How to solve the Russell's Paradox?

[Ben_Sinner]

How do we solve Russell’s Paradox and show that the Law of Non-Contradiction still remains true?

The Paradox is this:

Some sets are members of themselves; the set of all sets, for example, is a set, so it belongs to itself. But some sets are not members of themselves. The set of cats, for example, is not a cat, so it’s not a member of the set of cats. But what about the set of all the sets that are not members of themselves? If it is a member of itself, then it isn’t. But if it isn’t, then it is. It seems that it both is and isn’t.

Suppose the barber shaves everybody in town, except for all of those who shave themselves. Who shaves the barber? If he shaves himself, then he doesn’t shave himself; if he doesn’t, then he does.

 

[Reepicheep]

How do we solve Russell’s Paradox and show that the Law of Non-Contradiction still remains true?

The Paradox is this:

Some sets are members of themselves; the set of all sets, for example, is a set, so it belongs to itself. But some sets are not members of themselves. The set of cats, for example, is not a cat, so it’s not a member of the set of cats. But what about the set of all the sets that are not members of themselves? If it is a member of itself, then it isn’t. But if it isn’t, then it is. It seems that it both is and isn’t.

Suppose the barber shaves everybody in town, except for all of those who shave themselves. Who shaves the barber? If he shaves himself, then he doesn’t shave himself; if he doesn’t, then he does.

Or the barber is a woman.

Or the statement is untrue. The barber shaves all the men who do not shave themselves, and also shaves himself.
Or he shaves all those who do not shave themselves *except *for himself.
In general a set of statements that presents a paradox does so only if all the statements are as claimed, which might not be the base.

As for Russell’s Paradox, I don’t know enough about math to deal with it.

 

[Wesrock]

My understanding is that Russell’s paradox shows that we can describe things that are not sets. I believe the terms “types” and “classes” have been applied here. Essentially, mathematics had been (until then) working under the assumption that anything could be considered a set, but that was shown not to be true. It’s not that we had to solve the paradox under conventional rules, but realizing that the conventional rules mathematics had been working with needed modification.

 

[rossum]

How do we solve Russell’s Paradox and show that the Law of Non-Contradiction still remains true?

It is only true in certain circumstances, see Kurt Gödel’s incompleteness theorem.

Mathematics is an axiomatic system, and (logical) truth depends on the set of axioms we are using. Is “1 + 1 = 10” true? It is true using the axioms of binary arithmetic; it is false using the axioms of ternary arithmetic.

It you want the law of non-contradiction to be true, then assume it as one of your axioms. If you don’t want it to be true, then use a different logic. For example a four-valued logic: true, false, both, neither does not have the Law of Non-contradiction.

rossum

 

[Wesrock]

Suppose the barber shaves everybody in town, except for all of those who shave themselves. Who shaves the barber? If he shaves himself, then he doesn’t shave himself; if he doesn’t, then he does.

In my last post, I stated that Russell’s paradox had demonstrated that not all things that can be described are sets, and that is true.

Still, I wanted to touch again on this barber example, as it is very concrete and not abstract. Ultimately, the “set” as described cannot exist in reality precisely because it results in a contradiction. The barber cannot both shave himself and not shave himself. Just because I can offer a description does not make it a real possibility, something that can be actual. The statement simply needs to be more descriptive, or we need to assume the barber is excluded from the statement.

 

[Charlemagne_III]

In my last post, I stated that Russell’s paradox had demonstrated that not all things that can be described are sets, and that is true.

Still, I wanted to touch again on this barber example, as it is very concrete and not abstract. Ultimately, the “set” as described cannot exist in reality precisely because it results in a contradiction. The barber cannot both shave himself and not shave himself. Just because I can offer a description does not make it a real possibility, something that can be actual. The statement simply needs to be more descriptive, or we need to assume the barber is excluded from the statement.

Here is the solution to the paradox given by William Shakespeare.

“Much ado about nothing.”

 

[inocente]

Suppose the barber shaves everybody in town, except for all of those who shave themselves. Who shaves the barber? If he shaves himself, then he doesn’t shave himself; if he doesn’t, then he does.

If instead the sets are (a) those who shave themselves and (b) those who don’t, there is no ambiguity. This seems to indicate the issue isn’t with reality but with how we try to divide it up into categories.

 

[John_Martin]

How do we solve Russell’s Paradox and show that the Law of Non-Contradiction still remains true?

The Paradox is this:

Some sets are members of themselves; the set of all sets, for example, is a set, so it belongs to itself. But some sets are not members of themselves. The set of cats, for example, is not a cat, so it’s not a member of the set of cats. But what about the set of all the sets that are not members of themselves? If it is a member of itself, then it isn’t. But if it isn’t, then it is. It seems that it both is and isn’t.

Suppose the barber shaves everybody in town, except for all of those who shave themselves. Who shaves the barber? If he shaves himself, then he doesn’t shave himself; if he doesn’t, then he does.

If the barber pays himself for his own shave, then he is being shaved by the barber, not by himself. Did he pay the barber? The “barber” is a vocational function, the function of which costs money. So who shaved the barber, a private individual at no cost, or a professional charging a fee for services?

The “set of all sets that are not members of themselves” is not a member of the “set of all sets that are not members of themselves”, but the “set of Cats” is in that set.

I do not see a paradox in this, except for someone trying to say that the “set of all sets that are not members of themselves” is in itself (which it is not, therefore, no paradox)

The “set of all sets that are not members of themselves” is a member of the “set of all sets”, as a set, but not mixed with its members, which are in two sets.
The set of cats is twice listed; once as a member of all sets, and embedded a second time in the “set of all sets that are not members of themselves”.

 

[e_c]

How do we solve Russell’s Paradox and show that the Law of Non-Contradiction still remains true?

The Paradox is this:

Some sets are members of themselves; the set of all sets, for example, is a set, so it belongs to itself. But some sets are not members of themselves. The set of cats, for example, is not a cat, so it’s not a member of the set of cats. But what about the set of all the sets that are not members of themselves? If it is a member of itself, then it isn’t. But if it isn’t, then it is. It seems that it both is and isn’t.

Suppose the barber shaves everybody in town, except for all of those who shave themselves. Who shaves the barber? If he shaves himself, then he doesn’t shave himself; if he doesn’t, then he does.

The disjunction “except” is simply implying a contradiction, so the proposed description is not truth-apt.

Solved.

 

[Gorgias]

How do we solve Russell’s Paradox and show that the Law of Non-Contradiction still remains true?

By reading up on the history of the problem.

Basically, it comes down to this: whether you call it a “vicious-circle principle”, or an “infinite regress”, it all hinges on the reasonableness of whether something can be asserted to be part of the answer before it’s identified as part of the domain of a problem. A formal statement of this requirement might look like Zermelo’s schema of separation:

∀A ∃B ∀x (x ∈ B ≡ (x ∈ A ∧ φ))