Re: Separation,Power and Countability.

On Jun 14, 10:48 am, zuhair <zaljo...@xxxxxxxxx> wrote:
On Jun 13, 7:45 pm, Rupert <rupertmccal...@xxxxxxxxx> wrote:





On Jun 14, 10:42 am, zuhair <zaljo...@xxxxxxxxx> wrote:

On Jun 13, 7:36 pm, Rupert <rupertmccal...@xxxxxxxxx> wrote:

On Jun 14, 10:34 am, zuhair <zaljo...@xxxxxxxxx> wrote:

On Jun 13, 7:23 pm, Rupert <rupertmccal...@xxxxxxxxx> wrote:

On Jun 14, 10:05 am, zuhair <zaljo...@xxxxxxxxx> wrote:

On Jun 13, 5:50 pm, Rupert <rupertmccal...@xxxxxxxxx> wrote:

On Jun 14, 7:57 am, zuhair <zaljo...@xxxxxxxxx> wrote:

Hi all,

This will be the last post of mine to this group.

In the wikipedia, the axiom schema of separation is wrote in this way:

1)if P is a formula in one variable that doesn't use the symbol x,
then AaExAy(yex<->(yea & P(y))) is an axiom.

Now what is exactly meant by 'formula in one variable that doesn't use
the symbol x' , does that mean that it might contain other variables
that use the symbol x and only one variable is permitted not to use
the symbol x, I don't think so.

Or does it mean 'formula in one free variable, and that this free
variable is not x'.

It means a formula in one free variable, and x does not occur in the
formula at all.

Anyhow this still appears to be different from 'formula in which x is
not free' ,since the later imply that there can be many free variables
in the formula but none of them is x.

I personally prefer this form of separation:

2)if P is a formula in which x and a are not free then any closure of
AaExAy(yex<->(yea & P(y))) is an axiom.

However it seems that the standard schema of separation is:

3)if P is a formula in which x is not free then any closure of
AaExAy(yex<->(yea & P(y))) is an axiom.

I think that 3 is what leads to P(w) being uncountable ( see below ).

You're wrong, these axiom schemas are all equivalent.

There is a problem in Z.
This problem is that Z has a countable number of formulae,
and since for any set A each formula when used in separation schema
would prove the existence of ONE and only ONE subset of A ( this can
be proved easily for extensionality ), then to prove the existence of
any two different subsets of A using separation schema, we need AT
LEAST two non equivalent formulae to be used in separation schema.
Since the total number of formulae in Z is countable, then for any set
A, P(A) would also be countable.

No, this is complete nonsense. There are all sorts of confusions here.

Given a model of Z and a set a in the model, there is no reason why
every subset of a in the model should be definable using parameters in
the model.

Oh really, that's your opinion. And it is wrong.

It is not wrong, and you will find it is also the opinion of everyone
else who knows set theory. As I say, you really should listen to
people who actually know the subject.

There is every reason why every subset of a in the model should be
definable using parameters in the model. Otherwise how do you suggest
we define it, using parameters outside the model?huh., or perhaps you
think it might not be definable,

Yes, exactly. Not every member of the model need be definable.

Yes, this is the nonsense here.

Every member of the model need to be definable.

Saying that a member of the model exist and it is not definable is
nonsense.

I'm afraid you don't know what you're talking about. This is a very
basic confusion. As I say, you really should listen to people who know
the subject.

I don't need to listen to nonsense. Why should one have a theory which
have objects that are not definable in it.
This makes no sense at all. I don't care weather you are experienced
or not. What you are saying just makes nonsense.

Zuhair

You will find that everyone who knows the subject agrees with me.

If you don't want to learn and understand the subject, that's fine,
but in that case stop pestering this newsgroup talking about things
you don't understand.

Even the constructivists.

Zuhair


If you want to take a nonstandard point of view in foundations, that's
fine: just specify what theory you're working in. However, you
obviously don't understand what can be done in which theory. You're
deeply confused about the subject. If you're not interested in trying
to overcome those confusions, then stop bothering us.





if so then how do you know that it
exist.

Why shouldn't a non-definable element exist?

The axiom schema of separation guarantees that for each

formula with parameters there exists a subset of a which is definable
via that formula. But the converse need not hold.

Another important point is that countability is not an absolute
notion. There is a difference between being countable in the model and
being countable in the real world.

What I mean by countability is injectability to Omega.

Ax( x is countable <-> Ef(f:x->w,f is injective) ).

Yes, I know, but if you relativize every quantifier to the model then
that makes a difference.

But on the other side it is said the P(w) has uncountable number of
subsets?

Now the EXISTANCE of any unique subset of A can be proved using
separation. My conviction is that any other way will not succeed in
proving the existence of a unique subset of A if methods having
separation in them couldn't manage to do so. Therefore I conclude that
for any set A , Z can only prove the existence of a countable amount
of subsets of A. Thus P(A) is countable.

However Cantor presented a prove that any injective function from
w to P(w) is not surjective, and thus |P(w)|>|w|.

Af( (f:w->P(w), f is injective) ->~{x|xew & ~xef(x)}eRange(f) )

Now what is the basis for existence of {x|xew&~xef(x)}
it comes from separation according to formulation 3, since

Af(f:w -> P(w) & f is injective)ExAy(yex<->
( yew & Ez( <y z>ef -> ~yez ))).

is an instance of a separation axiom according to formulation 3).

This comes from separation written according to 3). were a is allowed
to be a free variable in P. If 2) is used instead ,which forbids
that ,then Cantor's proof fails simply because w can't be free in P.

So I think that this is the cause of the problem.

According to that if we use 2) instead of 3), then we have no problem,
and we'll have P(w) as a countably infinite set as w is.

Bye Bye to uncountability.

Zuhair- Hide quoted text -

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Relevant Pages

  • Re: Separation,Power and Countability.
    ... not free',since the later imply that there can be many free variables ... would prove the existence of ONE and only ONE subset of A (this can ... any two different subsets of A using separation schema, ... The people on this list who actually understand set theory. ...
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  • Re: Separation,Power and Countability.
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