From: Robin Chapman
Subject: Re: uncountable ordinals without AC
Date: Fri, 10 Mar 2000 22:36:51 GMT
Newsgroups: sci.math
Summary: [missing]
In article <8abn7k$gn8$2@news.acns.nwu.edu>,
lerma@math.nwu.edu (Miguel A. Lerma) wrote:
> How do you prove that there are uncountable ordinals without
> using the Axiom of Choice?
The usual argument proves that there are uncountable well-ordered
sets. Let X be a fixed countable infinite sets, and let W be the
set of well-orderings of subsets of X. A bit more formally, W
is the set of pairs (A,R) where A is a subset of X and R is
a well-ordering on A (so as a relation a subset of A x A).
We put an equivalence relation on W by setting two ordering
equivalent if order-isomorphic, then ordering W' the set of equivalence
classes by setting one order to be less than another if it's
order-isomorphic to a proper initial segment of the second.
Then W is an uncountable well-ordered set.
> More generally, I would like to see a proof of the following
> statement in ZF without AC:
>
> (1) "For every set M there is a set Z of ordinals such that
> alpha is in Z iff |alpha| =< |M|."
>
> In other words, the collection of all ordinals equipotent to
> subsets of M form a set (if M = omega, then Z would be omega_1
> = the first uncountable ordinal).
>
> Sierpinski' 1947 paper proving GCH => AC uses (1), but it does
> not justify it.
Would replacing X with M give a proof that works here?
--
Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html
"`The twenty-first century didn't begin until a minute
past midnight January first 2001.'"
John Brunner, _Stand on Zanzibar_ (1968)
Sent via Deja.com http://www.deja.com/
Before you buy.
==============================================================================
From: lerma@math.nwu.edu (Miguel A. Lerma)
Subject: Re: uncountable ordinals without AC
Date: 11 Mar 2000 05:38:47 GMT
Newsgroups: sci.math
Robin Chapman (rjc@maths.ex.ac.uk) wrote:
:
: Would replacing X with M give a proof that works here?
After I posted my question I realized that (1) can be
proved with the replacement axiom schema. For each
element w of W = well-orderings of subsets of M, let
a(w) = the ordinal isomorphic to w. W is a set - in fact
a subset of P(M) x P(M x M), where P represents power set.
Hence by replacement Z = { a(w) | w in W } is a set.
Miguel A. Lerma
==============================================================================
From: Fred Galvin
Subject: Re: uncountable ordinals without AC
Date: Fri, 10 Mar 2000 16:54:14 -0600
Newsgroups: sci.math
On 10 Mar 2000, Miguel A. Lerma wrote:
> How do you prove that there are uncountable ordinals without
> using the Axiom of Choice?
>
> More generally, I would like to see a proof of the following
> statement in ZF without AC:
>
> (1) "For every set M there is a set Z of ordinals such that
> alpha is in Z iff |alpha| =< |M|."
>
> In other words, the collection of all ordinals equipotent to
> subsets of M form a set (if M = omega, then Z would be omega_1
> = the first uncountable ordinal).
You are asking about Hartogs' Theorem, which you can find in many
books. To pick one at random, it's on p. 50 of J. E. Littlewood, _The
Elements of the Theory of Real Functions_, 3rd edition, Dover
Publications, 1954, $1.35. Here is a sketch of the proof (for the M =
omega case) which I posted a couple of months ago in this newsgroup:
Subject: Re: Stupid set theory questions
From: Fred Galvin
Date: 1999/12/08
Newsgroups: sci.math
On 8 Dec 1999, Seraph-sama wrote:
> The "first uncountable ordinal" is the set aleph_1 = {1, 2, ..., w, w
> + 1, ..., 2w, 2w + 1, ..., w^2, w^2 + 1..., w^2 + w, w^2 + w + 1, ...,
> w^n, ...} = w^w. Correct? Is this really the first cardinality after
Incorrect, as other posters have already told you. Also, it's
customary to include 0 as the first ordinal; and the preferred
convention for ordinal multiplication (which is noncommutative) is to
write w+w = w2 and 2w = w.
By the way, since 0 is the first ordinal, it follows that 1 is the
2nd, 2 is the 3rd, and so on; then omega is the (omega+1)th (there is
no "omegath"), and so on; any ordinal alpha is the (alpha+1)th
ordinal. Most books (maybe all of them) get this wrong.
> aleph_0? If not, could somebody provide a proof (preferably
> constructive, if possible) of the existence of aleph_1? And, after
> that, the continuum hypothesis states that c = aleph_1 = w^w, right?
To construct the cardinal aleph_1 and the ordinal omega_1, it's enough
to construct any uncountable well-ordered set. (Let W be any
uncountable well-ordered set. If every element of W has only countably
many predecessors in the well-ordering of W, then W has order type
omega_1 and cardinality aleph_1, and the Von Neumann ordinals up to
omega_1 can be defined by transfinite induction on W. If some elements
of W have uncountably many predecessors, let a be the least such
element, and replace W by the set of predecessors of a.) You can
construct an uncountable well-ordered set without the axiom of choice.
(I believe this construction is due to the mathematician F. Hartogs.)
The idea is simple and natural, but there are many details to check.
I'll just give you the outline, and you can fill in the details
yourself, or look them up in a good book on set theory.
Let N be the set of all natural numbers. Consider the collection of
all ordered pairs (X,R) where X is a subset of N and R is a
well-ordering of X. (In a manner of speaking, this is the collection
of all countable well-ordered sets.) Partition this collection into
isomorphism classes; i.e., two pairs (X,R) and (Y,S) are in the same
cell of the partition if and only if they are isomorphic as ordered
sets. Let W be the set of all these isomorphism classes. (In effect,
the set of all countable ordinals.) Define an ordering of W as
follows: for any elements x, y in W, let x < y if and only if some
element (X,R) of x is a proper initial segment of some element (Y,S)
of y. Now you can prove that W is well-ordered by <, and then you can
prove that W is uncountable, and that every element of W has only
countably many predecessors, whence (W,<) has order type omega_1