Re: [Set Theory] Class of Ordinals well-ordered ?

From: Dave Seaman <[EMAIL PROTECTED]>
Newsgroups: sci.math
Subject: Re: [Set Theory] Class of Ordinals well-ordered ?

On Wed, 3 Dec 2003 03:58:29 -0800, William Elliot wrote:
> From: Dave Seaman <[EMAIL PROTECTED]>
> Newsgroups: sci.math
> Subject: Re: [Set Theory] Class of Ordinals well-ordered ?

>>If R is nonempty, then consider the intersection of all the members
>>of R. There are several things to show, but all are fairly easy:
> I'll presume R a set.

>>      (i) The intersection is well defined.
>>      (ii) The intersection is a set.
> Results of ZF.

>>      (iii) The intersection is an ordinal.
> Straight forward

 >>     (iv) The intersection is a lower bound for the members of R.
> Basic set theory.

 >>     (v) The intersection is a member of R.
 >I should have exchanged (iv) and (v).
Change made.

 >First, notice that "<" for ordinals means "is an element of", and
 >"<=" for ordinals means "is a subset of".
For ordinals a,b, a in b iff a proper subset b

 >Suppose alpha = intersect R.  Then by (iv) we have alpha <= x for
 >each x in R.  Suppose the inequality is strict: alpha < x for each x
 >in R.  Then it follows that (alpha+1) <= x for each x in R, and
 >therefore (alpha+1) is a subset of intersect R, contrary to the
 >assumption that alpha = intersect R.  Therefore, we must have alpha =
 >x for some x in R, Q.E.D.
Ok, interesting, well ordering of an ordinal without regularity.

>>I don't see where either regularity or choice comes into play here.
>>Ordinals are transitive sets with the property of being well-ordered
>>by the set membership relation.  This is a matter of definition, not
>>of regularity.

Have you not shown that a transitive set that's only linearly orderd
by 'in' is an ordinal?

 >> Futhermore if R(x) is a predicate about ordinals, to find the
 >> least ordinal s with R(s), then I find least of { s <= r | R(s) ?
 >Yes.

 >I don't see where regularity is needed to define the intersection.
I've checked to see where regularity is used.
It's used to show if eta in ordinal beta, then eta ordinal.

S is transitive when for all x in S, x subset S ?
Ok, I've amended the proof so it doesn't use regularity.

Now without regularity, can you show for ordinals eta,beta
eta in beta ==> eta /= beta ?  I'll even settle for a proof of
        ordinal eta ==> eta not in eta
without regularity.

If not, then I can not use
        For ordinals a,b, a in b or a = b iff a subset b
instead of
        For ordinals a,b, a in b iff a proper subset b

Likely I can, but then how do I show eta < S(eta) instead of
eta <= S(eta).  Again I need show eta not in eta otherwise
eta \/ {eta} = eta.

----