Russell's paradox question

Question

Tao's analysis book uses following example for Russell's paradox: $$P(x) \Longrightarrow `` x\text{ is a set, and }x \notin x"\\ \Omega := \{x : P(x)\text{ is true} \} = \{x : x\text{ is a set and }x \notin x\}$$ then conclude that $­\Omega \in \Omega$ and $­\Omega \notin \Omega$ happened same time. so it is a paradox. My question is if $x$ is a set, you can't write $x \notin x$ right? only element belong to a set, isn't the statement wrong at the first place? or this could not be called as a statement.

Answer 1

Speaking "pre-theoretically" - that is, in naive set theory, before we fix a formal system of axioms - there is no reason that a set can't be an element of itself! And in fact, even in formalized set theory it can be okay: in the usual set theory, $ZFC$ (or its variants), this is ruled out by the axiom of foundation, but there are set theories such as $NF$ ("New Foundations") or the "anti-Foundation" set theories (Boffa, Aczel, . . .) which do allow sets to be elements of themselves.

In fact, in $NF$, the "set of all sets" does exist, and is an element of itself! $NF$ escapes Russell's paradox by not allowing full comprehension, or even full separation: just because $A$ is a set, doesn't mean (in $NF$) that the collection of all $x\in A$ satisfying some formula is a set.

The point is that, surprising as it may be, there is no reason to outlaw a set being an element of itself from the get-go. If we think of sets as properties, then we should in fact expect this: after all, the property "is a property" is a property! :P

Answer 2

Sets are mathematical objects which are collections of mathematical objects. In particular, sets can be elements of other sets.

There is no reason for discrimination against sets. However, what Russell's paradox shows is that not every "expressible collection" of mathematical objects is a set. This is a naive belief which turns out to be false.

Answer 3

The problem was not set self-membership. If we assume $$Set(r) \land \forall x:[x\in r \iff Set(x) \land x\notin x]$$ and apply nothing more than the rules of logic (no set theory), then we can obtain the contradiction $$r\in r \land r\notin r$$

where $Set(r)$ means $r$ is a set.

Therefore, we conclude that no such set $r$ can exist. (A similar argument holds without the $Set$ predicate.)

In an early attempt to axiomatize set theory which used the same rules of logic, there was a very powerful axiom (too powerful as it turned out) that for every unary predicate $P$, there existed a set $S$ such that $\forall x:[x\in S \iff P(x)]$. Of course, $P(x)\equiv Set(x) \land x\notin x$ led to the above contradiction. Using this early set theory, you could both prove and disprove the existence of the set $r$. Clearly, this powerful axiom had to be replaced by other less powerful axiom(s). Various patches were subsequently proposed. The most popular are the ZFC axioms of set theory which effectively banned set self-membership anyway. It should be noted that other popular set theories do not ban set self-membership.