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I've seen this assertion in a few comments around the site, and I found the answer to the $\rightarrow$ implication here.

Does anyone know a (hopefully simple) proof of the $\leftarrow$ implication?

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YoTengoUnLCD
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  • Prove it by induction on $\kappa = |A|$. Well-order $\kappa \times \kappa$ by first considering $\max(x,y)$ and then applying the lexicographic ordering. Check that all initial segments have cardinality $< \kappa$. – David Jan 19 '16 at 04:57
  • More precisely: given AC, every cardinal $\kappa =\aleph_{\alpha}$ for some $\alpha$, so prove by (transfinite) induction on $\alpha$ that the order type on $\kappa \times\kappa$ has length $\kappa$. The ordering is $(a,b) \prec (c,d) \iff $ either $\max(a,b) < \max(c,d)$, or $\max(a,b) = \max(c,d)$ and $(a,b)$ precedes $(c,d)$ lexicographically. – BrianO Jan 19 '16 at 05:24
  • This has more than a handful duplicates. Do you want one with ordinals, or one with Zorn's lemma? – Asaf Karagila Jan 19 '16 at 05:28
  • I'd prefer to avoid ordinals. I couldn't find duplicates. – YoTengoUnLCD Jan 19 '16 at 05:29
  • http://math.stackexchange.com/questions/1043350/show-that-an-infinite-set-c-is-equipotent-to-its-cartesian-product-c-times-c should be an answer; although the nice argument is with ordinals. I can't close as a duplicate since there are no upvoted answers, though. – Asaf Karagila Jan 19 '16 at 05:32
  • (Frankly, the proof via Zorn is incredibly convoluted and requires several additional proofs; the proof that for every infinite ordinal $|\alpha|^2=|\alpha|$ is relatively straightforward and beautiful.) – Asaf Karagila Jan 19 '16 at 05:42
  • @AsafKaragila Thanks for the edits to your answer, it made it much clearer. Do you have a link for the proof using ordinals at hand? I know very little about ordinals, but perhaps I understand it. – YoTengoUnLCD Jan 19 '16 at 16:12
  • Sure, http://math.stackexchange.com/questions/608538/godels-pairing-function-and-proving-c-cc-for-aleph-cardinals has both a summary of the ordinal-based proof (with an expanded link) as well a variant of Zorn's argument which avoids choice (namely we make an appeal to a Zorn-like principle, but we can prove that the partial order we're interested in has a maximal element without really using the axiom of choice). – Asaf Karagila Jan 19 '16 at 16:16

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