# Does the Russell Set exist?

I am currently reading "Naive set Theory" by Paul Halmos. In the second chapter, on the axiom of specification we show that the Universal Set does not exist. The proof is the following:

Lets define $R$ as: $R=\{x \in A\mid x \notin x\}$

Lets suppose that $R \in A$, there are two options:

1. $R \in R$. We reach an immediate contradiction, since, if $R \in R$ then $R \not\in R.$
2. $R \not\in R$. And again we reach a contradiction since, if $R \not\in R$, and $R \in A$, then $R \in R$.

We can conclude therefore that $R \not\in A$. And since the set $A$ is completely arbitrary, we conclude that no set contains everything, because no set contains $R$. Therefore, there is no universe.

But, if we suppose that $R$ does exist, and assuming the existence of the empty set, with the Axiom of Pairing, I should be able to create a set in which both $R$ and $\emptyset$, belong to. Let's call this new set $U$. But since we said that $R$ can not be a member of any set, then $R$ can not be a member of $U$.

The only solution to this problem I can think of, would be to restrict the Axiom of Pairing, such that it could be expressed as: For any two sets $X$, $Y$, if $X$ is a member of a set $N$, and $Y$ is a member of a set $M$, then there exists a set $U$, such that both $X$, $Y$ are members of it.

And since $R$ is not a member of any set, it wouldn't be a valid choice for the Axiom of Pairing.

Is the solution correct? Have I maybe completely misunderstood the axiom of Pairing? And even if we define the axiom of Pairing as above, $R$ is not a member of any set, but does it exist? Because if we conclude that it doesn't, we can ignore our proof that the Universe does not exist.

• $R$ exists, but it isn't a set. – vadim123 Jul 7 '15 at 3:06
• How can that be? Is it a class then? – Victor Chavauty Jul 7 '15 at 3:14
• After some research, i've read that the difference of a class and a set is that a Set can be a member of another Set, and a Class can not be a member of another Class. Therefore, $R$ could exist, but it would be a Class, since it is not a member of anything, as showed. But does that mean that we can define the "universal set" as a class? Or would we reach another paradox? – Victor Chavauty Jul 7 '15 at 3:20
• If you have a predicate $\phi(x)$ that is true or false for every set, then $\{x : \phi(x) \}$ is a class (and may or may not also be a set). The "universal set" (really, a proper class) comes from letting $\phi(x)$ be true for all $x$. There's really no question as to whether a class "exists"... if $\phi(x)$ is a well-formed formula in the language of set theory, then the associated class is meaningful, and moreover can be intersected with sets to yield sets. – mjqxxxx Jul 7 '15 at 6:11

The axiom of pairing is fine. That's not the issue. You're confusing between the Russell paradox and a construction of sets which "look like" the class defined in Russell's paradox.

Note the subtle difference here. The classical Russell paradox is that $\{R\mid R\notin R\}$ is not a set. But here we only use restricted comprehension (also known as separation). We are given a set $A$ and we create a new set $R$, and then we argue that it is not possible that $R\in A$.

But nowhere we argued that $R$ is not an element of any set. If you look at $A'=\{R\}$, then you can define $R'$ which may or may not be different from $R$ itself. But you still know with certainty that $R'\notin A$.

(Also I wouldn't say that the fact no set contains all sets means "there is no universe", rather "there is no universal set" is the right choice of words.)

• "There is no universal set" is the right choice of words. I was just quoting what the book actually says. It clearly means "universal set". But i apologize, because i still haven't understood it. You defined $A' = \{R\}$. If we say $A = A'$, then when we analize wether $R \in R$ or not, we would reach a contradiction again, wouldn't we? since $R \in R$ if and only if $R \in A and R \not\in R$ if we assume that $R \in A$, then the requisit would only be: $R \in R$ if and only if $R \not\in R$, reaching the same contradiction. So therefore, we can't make $A = A'$. Why not? – Victor Chavauty Jul 7 '15 at 23:53
• If you say that $A=A'=\{R\}$, you need to justify that. But I reckon that you cannot. Exactly because $R\notin A$, and $R\in A'$. The point is that given any set $X$, you can define a set $Y$ such that it is impossible that $Y\in X$. But $Y$ itself depends on the set $X$. If you change $X$, you get a different $Y$. So in particular the fact that $R\notin A$ does not mean that $R\notin\{\varnothing,R\}$. – Asaf Karagila Jul 7 '15 at 23:56
• I see. That is quite interesting. So, in simpler words, The impossibility here is not the "existence" of $R$ but rather the impossibility to make $A = A'$ ? – Victor Chavauty Jul 8 '15 at 0:06
• Why should all sets be equal? You are given a set $A$, from which you define $R$ and show that $R\notin A$. Therefore any set $A'$ such that $R\in A'$, cannot possibly be equal to $A$. – Asaf Karagila Jul 8 '15 at 0:10
• Of course. I completely understand now. Thank you very much! – Victor Chavauty Jul 8 '15 at 0:16