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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Thu Jul 20, 2006 10:56 am Post subject:
How many ordinals?



What and/or how many ordinals are given by ZFC
1) without axiom of infinite, without axiom of choice
2) without axiom of infinite, with axiom of choice
3) with axiom of infinite, without axiom of choice
4) with axiom of infinite, with axiom of choice?
Answers?
1) All the finite ordinals only.
2) ?
3) ?
4) ?
(As many as you like and more than you can ever use. ;)
Riddle of the day.
How much faster can cardinals count than ordinals can? 

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Dave L. Renfro science forum Guru
Joined: 29 Apr 2005
Posts: 570

Posted: Thu Jul 20, 2006 12:36 pm Post subject:
Re: How many ordinals?



William Elliot wrote (in part):
Quote:  Riddle of the day.
How much faster can cardinals count than ordinals can?

I'd say the ordinals are only faster for (relatively) short
periods and, as you get farther out along the ordinals, these
intervals where the ordinals are ahead of the cardinals become
blips that can be ignored.
The mapping beta > aleph_beta that takes the beta'th
ordinal to the beta'th cardinal is a normal function
(on a sufficiently large initial segment of the ordinals),
and hence has fixed points. In fact, it's fixed points
are orderisomorphic to every initial segment of the
ordinals, and so before too long it's fixed points start
to look like the limit ordinals when you're out around
the first epsilon number. Moreover, the fixed points have
fixed points as well, and these 2'nd order fixed points
have the same behavior. Same for the 3'rd order fixed points,
and so on, even for the omega'th order fixed points (use a
countable intersection of clubs to get them) and beyond.
After a while, the epsilon_0'th order fixed points start
to look like the limit ordinals out around epsilon_0. And
this is only the beginning, because we can jump past this
pedestrian pace of iterating the fixed point operation by
taking diagonal intersections (look it up), and on and on
and on ...
See the beginning of Section 4.33 in Azriel Levy's "Basic
Set Theory", Dover Publications, 1979/2002.
Use 'Search in this book' = "omitted by diagonal intersection"
http://books.google.com/books?vid=ISBN0486420795
I also posted this in sci.math once, but it seems to only
be in the Math Forum archive, not the google archive.
(Also, my URL's and some other things seem not to have
been transmitted faithfully.)
Thread: "Different size infinities?"
Date: October 29, 2004
http://mathforum.org/kb/thread.jspa?messageID=3414869
Dave L. Renfro 

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Jonathan Hoyle science forum Guru Wannabe
Joined: 04 Sep 2005
Posts: 260

Posted: Thu Jul 20, 2006 2:26 pm Post subject:
Re: How many ordinals?



William Elliot wrote:
Quote:  What and/or how many ordinals are given by ZFC
snip 
ZFC doesn't "give" ordinals. There are many, many models of Set Theory
consistent with ZFC. You are confusing the model with the theory.
For example, the existence of inaccessible cardinals is undecidable in
ZFC. So you can have a model in which the V = V_theta, where theta is
the smallest inaccessible cardinal. That is to say, in the universe of
V_theta, the only sets are those which have accessible cardinality.
Anything larger cannot exist as a set.
On the other hand, you can have a model of ZFC which contains
inaccessible cardinals, say V= V_kappa where kappa is the smallest
hyperinaccessible cardinal. Here you can have sets which are of
inaccessible cardinality. Or you can choose kappa = to a Mahlo
cardinal, or a measurable cardinal, or Shelah, or huge, or all the way
up the line.
All of these models are consistent with ZFC, or at least not proven to
be inconsistent with it. There does not appear to be an upper limit on
large cardinals, although as they get bigger, they border on breaking
out of the ZFC consistency shell. For example, the largest cardinals
known currently are Reinhardt cardinals which are inconsistent with the
Axiom of Choice, and thus cannot be a ZFC model. The largest known
after them are the "rankintorank" cardinals, which have not yet been
shown to be inconsistent with ZFC. 

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Chip Eastham science forum Guru
Joined: 01 May 2005
Posts: 412

Posted: Thu Jul 20, 2006 2:42 pm Post subject:
Re: How many ordinals?



William Elliot wrote:
Quote:  What and/or how many ordinals are given by ZFC
1) without axiom of infinite, without axiom of choice
2) without axiom of infinite, with axiom of choice
3) with axiom of infinite, without axiom of choice
4) with axiom of infinite, with axiom of choice?
Answers?
1) All the finite ordinals only.
2) ?
3) ?
4) ?
(As many as you like and more than you can ever use.

I suspect you may be interested in a result provable in
ZFS without axiom of infinity (the C stands for Choice).
Friedrich Hartogs proved a "lemma" on the way to
showing the Axiom of Choice is equivalent to the
trichotomy principle for cardinals. Given any set
X, there is of course the cardinal number of X,
which is naively the equivalence class of sets
equinumerous with X. Ordinal numbers may be
considered either as equivalences on wellorders
or (in the von Neumann approach) hereditarily
transitive sets (the "transitive" meaning here
that every member of a member is a member
of a transitive set).
What Hartogs showed is that, without assuming
the Axiom of Choice (or Axiom of Infinity, for that
matter), there exists a least ordinal which is not
"injective" to X, i.e. not less than or equal to X
in cardinality, for any set X whatever.
Without trichotomy, of course, we cannot claim
that this is the same as having cardinality greater
than or equal to X. But as trichotomy establishes
that, it establishes that any X can be wellordered.
Hence the Axiom of Choice is equivalent to the
principle of trichotomy for cardinal numbers (the
other direction being trivial).
Regarding the various "subsystems" of ZFC you
propose, I would highlight one implication of
Hartogs lemma concerning the assumption of
the Axiom of Infinity but not Axiom of Choice.
Given X as littleomega, there must be an ordinal
Y which is not injective to X. But since both X
and Y are wellordered, by transfinite induction
we must have X strictly less than Y. That is,
there exists a wellorderable cardinal strictly
beyond the countable littleomega. Of course
the argument can be repeated with X replaced
by Y, to get a chain of strictly increasing well
orderable cardinals.
regards, chip 

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Stephen J. Herschkorn science forum Guru
Joined: 24 Mar 2005
Posts: 641

Posted: Thu Jul 20, 2006 6:05 pm Post subject:
Re: How many ordinals?



William Elliot wrote:
Quote:  What and/or how many ordinals are given by ZFC
I assume you mean in ZF  Infinity. My answers are actually for ZF  
Inifinty  Foundation; I strongly suspect that Foundation has no effect
here.
Quote:  1) without axiom of infinite, without axiom of choice
2) without axiom of infinite, with axiom of choice
3) with axiom of infinite, without axiom of choice
4) with axiom of infinite, with axiom of choice?
Answers?
1) All the finite ordinals only.
2) ?
3) ?
4) ?
(As many as you like and more than you can ever use. ;)

Regarding (3) and (4), with Infinity, you get the same class of
ordinals with or without AC. You and I discussed this here some time
ago, William, though I cannot track down the thread right now. There
are other such threads, however.
Regarding (2), "All sets are finite" is consistent with ZFC 
Infinity. See Kunen, p. 123. In fact, isn't AC a theorem of ZF 
Infinity + "All ordinals are finite" ?

Stephen J. Herschkorn sjherschko@netscape.net
Math Tutor on the Internet and in Central New Jersey and Manhattan 

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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Fri Jul 21, 2006 12:51 pm Post subject:
Re: How many ordinals?



From: Dave L. Renfro <renfr1dl@cmich.edu>
Quote:  William Elliot wrote (in part):
Riddle of the day.
How much faster can cardinals count than ordinals can?
I'd say the ordinals are only faster for (relatively) short
periods and, as you get farther out along the ordinals, these
intervals where the ordinals are ahead of the cardinals become
blips that can be ignored.

Nay, the cardinals go 1,2,3,.. aleph_0, aleph_1. While the
ordinals go 1,2,3,.. omega_0, omega_0 + 1, ... omega_0 * 2 ...
So at aleph_0 = omega_0, they're neck and neck but then the cardinals jump
off to aleph_1 while the ordinals take forever and a half to get to
omega_1 = aleph_1. While they're still getting to the edge of the solar
system, the cardinals have raced off aleph_1, aleph_2,... aleph_omega0 to
the nearest stars and beyond to aleph_omega1 at the edge of the galaxy.
Quote:  The mapping beta > aleph_beta that takes the beta'th ordinal to
the beta'th cardinal is a normal function (on a sufficiently large
initial segment of the ordinals), and hence has fixed points. In
fact, it's fixed points are orderisomorphic to every initial segment
of the ordinals, and so before too long it's fixed points start to
look like the limit ordinals when you're out around the first epsilon

Isn't that first fixed point aleph_omega_omega_...?
So while the ordinals are still in a race with Voygers 1 and 2 to get to
the nearby stars, the cardinals have reached the speed of light.
Quote:  number. Moreover, the fixed points have fixed points as well, and
these 2'nd order fixed points have the same behavior. Same for the
3'rd order fixed points, and so on, even for the omega'th order fixed
points (use a countable intersection of clubs to get them) and
beyond. After a while, the epsilon_0'th order fixed points start to
look like the limit ordinals out around epsilon_0. And this is only
the beginning, because we can jump past this pedestrian pace of
iterating the fixed point operation by taking diagonal intersections
(look it up), and on and on and on ...

What if I prefer to see nothing beyond the speed of light, to leave the
intangible untouchable inaccessibles in the dungon of their own
remoteness?
Quote:  See the beginning of Section 4.33 in Azriel Levy's "Basic
Set Theory", Dover Publications, 1979/2002.
Use 'Search in this book' = "omitted by diagonal intersection"
http://books.google.com/books?vid=ISBN0486420795
I also posted this in sci.math once, but it seems to only
be in the Math Forum archive, not the google archive.
(Also, my URL's and some other things seem not to have
been transmitted faithfully.)
Thread: "Different size infinities?"
Date: October 29, 2004
http://mathforum.org/kb/thread.jspa?messageID=3414869

 

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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Fri Jul 21, 2006 12:52 pm Post subject:
Re: How many ordinals?



From: Jonathan Hoyle <jonhoyle@mac.com>
Quote:  For example, the existence of inaccessible cardinals is undecidable
in ZFC. So you can have a model in which the V = V_theta, where
theta is the smallest inaccessible cardinal. That is to say, in the
universe of V_theta, the only sets are those which have accessible
cardinality. Anything larger cannot exist as a set.
On the other hand, you can have a model of ZFC which contains
inaccessible cardinals, say V= V_kappa where kappa is the smallest
hyperinaccessible cardinal. Here you can have sets which are of
inaccessible cardinality. Or you can choose kappa = to a Mahlo
cardinal, or a measurable cardinal, or Shelah, or huge, or all the
way up the line.

As I said, as many as you wish and more than you can ever use.
But you are dreaming for by the LoewenheimSkolem paradox,
it's just a countable model.
Quote:  All of these models are consistent with ZFC, or at least not proven
to be inconsistent with it. There does not appear to be an upper
limit on large cardinals, although as they get bigger, they border on
breaking out of the ZFC consistency shell. For example, the largest
cardinals known currently are Reinhardt cardinals which are
inconsistent with the Axiom of Choice, and thus cannot be a ZFC
model. The largest known after them are the "rankintorank"
cardinals, which have not yet been shown to be inconsistent with
ZFC.

As I recall V = L, assuming the universe of sets is constructable
implies AxC. Does it not deny existence of inaccessibles? What
does it do regarding CH and GCH, the continuum hypothesis?
 

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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Fri Jul 21, 2006 12:53 pm Post subject:
Re: How many ordinals?



From: Stephen J. Herschkorn <sjherschko@netscape.net>
Quote:  William Elliot wrote:
What and/or how many ordinals are given by ZFC
I assume you mean in ZF  Infinity. My answers are actually for ZF
 Inifinty  Foundation; I strongly suspect that Foundation has no
effect here.
1) without axiom of infinite, without axiom of choice
2) without axiom of infinite, with axiom of choice
3) with axiom of infinite, without axiom of choice
4) with axiom of infinite, with axiom of choice?
Answers?
1) All the finite ordinals only.
2) ?
3) ?
4) ?
(As many as you like and more than you can ever use. ;)
Regarding (3) and (4), with Infinity, you get the same class of
ordinals with or without AC. You and I discussed this here some
time ago, William, though I cannot track down the thread right now.
There are other such threads, however.

Not that I recall, nor have I watched closely the construction of the
ordinals to say for sure that AxC wasn't used.
Quote:  Regarding (2), "All sets are finite" is consistent with ZFC 
Infinity. See Kunen, p. 123.

I'm shocked.
Quote:  In fact, isn't AC a theorem of ZF 
Infinity + "All ordinals are finite" ?

I'd think it'd be a theorem of ZF + all sets are finite.
or of ZF + all ordinals are finite
 

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