[FOM] complete atomless boolean algebras

Robert M. Solovay solovay at Math.Berkeley.EDU
Fri Aug 18 20:46:50 EDT 2006




On Thu, 17 Aug 2006, Robert Black wrote:

> I have two questions about complete atomless boolean algebras to
> which I find I can't (or can't easily) get answers from the
> Koppelberg Handbook, but I expect there are members of this list who
> can answer them instantly:
>
> 1) An infinite complete boolean algebra must have a cardinality k
> such that k=k^aleph_0. Is this the only cardinality restriction on
> complete *atomless* Boolean algebras? In particular, can the
> cardinality of a complete atomless Boolean algebra be inaccessible?

 	The complete Boolean algerbra relevant to my "All sets Lebesgue 
measurable" paper has cardinality an inaccessible and is homogeneous in 
your sense. I don't know of any place where these facts are established. 
But they are not difficult to prove.

 	An approximate description of this algebra is that one adds 
functions that collapse alpha to omega for each non-zero ordinal alpha 
less than the inaccessible kappa. It is proved in my paper that this 
algebra satisfies the < kappa chain condition. From this, the cardinality 
estimate follows easily.

 	To tell the truth, in my paper I talk about forcing posets rather 
than Boolean algebras. But the translation is easy and standard.

 	--Bob Solovay

> > 2) A Boolean algebra B is homogeneous (I think this is the standard
> word) iff for every nonzero p in B, the algebra induced on the x in B
> less than or equal to p by the partial order inherited from B is
> isomorphic to the whole of B. Trivially a homogeneous Boolean algebra
> with cardinality greater than 2 is atomless. Now consider *complete*
> homogeneous Boolean algebras. (There are such things, since unless
> I'm making an embarrassing mistake the regular open algebra over R is
> one.) Same question again: what cardinalities can complete
> homogeneous Boolean algebras have, and in particular can their
> cardinality be inaccessible?
>
> Robert
>
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