Separation,Power and Countability.

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'.

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 ).

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.

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

.

Relevant Pages

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