sci.logic
[Top] [All Lists]

Re: Explicit choice functions on subsets of R

Subject: Re: Explicit choice functions on subsets of R
From:
Date: 5 Oct 2006 08:57:34 -0700
Newsgroups: sci.logic
Ross A. Finlayson wrote:
> aatu.koskensilta@xxxxxxxxx wrote:
> > That's certainly true. It is consistent with ZFC that there is a
> > definable well-ordering of the universe (and thus also of the reals),
> > though.
>
> I wonder what you mean by that, Aatu.  There are no universes in ZF.

That there is a definable well-ordering of the universe means that
there is a formula R(x,y) in the language of set theory such that

 a) R is set-like, that is, for every set a, { x | R(x,a) } is a set
 b) every non-empty set has an R-least element

Such an R exists e.g. if V=L, if V=L[0^#] and so forth.

> It's consistent that the cardinality of the reals is Aleph_1, Aleph_2,
> ..., then they're each equivalent.  How's that for a continuum
> hypothesis?  Ha ha ha.

Very amusing.

--
Aatu Koskensilta (aatu.koskensilta@xxxxxxxxx)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
 - Ludwig Wittgenstein, Tractatus Logic-Philosophicus


<Prev in Thread] Current Thread [Next in Thread>
Privacy Policy