Re: Separation,Power and Countability.

On Jun 15, 2:50 am, Keith Ramsay <kram...@xxxxxxx> wrote:
On Jun 14, 9:36 am, zuhair <zaljo...@xxxxxxxxx> wrote:
|The word uncountable set is not acceptable to me,because the language
|of any set theory T should be countable, otherwise it will affect the
|recurssive ability of that theory and by then it will lead to the
|emergence of the shadowy sets i.e the indefinable sets, which is un
|acceptable.

Declaring uncountable or undefinable sets "unacceptable"
has no substance to it.

People often find the relationship between constructivism,
definability and countability confusing. Part of the problem
in this context is that part of the time "definable" is
being used to mean "first-order definable". The sets
definable by the axiom schema are all first-order definable
(with parameters). It's entirely arbitrary to imagine that
those are all that there are. If to each formula there
exists a set, by separation, then there is a function from
the formulas to the sets, but this function isn't itself
first-order definable. But that's ok! Don't allow yourself
to suffer from chronic first-order-itis, folks.

It may be the case that constructivists typically don't
accept the general power set axiom,

The power set axiom is not the problem, the problem is 'unrestricted
separation', and what I mean by that: separation which allows formulas
to be in more than one free variable.

What is the proof that P(w) is uncountable?
The proof that Cantor came with depends on 'unrestricted' separation.
Cantor's proof is that: for any injective function f from w to P(w)
there will always exist a subset of w that is not in the range of f,
and this subset is defined using separation as {x|xew & ~xef(x)}
The formula ~xef(x) is not a formula in ONE free variable!
and because of allowing formulas with more than one free variable to
be used in separation it was this allowance that proved the
uncountability of P(w).

What I am saying is that if we restrict separation such as NOT to
allow formulas having more than ONE free variable to be used, then
this proof of Cantor's will be groundless.
and in this way using this restricted version of separation we can
CONSISTENTLY add the 'axiom of definability' that states that:
AxEPAy(yex<->P(y)) were P is a formula in one free variable that is y
and that is not x. to the axioms of second order Z with separation
writtin as
APAaExAy(yex<->(yea & P(y)))
were P is a formula in ONE free variable y and that this one free
variable is not x;

What I am saying is that with these modifications done, one doesn't
need to restrict power to
AxEP(x)Ay(yeP(x) <-> EF(Az(zey<->(zex & F(z))))).
were F is a formula in one free variable z that is other than y.

why? because since separation is restricted, then Cantor's proof of
uncountability of P(w) become groundless and thus in this theory we
don't have a proof of uncountability of
P(w), and from axiom of definability we can prove that
P(w) is actually countable, so we don't need to restrict power.

Power has nothing to do with the problem of countability
and definability here, since even if we restrict power as above, then
without restricting separation we will still have P(w) uncountable,
since Cantor's proof will still be working, leading to a paradoxical
theory.

In nutshell: If one wants to add axiom of definability
to the axioms of second order Z, then all what we need is to restrict
separation! not power.

That's what I was saying here.

Zuhair



find various who don't accept impredicative definitions.
But there's not an inherent conflict here. It's consistent
to believe that there are uncountably many subsets of the
integers and that for one to exist means that it can be
defined (somehow).

No, this cannot be done, this is the contradiction, there is nothing
called (somehow). What we have is separation, and that's it. using
separation to prove the existence of subsets means that separation can
only prove the existence of countably many subsets, and your statement
would be inconsistent, if we have uncountably many subsets, then there
should be subsets not definable by separation,which is meaningless,
since undefinable sets are shadowy objects, and actually such an
abstract concept is useless, and need not be thought of whatsoever.

It's just that you can't also believe
that there is some one fixed language in which they're all
definable. But believing that all definitions of sets of
integers can be given in a fixed language, let alone a
fixed first-order language, is a very strange belief.

Keith Ramsay


.

Relevant Pages

  • Re: Restricting Separation
    ... y (i.e has two free variables). ... Now the above instance of separation uses the formula ... the power of any set is larger in cardinality than its set, ... It is right that the proof of uncountability depends upon having ...
    (sci.logic)
  • Re: Restricting Separation
    ... y (i.e has two free variables). ... the same proof using a formula in separation that do not have at least ... Parameters are needed for the proof of uncountability to go ... formula phi of separation we need for the proof of uncountability to ...
    (sci.logic)
  • Re: Restricting Separation
    ... y (i.e has two free variables). ... the same proof using a formula in separation that do not have at least ... Parameters are needed for the proof of uncountability to go ... formula phi of separation we need for the proof of uncountability to ...
    (sci.logic)
  • Re: Separation,Power and Countability.
    ... an axiom of separation that for any injective function f:w->Pit ... since this uncountability clearly contradict ... is whether they provide for the great amount of mathematics that even ...
    (sci.math)
  • Re: Separation,Power and Countability.
    ... However I still think that separation is the ... but f is now an injection from w to P/ ... If I don't revise separation, ... pose no problem with definability at all. ...
    (sci.math)
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