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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580 -I. Some choiceless results

January 21, 2009

The topic of this course is Combinatorial set theory, so even though we will study additional axioms, we will not emphasize forcing or inner model-theoretic techniques. Similarly, we will study some results of a descriptive set theoretic nature, but will not delve into the fine definability issues that descriptive set theory involves. We will assume the axiom of choice throughout (and I will assume basic knowledge of axiomatic set theory, cardinals and ordinals), but we begin by looking at some results that do not require the axiom of choice. 

1. Cantor’s theorem

Version 1. If f:X\to{\mathcal P}(X) then f is not surjective.

Proof. Let A=\{y\in X:y\notin f(y)\}. Then A\notin{\rm ran}(f). {\sf QED}

Version 2. If f:{\mathcal P}(X)\to X then f is not injective.

Proof. Let A=\{y\in X: \exists Z\,(y=f(Z)\notin Z)\} and set a=f(A), so a\in A, so there is some Z\ne A with f(Z)=f(A). {\sf QED}

Note that in version 1 we explicitly (i.e., definably) found a set not in the range of f. In version 2, we found a set A for which there is a set Z with (A,Z) witnessing a failure of injectivity, but we did not actually define such a set Z. I do not know whether this can be done; we will see later a different argument in which such a pair is defined.

2. The Tarski-Knaster theorem.

Theorem. Let (L,\le) be a complete lattice, and let f:L\to L be order preserving. Then the set of fixed points of f is a complete lattice (and, in particular, non-empty).

This is a handout on this result that I wrote for a set theory course I taught at Caltech.

3. The Schröder-Bernstein theorem.

Theorem. Assume that there are injections f:A\to B and g:B\to A. Then there is a bijection h:A\to B.

This is proved as a corollary of the Tarski-Knaster result, see the handout attached above.

Another nice way of proving the result is graph theoretic. We may assume that A and B are disjoint and form a directed graph whose nodes are elements of A\cup B and there is an edge from a to b iff either a\in A and b=f(a) or a\in B and b=g(a). Consider the connected components of this graph. Each component is either a cycle of even length, or a {\mathbb Z}-chain, or a {\mathbb N}-chain. In each case, one can canonically find a bijection between the elements of the component in A and the elements in B. Putting these bijections together gives the result.

Cantor’s proof of this result uses the axiom of choice (in the form: every set is in bijection with an ordinal).