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I am fairly sure, given examples $\Bbb{R}\times \Bbb{R},\Bbb{R}\times \Bbb{Q},\Bbb{Q}\times \Bbb{Q} $, that this is correct, but do not know how to prove it.

In my cited examples the proof has always used some specific property of $A,B$, leaving me clueless as to the general method of solving this. I am especially interested in the case were $|\Bbb{R}|>A>|\Bbb{Q}|,|B|=|\Bbb{Q}|$, though ideally the answer would be more general.

To be explicit, I'm asking for a proof of the statement in the question title, where $A,B$ are sets with infinite cardinalities.

Meow
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  • $|\Bbb{R}|\gt A \gt |\Bbb{Q}|$ would be contrary to the continuum hypothesis – Henry Oct 18 '14 at 17:11
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    This proof appears at least once on this site. – Asaf Karagila Oct 18 '14 at 17:12
  • @AsafKaragila I searched for a while and found similar questions that seemed similar but simply cited the result. – Meow Oct 18 '14 at 17:12
  • (1) That's usually a good thing to indicate, at least in a comment, when posting the question. (2) I'm fairly sure that I posted at least one proof for this fact, let me search through my 400+ posts in [cardinals] and get back to you. – Asaf Karagila Oct 18 '14 at 17:13
  • I couldn't find anything that tied up all the loose ends in a single answer, so I've posted one. I hope it'll be helpful. – Asaf Karagila Oct 18 '14 at 17:28

3 Answers3

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This requires the axiom of choice in all accounts.

First because the axiom of choice is equivalent to the fact that every two cardinals can be compared (otherwise $\max$ has no meaning); and secondly because we use the fact that $|A|^2=|A|$ for infinite sets, which is also equivalent to the axiom of choice.

Now, using these two facts we have that if $|B|\leq|A|$ then:

$$|A|\leq |A\times B|\leq |A\times A|\leq|A|$$

In either case you can't quite prove this "explicitly" without the axiom of choice. Even if you do know that $|A|<|B|$, it might still be the case that $|B|<|A\times B|$.

Related threads:

  1. Godel's pairing function and proving c = c*c for aleph cardinals
  2. For all infinite cardinals $\kappa, \ (\kappa \times \kappa, <_{cw}) \cong (\kappa, \in).$
  3. Can an infinite cardinal number be a sum of two smaller cardinal number? (This might talk about sums, but the same argument shows that defining multiplication by taking the maximal of the two implies choice.)
  4. For any two sets $A,B$ , $|A|\leq|B|$ or $|B|\leq|A|$
  5. [To be added... maybe?]
Community
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Asaf Karagila
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Let $A$ be the biggest of the two $$ |A|\le|A\times B|\le|A^2| $$

The last piece you miss is $|A|=|A^2|$, that is equivalent to Choice Axiom, and can be proved, for example, through Zorn's Lemma

Exodd
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  • How would you apply Zorn here? The obvious partial ordering by inclusion doesn't work directly. – Meow Oct 18 '14 at 17:18
  • I remember it was a strange set you take for Zorn, but can't seem to reconscruct now. http://math.stackexchange.com/questions/54892/about-a-paper-of-zermelo/54904#54904 – Exodd Oct 18 '14 at 17:28
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Hint:$A\times B=\cup \{f:\{0,1\}\to A\cup B:f(0)\in A,f(1)\in B\}$.

Now suppose $|A|>|B|$ :

$|A|=|A\times A|$ and $A\times A=\cup \{f:\{0,1\}\to A:f(0),f(1)\in A\}$.

Find a bijection between $A\times B$ and $A\times A$.

Haha
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