Zermelo-Fraenkel never outlawed Russell's set

Neither Zermelo nor Fraenkel nor any logician ever outlawed Russell's set because, according to the elementary logic which was in part elucidated by Russell and his partner, A.N. Whitehead, Russell's set doesn't exist and never did.

Willard Van Orman Quine has related [1] that the logic developed by Russell and Whitehead in Volume I of Principia Mathematica is essentially the Boolean logic that was already in use, sans the symbols "0", "1", and "=", which are replaced by "T", "F", and "≡" (equivalence). In other words, Boolean logic was enough to show that the existence of Russell's set was false. From what I gather, Principia's version of non-quantifier logic was meant to address some special philosophical concerns, but is otherwise 1-to-1 with Boolean logic.

In naive set theory, any definable collection is a set. Let R be the set of all sets that are not members of themselves. If R is not a member of itself, then its definition dictates that it must contain itself, and if it contains itself, then it contradicts its own definition as the set of all sets that are not members of themselves. This contradiction is known as Russell's paradox, although I would say it is not actually a paradox. In symbols, we have

If R = {x|x ~e x}, then R e R <--> R ~e R

Of course, Russell dealt with his contradiction with his theories of types (supplemented by an addition to logic termed "the axiom of reducibility") whereas the Zermelo-Fraenkel system uses a set-theoretic axiom to rule out such a possibility.

The ZF axiom of foundation prevents x from being in x.

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A useful Wikipedia discussion relates that Emil Post in 1921 substituted Russell's "cumbersome" theory of types with "truth functions" and their truth tables.

Ludwig Wittgenstein also undercut the theory of types in Tractatus Logico-Philosophicus , published in 1922 but written a few years earlier. Using numbered paragraphs, Wittgenstein wrote:

3.331 From this observation we get a further view – into Russell's Theory of Types. Russell's error is shown by the fact that in drawing up his symbolic rules he has to speak of the meanings of his signs.

3.332 No proposition can say anything about itself, because the propositional sign cannot be contained in itself (that is the whole "theory of types").

3.333 A function cannot be its own argument, because the functional sign already contains the prototype of its own argument and it cannot contain itself...

Wittgenstein also proposed the truth-table method. In Tractatus 4.3 through 5.101, the philosopher adopts an unbounded Sheffer stroke as his fundamental logical entity and then lists all 16 functions of two variables (5.101).

And in 1914, a 20-year-old Norbert Wiener offered the first logico-mathematical definition of an ordered pair, drastically reducing the need for much of the convoluted material in Principia.

Originally, Russell was trying to deal with a flaw in Frege's system of logic which did indeed imply the contradictory set -- and thus proved that Frege's axiomatic system of logic is false. Now if it turned out that any rational system of logic implied Russell's contradiction, then one would be facing a paradox.

So if one says that naive sets are defined by their elements, with no other rules, then naive set theory is false. That is, if a proposition is used to define R, we soon discover that that proposition is false, and so the existence of R is likewise false. This simple observation shows that definition is not enough; mathematical objects (or their set-theoretic equivalents) must be tested, where possible, by logic. Or, as Quine put it, naive set theory's common sense is "bankrupt."

All this doesn't mean Zermelo's axiomatization wasn't a good idea. But it does suggest that there was no need to worry about R's existence but only about whether rules of set theory implied its existence.

I certainly don't mean to suggest that the history of Russell's contradiction isn't highly involved. One may get some flavor of the intricacy from, say, Quine [1], the noted logician-philosopher who, like Russell, studied under Whitehead's supervision. But I do mean to say that what we now take for elementary logic proves that Russell's set does not, and never did, exist.

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0. Russell's set does not exist.

0a. Statement 0 is equivalent to proving the falsehood of
"If R = {x|x ~e x}, then R e R <--> R ~e R"

To prove

1. R is the set of all sets that are not members of themselves.

Assumption.

2. If R e R, then R ~e R

Definition of R

3. If R ~e R, then R e R

Definition of R

4. (R e R --> R ~e R) & (R ~e R --> R e R)

Same as 2 and 3

5. (R ~e R v R ~e R) & (R e R v R e R)

Same as 4

6. R ~e R & R e R

Same as 5

7. Assign the name P to statement 1; Assign Q to "R e R"; assign ~Q to "R ~e R".

8. P --> Q & ~Q

9. ~P

P must be false,

because a true proposition cannot

imply a falsehood, which is what a contradiction is.

10. Russell's set does not exist.

Same as 9.

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1. For example, see Quine's essay on Russell's partner, A.N. Whitehead, in The Philosophy of Alfred North Whitehead, 2d ed., Library of Living Philosophers, 1941, 1951.