How do sets of language used to formulate ZFC axioms escape Russell's paradox?

Question

We formulate sets using ZFC. Nonetheless, to write its axioms we already use the notion of sets. For instance, in formulating the Axiom of Extensionality, we write the following concatenation of symbols:

∀ u(u ∈ X ≡ u ∈ Y)=>X=Y.

The symbols ∀, u, X, ∀, ≡, ), (, =>, ∈, = are the set of symbols of a given language.

The constants, predicates and functions assigned to such symbols comprise the underlying set of a given structure.

As it can be concluded by the meta-theoretical statements above, the notion of sets has to be conceived naively, prior to ZFC, for the latter to be formulated.

Nevertheless, ZFC was conceived to avoid Russell's paradox, which is said to be present in the Naive Set Theory.

So, the question is: how does the naive notion of Sets used to formulate ZFC escapes Russell's paradox?

Answer 1

This question is actually a bit more general. For example, when one talks of non-well-founded set theory, e.g. Aczel's, then you need a "pre-theoretical" notion of [rooted] graph (not just set) just to state Aczel's anti-foundation axiom.

The way this is resolved, in both cases, is to make use of the distinction between language and meta-language that Tarski used to solve/avoid liar's paradox. In other words, the sets defined by the ZF[C] axioms and those (barely) used in the (first-order) logic meta-language that states ZF are not the same sets. Likewise, the graphs used to state Aczel's axiom are not the graphs that can be defined within a set theory (be it ZF or non-well-founded.)

The naive notion of sets used to state FO doesn't really need to escape Russel's paradox, because it's not axiomatized, or used for any other purpose other than to build up ZF. And in ZF you cannot state Russel's paradox.

In the Tarskian layering, you cannot assign [in the meta-language itself] truth values to [all] statements appearing in the meta-language; you can only assign truth values to statements made in the object-language, which by inclusion/identification can be considered contained in the meta-language. So even you could formulate $R = \{x\; |\; x \in x\}$ in the meta-language MZF in which you talk about truth values of statements in ZF, because this sentence fragment is not constructible in ZF itself, you cannot assign either true or false to a Russel-paradox-type statement $R \in R \iff R \notin R$ within the meta-language MZF; to assign a truth value to this, you need another meta-meta-language MMZF whose object-language (MZF) can express Russel's paradox (and would be useless/inconsistent). This is basically the content/consequence of the so-called Tarski's undefinability theorem (which may have actually been discovered by Gödel).

Answer 2

Russell's paradox arose from the axiom of abstraction: given a property $P$ there exists a set consisting of all objects with that property.

ZFC has no axiom of abstraction. Instead it only allows you to synthesize new sets from old sets in various ways. So the paradox simply doesn't arise since ZFC doesn't allow you to define a set that consists of all sets that are not members of themselves.

Answer 3

As you mentioned it before you have to take into account the metatheory. When we formulate the axioms of ZFC, formally we do not intend for them any interpretation. Hence, you could formulate the axioms of ZFC in PA by Gödel-Numbering the symbols and formulas of ZFC. This can also be done in ZFC as a metatheory. In the first case you just deduce things from the axioms and 1st order Logic and in the second case you do the same but also you can have models of some subset of the axioms of ZFC but not all of them at once. So, you avoid Russell's paradox by not working in the metatheory with the original naive Set Theory but with the formal one.