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**A solution to Russell's Paradox**

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Wed 30 Dec, 2009 05:49 am

Russell's paradox is the most famous of the logical or set-theoretical paradoxes. The paradox arises within naive set theory by considering the set of all sets that are not members of themselves. Such a set appears to be a member of itself if and only if it is not a member of itself, hence the paradox.

Some sets, such as the set of all teacups, are not members of themselves. Other sets, such as the set of all non-teacups, are members of themselves. Call the set of all sets that are not members of themselves "

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Russell does not offer a solution to the paradox, rather his theory of types avoids the paradox. The function (x e x) cannot be expressed within his type theory.

The answer to Russell's question, "Is the class of those classes that are not members of themselves a member of itself or not?" is No, because it does not exist.

There is a theorem of first order predicate logic, 1. ~EyAx(xRy <-> ~(xRx)),

that resolves the Russell Paradox and the Barber Paradox.

Because of 1, there is no y such that Ax(x e y <-> ~(x e x)),

and there is no y such that Ax(y shaves x <-> ~(x shaves x)).

1. ~EyAx(xRy <-> ~(xRx))

Proof:

1a. Ax(xRy <-> ~(xRx)) -> (yRy <-> ~(yRy)) ..when x has the value y.

2a. (yRy <-> ~(yRy)) is a contradiction for all y. (p <-> ~p) is contradictory.

3a. ~Ax(xRy <-> ~(xRx))

4a. Ay(~Ax(xRy <-> ~(xRx))).

5a. ~EyAx(xRy <-> ~(xRx))

QED.

That is, ~EyAx(x e y <-> ~(x e x)) is true. (it is an instance of 1.)

That is, ~EyAx(y shaves x <-> (x shaves x)) is true. (it is also an instance of 1.)

The Russell class is not a member of any class, including proper classes if such there be. And it has no members, nor does it have any properties.

Both Frege (axiom V of

That there is a class for every predicate, is false in their systems.

Zetetic11235

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Sun 3 Jan, 2010 05:13 pm

@Owen phil,

I just wonder if you have heard of ZFC or NF or NGB? They are all consistent axiom schemes for set theory. All of them On top of that, it seems to me that all of the results in your conclusion are already well known and your paper is exceedingly short, which leads me to ask: Where is this published?

Owen phil

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Mon 4 Jan, 2010 03:20 am

@Zetetic11235,

Zetetic11235;116723 wrote:

I just wonder if you have heard of ZFC or NF or NGB? They are all consistent axiom schemes for set theory. All of themavoidRussel's paradox in the sense that they haveadjustedtheir axioms so that it does not occur. You have suggested the same thing. I suppose it's good armchair logic. The only thing that seems odd is the quantification over the unbound variable. You come up with the result that there is no y satisfying Ax( xRy<->xRx), but it seems like this cannot be true in general. You are saying for instance that if R is >= then Ax (x>=y <->x>=x). Lets say our domain of discourse is over the ordinal numbers. Isn't it true that Ax (x>=0<->x>=x)? Both are tautologies, that is, X is always greater than or equal to zero, zero is the least element of Ord. X is always greater than or equal to itself. Since set theory is meant to provide a foundation to mathematics, I don't see how taking such an axiom in general on a binary relation is legitimate.

On top of that, it seems to me that all of the results in your conclusion are already well known and your paper is exceedingly short, which leads me to ask: Where is this published?

Zetetic11235,

"I just wonder if you have heard of ZFC or NF or NGB?"

Yes, I have heard of these systems.

Quine, (NF) uses EyAx(x e y <-> Fx) where F is stratified.

Zermelo, (ZFC) uses EyAx(x e y <-> (x e z & Fx)).

von Neumann, (NBG) uses "proper classes" which I have doubts about.

On this point I would suggest, EyAx(x e y <-> (EzAw(w e z <-> Fw) & Fx)).

Here the Russell class does exist, and it is equal to the null set.

That is, there is a class for every predicate is true.

Zetetic11235,

"You come up with the result that there is no y satisfying Ax( xRy<->xRx), but it seems like this cannot be true in general."

Did you make a typo here?

There is no y such that: Ax(xRy <-> ~(xRx)), is a theorem of FOPL.

You are correct about Ax(x >= 0 <-> x >= x) being tutologous and,

EyAx(x >= y <-> (x >= x)) is a tautology.

But, ~EyAx(x >= y <-> ~(x >= x)) is also tautologous.

Zetetic11235,

"On top of that, it seems to me that all of the results in your conclusion are already well known and your paper is exceedingly short, which leads me to ask: Where is this published?"

Thanks for the vote of confidence, but it is not published anywhere to my knowledge.

"Armchair logic" is what I try to do...logic is a hobby for me, how about you?

Zetetic11235

Reply
Mon 4 Jan, 2010 11:09 am

@Owen phil,

Owen;115524 wrote:

Russell's Paradox

FirstpublishedFri Dec 8, 1995; substantive revision Wed May 27, 2009

That made me think you were claiming that it had been published.

Sorry, I got the impression that you were trying to put forward that you had made a great, earth shaking discovery in FOPL and were putting it forward on a forum. This happens a lot; people with theories of everything, new theories, proofs of god's existence (which can be quite long and involved) and so forth. So I apologize for being so hostile and for not reading your post more carefully.

On a second look; your point is that in FOPL, ~EyAx(xRy<->~(xRx)) is true, and so Russel's paradox developed because Frege had essentially assumed something that is demonstrably false in FOPL, something that is well known but a good excersise to work out in a novel way. Still, I don't know if it's fair to say that Russell was 'avioding' the paradox.

WEll, you asked, so: As an aspiring mathematical logician I do math and logic (usually category theory/set theory and not so much pure logic) everyday for at least a few hours, so I am probably too quick to criticize and I was too quick to assume you thought you were presenting ground breaking material.

Owen phil

Reply
Mon 4 Jan, 2010 11:41 am

@Zetetic11235,

Zetetic11235;116909 wrote:

That made me think you were claiming that it had been published.

Sorry, I got the impression that you were trying to put forward that you had made a great, earth shaking discovery in FOPL and were putting it forward on a forum. This happens a lot; people with theories of everything, new theories, proofs of god's existence (which can be quite long and involved) and so forth. So I apologize for being so hostile and for not reading your post more carefully.

On a second look; your point is that in FOPL, ~EyAx(xRy<->~(xRx)) is true, and so Russel's paradox developed because Frege had essentially assumed something that is demonstrably false in FOPL, something that is well known but a good excersise to work out in a novel way. Still, I don't know if it's fair to say that Russell was 'avioding' the paradox.

WEll, you asked, so: As an aspiring mathematical logician I do math and logic (usually category theory/set theory and not so much pure logic) everyday for at least a few hours, so I am probably too quick to criticize and I was too quick to assume you thought you were presenting ground breaking material.

It was a quote from Stanford University about the Russell paradox.

There is nothing to be sorry about. My mistake, I didn't make it clear.

Russell's theory of types denies that (x is a member of x) is a predicate of x.

Thereby he avoids the paradox.

odenskrigare

Reply
Thu 7 Jan, 2010 12:51 pm

@Owen phil,

well we got a lot of crackpots hereI have to admit that the post title was a bit flippant and I thought I was going to see the VenomFangX of mathematical logic

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