $$|A\times A|=|A|$$ for all $$A$$ implies AC, and I'm pretty confident that it takes at least some choice to prove $$|A\times\omega|=|A|$$. For instance, it would seem to clearly fail if $$A$$ is Dedekind-finite: given the bijection $$\beta: A\times\omega\mapsto A$$, the set $$B=\{\beta(\langle a,0\rangle): a\in A\}$$ is a proper subset of $$A$$ equinumerous with it. But how much choice does it take? The natural assumption would be Countable Choice, but I don't see any sensible way of modifying the proof of Tarski's theorem I've seen to get CC from the bijection between $$A\times\omega$$ and $$A$$.

We can show, without choice whatsoever that $$|A|+|A|=|A|\implies|A|\times\aleph_0=|A|.$$

So the exact choice you need is "for every infinite $$A$$, $$|A|+|A|=|A|$$". This principle implies that there are no infinite Dedekind-finite sets, but it does not follow from $$\sf DC_\kappa$$, for any $$\kappa$$, nor it implies even $$\sf AC_\omega$$. It is, indeed, part of the "strange principles of choice" family that is just... out there.

• This makes sense! Is it known what the implications between that principle and the more familiar 'sub-choice' axioms are? Dec 2, 2020 at 1:07
• No, not really. Dec 2, 2020 at 1:11