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Aug 3, 2015 at 4:03 history edited Asaf Karagila CC BY-SA 3.0
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Aug 3, 2015 at 4:02 comment added Asaf Karagila @Harry: The question itself asks about cardinal arithmetic in presence and failure of "there is a choice of representatives from every set of cardinals". I don't think that answer appears in any paper or book. However questions about disjoint union of sets and the obtained cardinality do (which is morally the same question). It turns out that asking whether "the sum is well-defined" and "repeated sums is product" is the same thing, which is what I pointed out, and remarked both are little researched into. In general it's hard to work with "little bit" of cardinal structure.
Aug 3, 2015 at 2:50 comment added Harry Altman Just a clarifying question on that equivalent condition -- is it saying that if you sum an $I$-indexed collection of sets, all in bijection with a fixed set $A$, then the result has cardinality $|A||I|$? (I assume this is what you must mean as the other obvious interpretation would appear to be trivially true regardless of any assumption like this, but I figured I should ask.)
Aug 2, 2015 at 15:56 comment added user5810 "non-empty either" $\: \mapsto \:$ "also non-empty" $\;\;\;\;$
Aug 2, 2015 at 12:44 comment added Asaf Karagila You're welcome. The existence of such set is actually a trick I collected in my very first course in [axiomatic] set theory. If you want $A\times 2\approx A$, just replace $A$ by $(A\times\omega)$, and if you want $A\times A\approx A$, just replace it by $A^\omega$. This also allows you, often when needed, to assume that you are working with Dedekind-infinite sets.
Aug 2, 2015 at 12:32 comment added Wojowu Thank you very much, I'll look into these references later. I wouldn't say that products "trivially" imply choice, since existence of such $Y$ is something that I honestly probably couldn't have figured out :P
Aug 2, 2015 at 12:31 vote accept Wojowu
Aug 2, 2015 at 11:51 history answered Asaf Karagila CC BY-SA 3.0