# Math Help - Set Foundations and Russel's Paradox

1. ## Set Foundations and Russel's Paradox

=’ not equal to, ϵ’ not a “member” of

1) x=y: undefined except x=x (axiom), x=’x false (logic).
2) xϵy: Principal Primitive Undefined concept*
3) x and y undefined.

Theorem: xϵx iff x=x.
Proof: x and x are identical so there can be no other relation between them.

Theorem: xϵ’x iff x=’x.
Proof:
If xϵ’x and x=x then xϵx. Therefore xϵ’x → x=’x
If x=’x and xϵx then x=x. Therefore x=’x → xϵ’x

Conclusion: xϵx is x=x. xϵ’x is x=’x which is a false statement.

Ex: Suppes obtains Russel’s Paradox by using the function φ(x) is (xϵ’x) which is (x=’x) which is false and inadmissible in the”axiom of abstraction,” Suppes pg 6.

Any derivation which leads to Russel’s Paradox in the form “(xϵx) iff (xϵ’x)” should be “(x=x) iff (x=’x)” where (x=’x) is false and inadmissible so there is no paradox.

Patrick Suppes Axiomatic Set Theory, pg 5. Dover, 1972
*Paul Halmos Naive Set Theory, pg 2, Van Nostrand, 1960
Willard Quine Set Theory and its Logic, Harvard, 1969

Personally, I can bareley scratch out a few nuggets from the first few pages of these references.

2. ## Re: Set Foundations and Russel's Paradox

Given ϵ and =, there are two possibilities: ϵ and = are the same, explored in previous post, or, the standard case:

ϵ is not the same as =.
In this case, xϵx is impossible because x=x.
Then Russels Paradox, (xϵx) iff (xϵ’x), doesn’t exist because there is no x st (xϵx).

Think of it like this: A Proposed Paradox (x<x) iff (x<’x) is impossible because (x<x) doesn’t exist for all x because < and = are different (mutually exclusive) and x=x.

I put this result in a separate thread because of its fundamental implication for Axiomatic Set Theory. (It didn't occurr to me when I started this thread)

3. ## Re: Set Foundations and Russel's Paradox

Originally Posted by Hartlw
Given ϵ and =, there are two possibilities: ϵ and = are the same, explored in previous post, or, the standard case:

ϵ is not the same as =.
In this case, xϵx is impossible because x=x.
Why would tha follow? The fact that " $\in$" is not the same as "=" doesn't mean the are contrary. "is blue" is not the same as not the same as "is round" but "x is blue" does not contradict "x is round".

Then Russels Paradox, (xϵx) iff (xϵ’x), doesn’t exist because there is no x st (xϵx).

Think of it like this: A Proposed Paradox (x<x) iff (x<’x) is impossible because (x<x) doesn’t exist for all x because < and = are different (mutually exclusive) and x=x.

I put this result in a separate thread because of its fundamental implication for Axiomatic Set Theory. (It didn't occurr to me when I started this thread)

4. ## Re: Set Foundations and Russel's Paradox

Set Membership, Equality, and Russells Paradox

EDIT: In response to post #3, maybe this helps: xϵx means whatever is on the left side of ϵ is the same as whatever is on the right side of ϵ, ie x=x.
So xϵx really means x=x and so there is no such thing as xϵx other than x=x..

5. ## Re: Set Foundations and Russel's Paradox

OK. Let's start at rock bottom of Axiomatic Set Theory:

Suppes (and others I believe) gives the Primitive Atomic Formulas: (uϵv) and (u=v).

Substitute u = v into both formulas and you get vϵv and v=v are the same thing.

6. ## Re: Set Foundations and Russel's Paradox

Originally Posted by Hartlw
=

Any derivation which leads to Russel’s Paradox in the form “(xϵx) iff (xϵ’x)” should be “(x=x) iff (x=’x)” where (x=’x) is false and inadmissible so there is no paradox.
I trust that you are aware that all modern versions of set theory are designed to avoid Russell's paradox?

.

7. ## Re: Set Foundations and Russel's Paradox

Originally Posted by zzephod
I trust that you are aware that all modern versions of set theory are designed to avoid Russell's paradox?

.
You are right. The paradox is derived in Suppes from the pre-Russel axiom scheme. Russell published his Paradox in 1901 shortly after Frege published his Foundations of Arithmetic. The axioms were subsequently revised to avoid Russells paradox. Since Russells Paradox doesn’t exist, the revision was unnecessary and raises the question about axioms which avoid a non-existent paradox.

Russells Paradox boils down to the validity of (xϵx). Responding also to HallsofIvy post #3 I note:

xϵy says x is a member of the set y.
Let x=y, and y={A}. Then xϵy becomes yϵy, or {A}ϵ{A}, which is incorrect.
{A}ϵ{{A}} is correct.

So it’s not just a matter of xϵx meaning x=x, xϵx is incorrect (doesn’t exist, meaningless).

It’s not a trivial distinction. xϵy occurrs constantly in books on set theory without the qualification x='y.

8. ## Re: Set Foundations and Russel's Paradox

Primitive Atomic Formulas for Axiomatic Set Theory are (vϵw) , (v=w), from which
vϵv
and that is no problem if you don’t define ϵ and =.

Howeever, the subject is SET theory, so at some point Set Membership and Equality have to mean something*, and then there is no such thing as (vϵv).