Re: Set theory ZFC is inconsistent.


Rupert wrote:
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Yes this is right

Okay, so what do you want to do now?


Definition. <ω>:={X |(Er)(Ap)Q(r,p,X)}, [ω]:=ω union <ω>.

<ω> is actually the set of all arithmetically definable sets. Okay,
fine. What now?

From transfinite induction by usual way I can build transfinite
sequence of the sets
GL0=[ω], GL1=[GL0],...,GLα.

So it's really just a version of Goedel's constructible hierarchy, only
you're starting with the definable subsets of the structure <ω,є>
rather than the definable subsets of the structure <V_ω,є>.

Theorem. (Eβ) such that β<ω1 and GLβ+1=GLβ

This theorem is clearly false.

No. I dealing with definable-demonstrable sets:
(Axp){ [ xp belong X <--> F[r](xp)]
&[(En)Pr(n,G(r,p))]}

I have noticed a more fundamental problem. For your definition by
transfinite recursion to work each ordinal will have to be definable,
and you will have to specify how you are going to select a definition
for each ordinal.

You wish to tell, that definition for demonstrable-constructible
hierarchy GLβ should be
expressed by the formula of the ZFC- language???

Well, if we pass from GLβ to GL(β+1), we're dealing with formulas F_r
whose quantifiers are relativized to GLβ. So GLβ has got to be
definable. And you're going to have to come up with some way of
selecting which definition you want to use.


Theorem. AC valid for all sets GLα, α≤β where GLβ = GL(β+1),

As things stand we haven't defined the GLα. When we pass from GLβ to
GL(β+1), we're dealing with formulas F_r whose quantifiers are
relativized to GLβ. We have to specify which definition of GLβ we're
using, otherwise it's not well-defined what GL(β+1) is.


GL0=[ω], GLα=[union{γ<α| GLγ }]

What about GL(α+1)? That is defined in terms of formulas F_r in which
the quantifiers are relativized to GLα. So you've got to specify which
definition of GLα you're going to use.

I am not assured, that have correctly understood your question.
Please explain in more detail.

The definition of GL(α+1) makes a reference to formulas F_r which have
their quantifiers relativized to GLα. We need to know which formulas
these are. So we need to know what definition of GLα we're using.
Until you've specified that, you haven't defined the hierarchy properly.


You ask about ZFC-formula which defined GLα?

Yes. Specifying a definition of GLα is equivalent to specifying a
definition of α. It is choosing the definition for the ordinal that is
the issue.

You wish to tell, what the ZFC-formula by which I can define GLα does
not exist?

The issue is specifying a definition of each ordinal α.

Why you think, that this problem more difficult than in case of Godel's
hierarchy Lα ?


I can dealing with formulas F_r whose quantifiers are everytime
relativized to a big set ω1.
Definition. <ω>:={X⊂ω |(Er)(Ap)F(r,p,X)}, [ω]:=ω union <ω>.
GL0=[ω], GLα=[union{γ<α| GLγ }]
β= sup{α |GLα ≠ GL(α+1)}

All right. Now Q(r,p,X) is equivalent to (Axp){ [ xp belong X <-->
F[r](xp)] &[(En)Pr(n,G(r,p))]}. And G(r,p)=G(F[r](xp)) with xp free.
Now, an open formula is provable if and only if its universal closure
is provable. So (En)Pr(n,G(r,p)) says that it is provable that every xp
satisfies F[r]. Is that really what you want it to say?

Yes.

Every xp in the entire universe? Or just in some set?

I've also noticed another issue. GL0 will not be a subset of ω1, so it
doesn't make much sense to talk about the subsets of GL0 that are
definable in ω1. You could get around this problem by using V_ω1
instead.

Yes of course, you can using V_ω1 or V_ω_α, where α=1,2,...


Every xp in the entire V_ω1.

Okay. If we add an axiom schema of the form Prov_ZFC(S)->S for every S
to ZFC, then we can prove that your theorem that the hierarchy
terminates at some countable ordinal is false. We probably can't prove
it's false in ZFC alone. But I very much doubt that we can prove it
true.

You wish to tell, that the hierarchy GLα doesn't terminates at some
countable ordinal α?

I've changed my mind. I've noticed other issues with the definition of
the hierarchy. You've got <ω>:={X subset ω |(Er)(Ap)Q(r,p,X)},
[ω]:=ω union <ω>. And Q(r,p,X) is equivalent to (Axp){ [ xp belong X
<--> F[r](xp)] &[(En)Pr(n,G(r,p))]}. And G(r,p) is the Goedel number of
the formula asserting "for all x_p in V_ω1, F_r(x_p)". So you're
requiring that in fact F_r only be satisfied by those x_p in X, whereas
on the other hand that it can be proved that it is satisfied by all x_p
in V_ω1. There is no reason to think that will ever happen. So I think
in fact the hierarchy terminates at GL0.

You wish to tell, that the ω+1 doesn't belong to GL0? &

That's right. There's no formula defining ω+1 such that it can be
proved in ZFC that every set in ω1 satisfies this formula.

Formula (Axp){ [ xp belong X <--> F[1](xp)]&[(xp belong ω1)]
&[(En)Pr(n,G(r,p))]}
doesn't meant that every set in ω1 satisfies this formula.

(En)Pr(n,G(r,p)) does mean that it is provable in ZFC that every set in
V_ω1 satisfies the formula F_r, according to what you have said so
far. Perhaps you want to redefine G(r,p).


(Axp){ [ xp belong X <--> F[1](xp)]&[(xp belong ω1)]
&[(Eq)(En)Pr(n,G(r,q))]}???

.

Relevant Pages

  • Re: Set theory ZFC is inconsistent.
    ... Okay, so what do you want to do now? ... So it's really just a version of Goedel's constructible hierarchy, ... and you will have to specify how you are going to select a definition ... we're dealing with formulas F_r whose quantifiers are ...
    (sci.logic)
  • Re: Set theory ZFC is inconsistent.
    ... Okay, so what do you want to do now? ... So it's really just a version of Goedel's constructible hierarchy, ... and you will have to specify how you are going to select a definition ... we're dealing with formulas F_r whose quantifiers are ...
    (sci.logic)
  • Re: Set theory ZFC is inconsistent.
    ... Okay, so what do you want to do now? ... So it's really just a version of Goedel's constructible hierarchy, ... and you will have to specify how you are going to select a definition ... we're dealing with formulas F_r whose quantifiers are ...
    (sci.logic)
  • Re: Set theory ZFC is inconsistent.
    ... Okay, so what do you want to do now? ... So it's really just a version of Goedel's constructible hierarchy, ... and you will have to specify how you are going to select a definition ... we're dealing with formulas F_r whose quantifiers are ...
    (sci.logic)
  • Re: Set theory ZFC is inconsistent.
    ... Okay, so what do you want to do now? ... So it's really just a version of Goedel's constructible hierarchy, ... and you will have to specify how you are going to select a definition ... we're dealing with formulas F_r whose quantifiers are ...
    (sci.logic)
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