Re: Is zero even or odd?

From: George Cox (george_coxanti_at_spambtinternet.com.invalid)
Date: 12/23/04 

Date: Thu, 23 Dec 2004 22:00:35 +0000 (UTC)

John Woodgate wrote:
>
> I read in sci.electronics.design that Nicholas O. Lindan <see@sig.com>
> wrote (in <YhFyd.11474$Z47.2358@newsread2.news.atl.earthlink.net>) about
> 'Is zero even or odd?', on Thu, 23 Dec 2004:
> >"Dave Seaman" <dseaman@no.such.host> wrote
> >> On Thu, 23 Dec 2004 12:22:02 -0500, John W. Kennedy wrote:
> >> > Aleph-1 is at least aleph-null^aleph-null.
> >>
> >> No, it's the other way around. Since aleph_1 is by definition the
> >> smallest uncountable cardinal, and since the reals are uncountable, it
> >> follows that c (= 2^aleph_0 = aleph_0^aleph_0), the cardinality of the
> >> continuum, cannot be less than aleph_1. On the other hand, it could be
> >> that c is quite huge among the alephs.
> >
> >I'm lost.
>
> You need to study the math of infinities. Aleph-null is the smallest
> infinity, and whatever you do to it with finite numbers doesn't change
> it. Many operations with itself, even, don't change it. But raising it
> to its power, {-}o^({-}o), creates a new infinity with different
> properties. Although it's called aleph-one, no-one knows whether it is
> the *next* infinity after aleph-null, or whether there are other
> infinities in between.

Whether (aleph_0)^{aleph_0} = aleph_1 or not isn't decided in the usual
set theories.

aleph_1 is the next infinity after aleph_0 (given the axiom of choice),
the question is, is 2^{aleph_0} the next infinity after aleph_0? (And
generally, is 2^{aleph_{alpha}} the next infinity after aleph_{alpha}?)

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