Let me add my information to this confusing brew:
hal.domain.name.hidden wrote:
>
> Russell Standish, <R.Standish.domain.name.hidden>, writes:
> > My memory is fading somewhat about transfinite cardinal
> > numbers. However, it seems to me that c \leq \aleph_1. \aleph_1 is the
> > cardinality of the set of all sets of cardinalilty \leq\aleph_0. Since c
> > is the cardinality of the set of all subsets of N, which is a subset
> > of the set of all sets of cardinality \leq\aleph_0.
This is wrong, from what I know. I agree with Hal below that aleph-1
is defined to be the "next" cardinal after aleph-0. That is, by
definition, there is no transfinite number with cardinality > aleph-0
and < aleph-1.
> >
> > What has never been proven is that c=\aleph_1, although it is widely
> > suspected.
In fact it has been shown that c == aleph-1 is not provable by the
axioms of Zermelo-Fraenkel set theory. This is known as the
continuum hypothesis (CH). CH is also not disprovable,
which means that it is independent of those axioms. Thus it is
possible to construct set theories which assume that ~CH, and these
are known as non-Cantorian set theories.
On the other hand, in standard set theory, assuming the CH does no
harm.
[More below]
>
> This is pretty much over my head. As I understand it aleph 1 is
> defined to be the next cardinal number after aleph 0, and can be
> shown to be the cardinality of the set of all countable ordinals
> (1...w...w+1...2w...w^2...). Since the elements of this set are all
> ordered sets (i.e. ordinals), while the subsets of N don't have an
> ordering requirement, this gives more flexibility to c and so you can't
> compare them as simply as you have shown here.
>
> See http://www.ii.com/math/ch/ for a detailed discussion of these
> matters.
>
> Actually on further thought I think I was wrong to suggest that the number
> of TM programs is c, since that would allow for infinite length programs,
> which is perhaps outside the spirit of a TM. If we require only finite
> length programs then the number of TMs is aleph 0 since we can enumerate
> all the programs, and that would be the number of universes as well.
> Not so many after all.
>
> Hal
I think you are right that the cardinality of the set of all programs
is aleph-0.
But neither of you (nor anyone else) has addressed my reason for
conjecturing that the set of branches in our structure must be
aleph-(aleph-0), which is based on the SSA.
To tell you the truth, I'm certainly not convinced of it, but I
think it's worth considering. To discard the conclusion, I would
think that you'd have to assume "the identity of indiscernables".
My reasoning is illustrated if you only assume, for the moment,
that some observable (say, x) can take a continuum of possible
values when measured. Forget about the Plank length, for now.
That would mean that the set of all possible humans would have
cardinality c (at least). Thus it would be impossible to map
that set onto the set of all programs.
But if you believe in the computationalist hypothesis, then you'd
have to assume that at some point, a simulation of a human becomes
"close enough" to be identical. That is (to oversimplify) when
each particle is simulated to within a Plank length, then the
simulation becomes indiscernable from the original, and thus
identical. If this is true, then I would no longer expect the
physical laws to give rise to ever-increasing cardinality of
universes, since that could never increase the cardinality of
the set of humans past aleph-0, anyway.
--
Chris Maloney
http://www.chrismaloney.com
"Knowledge is good"
-- Emil Faber
Received on Tue Jul 13 1999 - 22:33:23 PDT