[FOM] Another question about ZF without Choice

Andres Caicedo caicedo at diamond.boisestate.edu
Mon Feb 2 22:35:37 EST 2009


  Hi Tom,

> For a simple (one-page) proof of the result sought below in ZF without 
> Choice see p. 550 of the article in The American Mathematical Monthly 
> (MAA),vol 109, No. 6, (June-July, 2002) by Leonard Gillman (pp544-553)
>
> "Two Classical Surprises Concerning the Axiom
> of Choice and the Continuum Hypothesis"
>
> best regards,
> Tom Dunion

  The question (whether aleph(X) injects into P(P(X))) is somewhat more 
subtle than this. In Gillman's paper two results that seem related to my 
question are shown. The first is theorem 1, that aleph(X) injects into 
P(P(P(X))). The second is lemma 3, which gives that if aleph(X) injects 
into P(P(P(X))), then it also injects into P(P(X)). Unfortunately, lemma 3 
also implies that X is well-orderable, and hence it is not a theorem of 
ZF, which makes sense, since it is easy to give counterexamples to lemma 3 
in ZF, and Gillman's proof uses GCH.

  This is what I know at the moment about the problem (assume X is 
infinite): Let aleph_*(X) be the supremum of the ordinals onto which X can 
be mapped, so aleph(X)\le aleph_*(X) and strict inequality is possible in 
the absence of choice.

  * P(aleph(X)) injects P(P(X^2)).

  * In particular, P(alpha^+) injects into P^2(alpha) for alpha an infinite 
ordinal.

  * Also, since aleph(Y)=aleph(Y^2) for any infinite Y, if X is a square 
(i.e., equipotent to Y^2 for some Y), then P(aleph(X)) injects into 
P^2(X).

  * If alpha is below aleph_*(X), then P^2(alpha) injects into P^2(X).

  * In particular, P(aleph(X)) injects into P^2(X) if aleph(X) is a 
successor cardinal or aleph(X)<aleph_*(X).

  * If aleph(X)=kappa and kappa is the kappa-th initial ordinal, then 
P(aleph(X)) injects into P^2(X).

  * In particular, P(aleph(X)) injects into P^2(X) if X is Dedekind-finite.

Hence, if there is an X with aleph(X)=aleph(P^2(X)), then necessarily 
(X is not a square and) aleph(X)=aleph_*(X)=aleph_lambda for some infinite 
limit ordinal lambda < aleph_lambda.

  All these are easy observations. At this point, I think the question is 
open, but I would be grateful for any suggestions or references.

  Best,
  Andres


More information about the FOM mailing list