# [FOM] Historical Queries on AC

Thomas Forster T.Forster at dpmms.cam.ac.uk
Wed Jan 9 05:32:56 EST 2008


On Tue, 8 Jan 2008, joeshipman at aol.com wrote:

> 1) In 1938, Tarski (Fund. Math. vol. 30) showed that AC follows from
> the axiom that there is a Universe containing any set (in other words,
> that arbitrarily large inaccessible cardinals exist). Of course, the
> consistency (rather than the truth) of AC doesn't need the full
> Universes axiom, just one inaccessible limit of inaccessibles (because
> that set will satisfy ZF and the Universes axiom).

You mean that if there is a proper class of strong inaccessibles then AC
follows..?   Surely you can't mean that!?!

> 3) Sierpinski's proof is is stronger than just "GCH-->AC", it actually
> shows that for a set A to be well-orderable one needs only that there
> are no intermediate cardinals anywhere in the sequence A < P(A) <
> P(P(A)) < P(P(P(A))) < P(P(P(P(A)))).  Has anyone improved this to
> require a smaller set of no-intermediate-cardinal assumptions?
>
> 4) Analogously to 3), what is the best known result on how many levels
> of Universes above A are necessary in order to well-order A?

Say $\alpha$ adj $\beta$ if there is no cardinal intermediate between the
two.  Then alpha adj 2^alpha adj 2^2^alpha implies that 2^alpha is an
aleph.  This last refinement is due to Ernst Specker i think. To do it he
used a lemma that says that if x> 5 then x^2 < 2^x.  (The proof is quite
hard: try it!)  As far as I know it is still an open question whether or
not alpha adj 2^alpha implies that alpha is an aleph.  That sounds
surprising, and i expect to be told that it was solved at some point when
i wasn't paying attention...

The person who knows all the detailed history of this is
John Truss in Leeds.

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