Re: Size Theory: corrected.
- From: Zaljohar@xxxxxxxxx
- Date: Sun, 6 Apr 2008 11:16:34 -0700 (PDT)
On Apr 5, 9:49 pm, Butch Malahide <fred.gal...@xxxxxxxxx> wrote:
On Apr 2, 10:31 am, Zaljo...@xxxxxxxxx wrote:
I will present here a modification on Axiom8, that I think it will
overcome the contradiction pointed by Butch Malahid.
[. . .]
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:
[. . .]
Axiom1: Z is standard infinite & Z is Dedekindian finite
[. . .]
Axiom3: Sa=Sb <-> Ef(f:a->b, f is bijective)
Axiom4: Sa<Sb -> ~Sb<Sa
[. . .]
Axiom6:[Ef(f:a->b,f is injective) &
Af((f:a->b,f is injective) -> ~ f is surjective)] -> Sa<Sb
Define: a comparable_to b <->
(Ef(f:a->b,f is injective) or Eg(g:b->a,g is injective))
[. . .]
Define:
x is overlapping <-> Ayz ((yex&zex&~y=z) ->
(y proper subset_of z or z proper subset_of y))
Axiom8: ~ c comparable_to Ux ->
[(Ay(yex-> Sy<Sc) & x is overlapping) -> SUx <= Sc]
were '<= ' denote 'smaller than or equal'.
I will sketch the proof of a contradiction from your Axioms 1, 3, 4,
6, and 8. I don't know what symbol you use for the the Cartesian
product of two sets, so I will use
A*B = {(a,b) | a in A & b in B}.
By Axiom 1, there is an infinite D-finite set. From this it follows
(as shown previously) that there exist a D-finite set D and a sequence
of sets D_n such that D = U{D_n | n in w} and, for each n in w, D_n is
a subset of D_{n+1} and a proper subset of D.
Let E = D*D, a D-finite set. Choose i in D and let
F = [D*(D-{i})]uw. The sets E and F are not comparable: there is no
injection from E to F, and there is no injection from F to E.
For each n in w, let E_n = D*D_n; there is an injection from E_n to F,
but there is no injection from F to E_n. It follows by Axiom 6 that
SE_n < SF, and then by Axiom 8 that SE <= SF.
For each n in w, let F_n = [D*(D-{i})]un; there is an injection from
F_n to E, but there is no injection from E to F_n. It follows by Axiom
6 that SF_n < SE, and then by Axiom 8 that SF <= SE.
By Axiom 4, from SE <= SF <= SE we get SE = SF; but this contradicts
Axiom 3, since there is no bijection between E and F.- Hide quoted text -
- Show quoted text -
Good! you are right. the theory is inconsistent.
hmmm..., it seems there is something essentially wrong in this theory.
I can simply cut this argument by further restricting axiom8 to
put the condition : c is infinite dedekindian finite , i.e axiom 8
will be:
Axiom8: [~ c comparable_to Ux & c is infinite dedekindian finite] ->
[(Ay(yex-> Sy<Sc) & x is nested) -> SUx <= Sc]
But even then it would like a kind of a biased fix, and I am sure that
you will have another argument of contradiction.
I was actually thinking of about something similar to what you said,
but from another angle, that is of the definition of equality, why
should it be by the existence of a bijection, I wanted to make that a
definition of equality between comparable sets only, but by then how
do we know that w < Z , another cirterion should be adopted . Anyhow.
Zuhair
.
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