# what does “a set of sets that are not members of themselves” of Russell’s Paradox mean

Russell’s Paradox begins with a statement of "Let $R$ be the set of sets that are not members of themselves", i.e. $R=\{S\mid S\notin S\}$.

I'm a little bit confused with the statement, for example, let $S=\{1,\{2,3\}\}$, of course $S\notin S$ since $S$ doesn't have an element $\{1,\{2,3\}\}$, $S$ only have two elements which are $1$ and $\{2,3\}$, So $R$ only have one element i.e. $R=\{\{1,\{2,3\}\}\}$, and again, of course $R\notin R$, I must have something wrong here but I don't know where I go wrong.

• That you are unable to write down a set that contains itself doesnt mean that it can't exist. – Tim B. Jul 23 '16 at 9:21
• @LeBtz How can a set be a member of itself, could you please give an example? – whoisit Jul 23 '16 at 11:29
• @LeBtz: From ZF we can prove that a set is never a member of itself. – Asaf Karagila Jul 23 '16 at 11:43
• @AsafKaragila I know, that is why I wrote that Cantor's axioms dont allow it by which I meant ZF. To be more specific: I wanted to emphasize that the reason that there is no such set is not due to not being able to write down an example but for another reason which is: the axioms of ZF. – Tim B. Jul 23 '16 at 11:44
• An example of a set which is a member of itself is the set of all sets, i.e. $\{x\mid x=x\}$. It doesn't exist in ZF set theory but does eg. in NF. – nonpop Jul 24 '16 at 0:42

Sets are collections of mathematical objects which are themselves mathematical objects. That means that in principle, a set can be a member of itself. We can therefore ask which sets are not members of themselves, and that is a valid question from a mathematical point of view.

Russell's paradox shows us that the collection of sets which are not members of themselves is not a set. It is a collection we can define, but it cannot be a set. And that was the main purpose of the paradox, to dispel the notion that every collection we can define forms a set.

Your example is not good, though, because the paradox applies to the entire mathematical universe. It encompasses all the sets out there. Not just that specific $S$.

Formally speaking, modern set theory has an axiom which prevents a set from being a member of itself. But it is possible to replace that axiom by one of several axioms which guarantee the existence of sets which are members of themselves (e.g. $x=\{x\}$, and more), and these new axioms do not introduce new contradictions to mathematics.

• How can a set be a member of itself, could you please give an example? – whoisit Jul 23 '16 at 11:29
• Naively it cannot. So I can't give you a specific example. That was the motivation behind the axiom of foundation (also known as regularity). The sets we are able to literally write down are not going to to be members of themselves, or members of their members and so on. This is not to say, however, that this notion is entirely impossible from an axiomatic point of view. – Asaf Karagila Jul 23 '16 at 11:40

You can have relations that are self-referential. That is not the problem. With the equality relation, for example, we have $x=x$ for all $x$. For any binary relation $R$ (including $\in$), however, we can prove, using only the rules of ordinary logic (independent of any set theory) that:

$$\neg \exists x: \forall a:[aRx \iff \neg aRa]$$

The problem arises from the late 1800's when mathematicians were trying to formalize the notion of what "set" means, what are numbers, what is logic? Georg Cantor proposed Naive set theory where "a set is described as a well-defined collection of objects. These objects are called the elements or members of the set. Objects can be anything: numbers, people, other sets, etc."

Anything that could be described in words was a set, for example the Universal Set, U, meaning the set that contains ALL sets. Since U is a set it must also be a member of the Universal Set so $U \in U$. That was a bit disturbing, even pathological but hey, we can describe it so U must be a set in Naive Set Theory. And since U contains itself, there might be other sets that also that contain themselves.

In 1901 Bertrand Russell proposed that we look at the set $A$ that contains only the "non-pathological" sets, $A = \{ S : S \notin S\}$ That is "the set of all sets that don't contain themselves." $A$ would be a great set to base mathematics upon except for one problem, is $A \in A$ ? If it is, then by the definition of $A$, $A \notin A$. We also get a contradiction if we assume $A \notin A$. So $A$, and by extension, Naive Set Theory, is logically inconsistent.

Because of Russell's Paradox, mathematicians realized they needed a new theory to restrict what a "formal" "set" means in such a way to exclude $U$ and $A$ from being mathematical "sets". Of the several theories created, the most popular today is Zermelo-Fraenkel set theory or ZF. But that's a topic for a different question.

I don't know of any set that is a member of itself, and the assumption that no such sets exist seems very reasonable to me.

Now, if we assume that no set is ever a member of itself, then Russell's paradox takes a simpler form: Let $R=\{x:x=x\},$ the set of all sets. Since all sets are members of $R,$ we have $R\in R.$ Since no set is a member of itself, we also have $R\notin R.$ Contradiction!

However, mathematicians and logicians like to base their proofs on the fewest and weakest possible assumptions. Someone might answer the simple argument I just showed you by saying, "How do you know that no set can be a member of itself? There is the source of your contradiction, that unjustified assumption! Some sets are members of themselves, and the set of all sets is obviously one of them!"

Russell's definition $R=\{x:x\notin x\}$ is a clever trick to make the contradiction work even if such outlandish sets exist.