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For everything below we assume A is infinite

For sets A and B, let $A \equiv B$ denote there is a bijection between A and B. I know $\forall A, A \times A \equiv A \iff \text{Axiom of Choice is true}$

However, for particular A, such as the natural numbers, there is a way to list out $\mathbb{N} \times \mathbb{N}$ as a square and pictorially zig-zag to create a bijection without AC. So I wonder, can we generalize this to all other well-ordered sets? Perhaps with transfinite induction? I'm not exactly sure how to do the limit ordinal step especially when the limit ordinal is a cardinal.

I refrained from using $|A|$ i.e. the cardinality symbol as for a set to has a cardinality is equivalent that it can be well ordered

wsz_fantasy
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    You need to say that $A$ is infinite somewhere... – Arturo Magidin Jun 02 '23 at 23:22
  • @ArturoMagidin Thank you. I will probably fix my sentences when I got a chance, as for now, just assume all the A mentioned are infinite. – wsz_fantasy Jun 02 '23 at 23:29
  • @spaceisdarkgreen Can you elaborate on what $<_{cw}$? mean – wsz_fantasy Jun 03 '23 at 00:05
  • @wsz_fantasy They define it in the post. You can also read the proof of this in Jech's book, and many other set theory references. This is a little stronger than what you're asking for, of course. Also see here https://proofwiki.org/wiki/Non-Finite_Cardinal_is_equal_to_Cardinal_Product – spaceisdarkgreen Jun 03 '23 at 00:09

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