Russell's paradox and the barber example

Question

In his reply, Peter Smith provided a clear explanation of the barber's paradox. However, I am having some hard time to understand the link between the established theorem and Russell's paradox. In other words, I do not know how to make the sets U and X fit into the established theorem:$$\neg\exists x\forall y(Rxy \leftrightarrow \neg Ryy)$$

I will be happy to have some help on that.

Answer 1

interpret $Rxy$ as '$x$ contains $y$', I,e '$y$ is an element of $x$'

So where does the $U$ come in?

Well, if there was a universal set of all sets, then we could define a set $X$ as follows:

$$X = \{ Y \in U | Y \not \in Y \}$$

Think of $X$ as 'the set of all normal sets' where a 'normal' set is one that doesn' t contain itself as one of its elements.

But of course since $X$ is a set, and since $U$ is supposed to be the set of all sets, we have $X \in U$, and thus it would be true that:

$$\exists X \in U \ \forall Y \in U (Y \in X \leftrightarrow Y \not \in Y)$$

But since that leads to a contradiction, there cannot be a universal set.

Answer 2

Russell's Paradox itself had nothing to do with the so-called universal set $U.$ It arose from the supposed existence of a set $X$ such that $\forall y: [y\in X \iff y\notin y]$ as required by an early attempt to axiomatize set theory by G. Frege. There is no need to refer to $U$. Specifying $y=X$, as we are permitted by the rules of logic, we obtain the contradiction $X\in X \iff X\notin X.$ It was a genuine paradox since it resulted from faulty axioms. Frege's axioms said that such an $X$ should exist, but its existence results in a logical contradiction, therefore it cannot exist. This is not the case with the Barber "Paradox." It is just a theorem of the rules of logic: $\neg \exists b:\forall y: [bSy \iff \neg ySy].$ It has nothing to do with any sets.