Re: Z-Infinity
- From: Zaljohar@xxxxxxxxx
- Date: Sun, 13 Apr 2008 09:30:19 -0700 (PDT)
On Apr 10, 2:15 pm, lwal...@xxxxxxxxx wrote:
On Apr 8, 5:11 am, Zaljo...@xxxxxxxxx wrote:
Size Theory is set of all sentences entailed ( from first order logic
with identity '=' and epsilon membership 'e' and the primitive
constant Z , and the primitve two place relation symbole '< ' to
denote 'smaller than' ,and the primitive one place function symbole
'S' to denote 'size') by the non
logical axioms of ZF and the following non logical axioms:
I've been thinking of a similar theory myself for a while. I
wonder whether this theory can be consistent.
My pet theory is to be an extension of ZF-Infinity. But instead
of an infinite set omega, I have a new type of infinite set, which I
call "alpha" (the first Greek letter, as omega is the last), which is
similar in many ways to Zuhair's Z.
Rather than attempt to write any new axioms, let me simply state
the properties that alpha is supposed to have.
-- Alpha is an "ordinal," or at least resembles an ordinal in that
it's
a transitive set, all of whose elements are themselves transitive.
-- Alpha is in some ways "infinite," but in other ways "finite." In
particular, it's D-finite, the weakest type of finite.
-- Alpha contains as elements 0, 1, 2, 3, etc.
-- Alpha contains what appears to be a infinitely descending e-chain,
alpha_1ealpha, alpha_2ealpha_1, alpha_3ealpha_2, etc.
Now you're probably thinking, hold on a minute! This already seems
very inconsistent! I said ZF-Infinity, so this includes the Axiom of
Foundation/Regularity, which forbids infinitely descending chains,
doesn't it?
Well, the Axiom of Regularity literally states that every set must be
disjoint with one of its elements. And since 0ealpha, 0 is the
element of alpha with which is distinct.
If you think about it, Regularity alone doesn't forbid the existence
of
an infinite chain per se! The proof of the nonexistence of an infinite
chain actually requires the Axiom of Infinity, by starting with omega
and applying the Replacement Schema to produce a set
{alpha, alpha_1, alpha_2, alpha_3, ...}
and it's this set that fails to be disjoint with any of its elements!
But
since I (conveniently) left out the Axiom of Infinity, we can't
conclude
that this is an ill-founded set. Instead, this is really a proof by
contradiction, in this theory, of the nonexistence of omega (and
therefore ~Infinity).
But what about the Separation Schema? Since alpha is supposed to
contain 0, 1, 2, 3, ..., why can't we simply apply Separation to alpha
with P being the predicate "is a finite ordinal," to obtain omega (and
therefore a contradiction)? Well, the answer is that we do -- and then
we conclude that there must exist a (nonstandard) "finite ordinal"
_other than_ 0, 1, 2, 3, ...! Indeed, if "finite" means D-finite, then
alpha
is itself a (nonstandard) finite ordinal -- but this may work even for
other definitions of finite.
So alpha is supposed to work almost the same way as Z does in
Zuhair's theory. The main difference is that in Zuhair's theory, he
shows that Z+omega exists, but not Z-omega. In my theory, there
should be a little more symmetry. If a < b, then I want both a+b and
b-a to exist, always.
Also, as Butch Malahide points out, there may be a little problem
concerning 2^2^Z = S(P(P(Z))), since it is D-infinite. So what about
my own P(P(alpha)). Butch's proof shows that the powerset of the
powerset of every infinite set has a subset whose cardinality is the
same as that of the set of all finite ordinals -- but since in my
theory, that set is _not_ omega, one hasn't proved that P(P(alpha))
has a subset of cardinality omega, or that P(P(alpha)) is D-infinite.
So what I'm wondering is, is this theory even possible? Is there
any axiom using first-order logic, e, and possibly alpha as another
primitive, I can write so that alpha has all of the properties I've
ascribed to it in this post? Or is it impossible? What I'm writing
seems to be a nonstandard model of ZF-Infinity+~Infinity, where
there are nonstandard natural numbers other than the usual ones.
Perhaps all I'm writing is really just a form of Internal Set Theory,
and so there's nothing I can do with alpha that I can't already do
in Internal Set Theory.
Interesting! but this theory is much more complicated than the one I
am presenting here. Size theory is clear and uncomplicated. However I
understand that your theory demands more than that of mine since it is
more symmetrical than mine as regards subtraction and addition.
Zuhair
.
- References:
- Z-Infinity
- From: Zaljohar
- Re: Z-Infinity
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- Z-Infinity
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