# Re: Cardinality of the MW

From: <hal.domain.name.hidden>
Date: Tue, 13 Jul 1999 18:54:23 -0700

Russell Standish, <R.Standish.domain.name.hidden>, writes:
> 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.
>
> What has never been proven is that c=\aleph_1, although it is widely
> suspected.

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
Received on Tue Jul 13 1999 - 19:00:40 PDT

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