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The axiom of choice is equivalent to the following statement: if $X$ is an infinite set then the cardinal of $X^2$ is the same as that of $X$. Is there an elementary proof of this statement?

Asaf Karagila
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Rodney Coleman
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  • See this wikipedia article. – Hanul Jeon Nov 22 '13 at 10:01
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    In fact the heading should be "cardinal of $X^2$. – Rodney Coleman Nov 22 '13 at 10:02
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    This has been asked some few times before. Let me find links. Please read those links and point out whether or not this is an "elementary proof" enough, or not. Regardless to that, let me point out that (1) the definition of elementary is not uniform amongst people, and you may mean "without advanced machinery" (such as? ordinals are elementary to set theory, for example), or you may mean "very easy to comprehend"; (2) not every fact which can be elementarily understood have a proof which is elementary. – Asaf Karagila Nov 22 '13 at 10:06
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    http://math.stackexchange.com/questions/174730/is-the-axiom-of-choice-needed-to-show-that-a2-a/ and http://math.stackexchange.com/questions/56466/for-every-infinite-s-s-s-times-s-implies-the-axiom-of-choice/ and http://math.stackexchange.com/questions/54892/about-a-paper-of-zermelo – Asaf Karagila Nov 22 '13 at 10:07

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The term elementary is a bit vague. It can be interpreted in two ways.

  1. Very simple proof, that can be understood by anyone which understands the basic definitions of the axiom of choice, and cardinality.

  2. Proof using only elementary concepts from set theory. This is another ambiguity, do ordinals count as elementary concepts? the equivalence between the well-ordering theorem and the axiom of choice?

The proof is generally divided into two and a half parts. Showing that without the axiom of choice, if $\delta$ is an infinite ordinal then $|\delta|=|\delta\times\delta|$ and that assuming the axiom of choice every infinite set is equipotent with an infinite ordinal. Then the proof of the other direction, if every infinite set is equipotent with its square, then every infinite set is equipotent with an ordinal.

In fact, you can escape the theory of ordinals, but you will end up developing some of its basic parts under the guise of sets which can be well-ordered (without saying that completely). Such proof you can find in Francois Dorais' answer here.

Generally speaking, if you understand the basic ideas behind ordinals, the proof should be accessible and elementary. If you don't consider ordinals to be elementary, you can still somehow get around them but the proof becomes somewhat awkward.

Let me finish by saying that in mathematics theorems whose statement is elementary does not necessarily have elementary proofs.

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Asaf Karagila
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