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From: <hal.domain.name.hidden>

Date: Tue, 13 Jul 1999 18:54:23 -0700

Russell Standish, <R.Standish.domain.name.hidden>, writes:

*> My memory is fading somewhat about transfinite cardinal
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*> numbers. However, it seems to me that c \leq \aleph_1. \aleph_1 is the
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*> cardinality of the set of all sets of cardinalilty \leq\aleph_0. Since c
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*> is the cardinality of the set of all subsets of N, which is a subset
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*> of the set of all sets of cardinality \leq\aleph_0.
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*>
*

*> What has never been proven is that c=\aleph_1, although it is widely
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*> 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

Date: Tue, 13 Jul 1999 18:54:23 -0700

Russell Standish, <R.Standish.domain.name.hidden>, writes:

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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