580 -Some choiceless results (3)

January 27, 2009

[Updated December 3. The previous proof that there is a canonical bijection \alpha\sim\alpha\times\alpha for all infinite ordinals \alpha was seriously flawed. Thanks to Lorenzo Traldi for pointing out the problem.]

5. Specker’s lemma.

This result comes from Ernst Specker, Verallgemeinerte Kontinuumshypothese und Auswahlaxiom, Archiv der Mathematik 5 (1954), 332-337. I follow Akihiro Kanamori, David Pincus, Does GCH imply AC locally?, in Paul Erdős and his mathematics, II (Budapest, 1999), Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, (2002), 413-426 in the presentation of this and the following result. The Kanamori-Pincus paper, to which we will return next lecture, has several interesting problems, results, and historical remarks, and I recommend it. It can be found here.

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580 -Some choiceless results (2)

January 25, 2009

There are a few additional remarks on the Schröder-Bernstein theorem worth mentioning. I will expand on some of them later, in the context of descriptive set theory.

The dual Schröder-Bernstein theorem (dual S-B) is the statement “Whenever A,B are sets and there are surjections from A onto B and from B onto A, then there is a bijection between A and B.”

* This follows from the axiom of choice. In fact, {\sf AC} is equivalent to: Any surjective function admits a right inverse. So the dual S-B follows from choice and the S-B theorem. 

* The proofs of S-B actually show that if one has injections f:A\to B and g:B\to A, then one has a bijection h:A\to B contained in f\cup g^{-1}. So the argument above gives the same strengthened version of the dual S-B. Actually, over {\sf ZF}, this strengthened version implies choice. This is in Bernhard Banaschewski, Gregory H. Moore, The dual Cantor-Bernstein theorem and the partition principle, Notre Dame J. Formal Logic 31 (3), (1990), 375-381. 

* If j : {}x \to y is onto, then there is k:{\mathcal P}(y)\to {\mathcal P}(x) 1-1, so if there are surjections in both directions between A and B, then {\mathcal P}(A) and {\mathcal P}(B) have the same size. Of course, this is possible even if A and B do not.

Open question. ({\sf ZF}) Does the dual Schröder-Bernstein theorem imply the axiom of choice?

* The dual S-B is not a theorem of {\sf ZF}.

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