# Set of All Sets (Russel's) Paradox

• Feb 25th 2013, 11:17 AM
Hartlw
Set of All Sets (Russel's) Paradox
Set of all sets (Russel’s) paradox is not a paradox because there is no such thing as the set of all sets.

Proof: Let A, B be sets. Then C={A,B,C} is not a set because it is undefined (circular definition).
• Feb 25th 2013, 11:33 AM
jakncoke
Re: Set of All Sets (Russel's) Paradox
Why can't there exist a set containing all sets? The definition of set under the Cantor definition is a set is a collection of distinct objects.

Russells paradox says if R is a set containing all sets which are not members of themselves. Then R is also included in this set, For R contains sets which are not members of themselves.

At the same time, R is a member of itself.

• Feb 25th 2013, 11:56 AM
emakarov
Re: Set of All Sets (Russel's) Paradox
Quote:

Originally Posted by Hartlw
Proof: Let A, B be sets. Then C={A,B,C} is not a set because it is undefined (circular definition).

What does this proof have to do with the set of all sets?
• Feb 25th 2013, 01:11 PM
Hartlw
Re: Set of All Sets (Russel's) Paradox
Quote:

Originally Posted by emakarov
What does this proof have to do with the set of all sets?

C={A,C} → C={A, {A,C}} → C={A, {A, {A, {A,C}}}} →………
There is no such thing as the set C (circular definition).
• Feb 26th 2013, 12:07 PM
Hartlw
Another point of view. (ϵ’ means does not belong to)

if R = {x│x ϵ’ x } then R ϵ R <-> R ϵ’ R (wiki)

This is not a Paradox because (Theorem) there is no such thing as R ϵ R.
Proof: Rmember ϵ Rset implies R unequal to R.
• Feb 26th 2013, 10:21 PM
zzephod
Re: Set of All Sets (Russel's) Paradox
Quote:

Originally Posted by Hartlw
C={A,C} → C={A, {A,C}} → C={A, {A, {A, {A,C}}}} →………
There is no such thing as the set C (circular definition).

Please give us the axioms for set theory you think you are using so we can see for ourselves that your circular definition is excluded.

.
• Feb 27th 2013, 07:09 AM
Hartlw
Re: Set of All Sets (Russel's) Paradox
Quote:

Originally Posted by zzephod
Please give us the axioms for set theory you think you are using so we can see for ourselves that your circular definition is excluded.

.

Axiom: You can’t define something in terms of itself.

If that doesn’t satisfy you, and you accept C={A,C} as a legitimate set, I suggest you substitute the set C={A,C} into the various systems of Axiomatic set theory (which don’t define set) and if you find an inconsistency you may wish to revise that particular axiom. If you don’t find an inconsistency, which I suspect you won’t since the axioms are aware of Russels Paradox, then the only conclusion you can come to is that the axiomatic set has to be revised to reject Russel’s paradox.

Russel published his “Paradox” shortly after Frege published his Foundations of Arithmetic, violating one of Frege’s axioms, which he subsequently revised to account for it. If MHF and this thread were around at the time, he could have saved himself the trouble:
Patrick Suppes Axiomatic Set Theory, pg 5. Dover, 1972

EDIT: If you don't accept the axiom of definition, I can suggest a few books on logic for you to work through.
• Feb 27th 2013, 11:49 AM
zzephod
Re: Set of All Sets (Russel's) Paradox
Quote:

Originally Posted by Hartlw
Axiom: You can’t define something in terms of itself. ...

Is that a theorem of ZF?

.
• Feb 27th 2013, 12:19 PM
jakncoke
Re: Set of All Sets (Russel's) Paradox
If you can't define something in terms of it self, then recursive functions wouldn't be well defined. The natural numbers wouldn't exist. So wouldn't alot of other things in mathematics.
• Feb 28th 2013, 05:13 AM
Hartlw
Re: Set of All Sets (Russel's) Paradox
Quote:

Originally Posted by jakncoke
If you can't define something in terms of it self, then recursive functions wouldn't be well defined. The natural numbers wouldn't exist. So wouldn't alot of other things in mathematics.

In a recursive function you define something new in terms of something that exists. Given x(n), you define x(n+1). You don’t define xn in terms of xn.