How do the consequences of Russell's paradox extend beyond universal comprehension principle as far as the set of all sets problem?

Question

I think I understand the way in which Russell's paradox shows that the following principle is wrong:

" for every predicate, there is a set having as elements all the objects that satisfy this predicate"

Russell's picks up a predicate ( namely the predicate " x is not an element of itself" ) and shows that the corresponding " set" would have contradictory properties, which means that " the set of all x such that x is not an element of x" does not exist.

This counts as a counter example to the alledged " principle".

My understanding of Russell's paradox does not go further than this.

Now, if I am correct, it is often said that Russell showed, with this paradox, that the " set of all sets" does not exist.

What is actually the relation between Russell's paradox and the non-existence of the set of all sets?

It seems difficult to me to answer that the relation consists in the fact that precisely a set has the property of not belonging to itself. For it seems to me that Russell's paradox forbids to define a set in this way.

Answer 1

Russell's paradox does not in itself prevent a set of all sets from existing. There are set theories that do contain a universal set and are not known to be inconsistent, such as Quine's NF.

It is only together with Zermelo's subset selection axiom (the core idea behind ZFC, which claims that $\{x\in A\mid \varphi(x)\}$ is a set whenever $A$ is), that it has this effect. If a universal set existed, then the subset selection axiom would effectively provide a universal comprehension principle, and then Russell's paradox would produce a contradiction.

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it is often said that Russell showed, with this paradox, that the "set of all sets" does not exist.

While it is probably not difficult to find a claim such as this in print, it is an ahistoric oversimplification. Russell published the paradox several years before Zermelo proposed his axiom system, so the ingredients for making the jump from "unrestricted set comprehension doesn't work" to the specific claim "there is no universal set" was not present at that time.

Answer 2

The thing is that the principle is wrong, but we'd like it to be true : you want to be able to consider the set of things satisfying a certain property, it's one of the bases of constructing new sets.

To make that possible without introducing (or so we hope) contradictions, the ZF axioms have the so-called comprehension schema, which says that if you have a formula $\varphi (x)$ and a set $A$ then you can consider the set of all sets satisfying $\varphi$ which are also in $A$; which is usually denoted $\{x\in A\mid \varphi(x)\}$

Now if $A$ is a set containing all sets, then $\{x\in A\mid \varphi(x)\}$ is precisely the set of all sets that satisfy $\varphi$ , and so having a set of all sets would imply having the naive principle which we know is contradictory : therefore there is no set containing all sets.

This is how Russell's paradox relates to the existence of a set of all sets; although at first sight it just implies that there is no set of all $x$ such that $x\notin x$ (which is a way more general fact and is very little related to set theory, as Maura Allegranza points out in the comments)

Answer 3

Russell's Paradox arose from an inconsistency in Frege's early attempt to axiomatize Cantor's set theory. As you point out, by using his axiom of unrestricted comprehension, it was possible to prove:

$$\exists r: \forall x: [x\in r \iff x\notin x]$$

Using only the rules of logic in Frege's system, however, it was also possible to prove the negation:

$$\neg\exists r: \forall x: [x\in r \iff x\notin x]$$

We kept his rules of logic, but got rid of unrestricted comprehension and are still able to prove the latter.

It wasn't necessary to ban set self-membership to resolve the Paradox. You can substitute any binary relation E to obtain:

$$\neg\exists r: \forall x: [E(x,r) \iff \neg E(x,x)$$