[Math] Is the axiom schema of specification sufficient for solving Russell’s paradox? If so, why

axiomselementary-set-theoryparadoxes

This is basically a two part question, as the title indicates. I understand why unrestricted comprehension will produce paradoxes like the Russell set, but I'm less clear on the question how the axiom of separation (or specification) solves this issue (and even whether it is able to solve it). Is the point that any set B defined in terms of having exactly those sets as members that are not members of themselves can never itself be a member of some arbitrary set A? In that case does this simply show that the Russell set is necessarily excluded from the things that one can meaningfully say about sets?

Any help on any aspect of this question would be appreciated!

Best Answer

See Russell's Paradox :

Zermelo replaces NC [Naïve Comprehension principle] with the following axiom schema of Separation (or Aussonderungsaxiom):

$$∀A ∃B ∀x (x \in B \iff (x \in A \land \varphi)).$$

Again, to avoid circularity, $B$ cannot be free in $\varphi$. This demands that in order to gain entry into $B$, $x$ must be a member of an existing set $A$. As one might imagine, this requires a host of additional set-existence axioms, none of which would be required if NC had held up.

How does Separation avoid Russell's paradox? One might think at first that it doesn't. After all, if we let $A$ be $V$ – the whole universe of sets – and $\varphi$ be $x ∉ x$, a contradiction again appears to arise. But in this case, all the contradiction shows is that $V$ is not a set. All the contradiction shows is that “$V$” is an empty name (i.e., that it has no reference, that $V$ does not exist), since the ontology of Zermelo's system consists solely of sets.

Consider a set $z$, we can still form the set $R = \{ x \in z : x \notin x \}$;

This only implies that :

if $R \in z$, then $R \in R \ \text { iff } \ R \notin R$, which means (reductio ad absurdum) that $R \notin z$.

Thus :

there is no universal set : $\forall z \exists R(R \notin z)$.

It can be worth to note that the "non-existence" of the Russell's set $R$ can be proved by logic alone :

1) $\exists y \forall x(A(x,y) \iff \lnot A(x,x))$ --- assumed [a]

2) $\forall x(A(x,c) \iff \lnot A(x,x))$ --- with $c$ a new constant

3) $A(c,c) \iff \lnot A(c,c)$ --- by instantiation.

The last line gives us a contradiction, because : $\vdash A(c,c) \iff A(c,c)$; thus, we conclude with :

$\vdash \lnot \exists y \forall x(A(x,y) \iff \lnot A(x,x))$ --- (*)

discharging the assumption [a].

Now, if we apply (*) to the language of set theory with the binary predicate $\in$ in place of $A$, we get :

$\lnot \exists y \forall x((x \in y) \iff (x \notin x))$.

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