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 $A$.


1 Answer 1


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.

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

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