- #1

- 6

- 0

The Axiom Schema of Unrestricted Comprehension as I understand it is stated as follows (forgive my lack of proper symbolism here):

There exists a set y such that for all x, x e y <==> P(x) where P(x) is any property of x and "e" represents the membership predicate.

To derive Russell's Paradox, we let x=y (which we can do by the quantifier logic expressed in the axiom), and let P(x) be (not)x e x. We then have:

y e y <==> (not) y e y.

Now, the Axiom Schema of Separation, which supposedly does not give you a way to get to the paradox, is as follows:

There exists a y such that for all x, x e y <==> x e z ^ P(x) where P(x) is again a property of x (though this time with the restriction that the variable z is not free in P(x)), z is free, and "^" denoted the appropriate logical connective.

My question is: Letting x=y and P(x) be (not) x e y we then have:

1) y e y <==> y e z ^ (not) y e y

==> 2) (y e y ==> y e z ^ (not) y e y) By definition of "<==>"

==> 3) ((y e y ==> y e z) ^ (y e y ==> (not y e y))

==> 4) (y e y ==> (not y e y)

In going from 2 to 3 I've used "(A==>B^C)==>(A==>B ^ A==>C)

In going from 3 to 4 I've used "(A^B==>A)"

And so we have again produced a contradiction of sorts. How are we to interpret the above result? Is this a justifying construction of the empty set, as we know that by extensionality any set defined by separation is unique? Are we instead to ignore the case of separation involving phrase "y e y" completely since we know that by the Axiom of Regularity it is true that for all nonempty sets y we have (not) y e y?

All help is appreciated. Thank you for your time.

Sincerely,

Adam Fiddler