The Barber

Form now the assemblage of all classes which are not members of themselves. This is a class; is it a member of itself or not?
- Bertrand Russell, Introduction to Mathematical Philosophy (1919)

Does a barber in a certain town who shaves all and only those who do not shave themselves, shave himself?

Russell's famous Barber Paradox was very important to the subject of set theory. It shows that it is possible to derive a contradiction using just the ideas of a set and being a member of a set.

There is a barber in the town of Seville who shaves all and only those men who do not shave themselves. The question arises-- does the barber shave himself or not? If he does shave himself, then, since he shaves only those who do not shave themselves, then he does not shave himself. And if he does not shave himself, then since he shaves all those who do not shave themselves, then he does shave himself. That is, if he does, he doesn't, and if he doesn't, he does!

This same puzzle can be put in the language of sets. Consider the set of all edible things. This set is itself edible, and so the set of all edible things is a member of itself. Now consider the set of all cats. This set is not a cat, and so the set of all cats is not a member of itself. Now consider the set of all sets that are not members of themselves. This set would include the set of cats, the set of trees, and so on. But does this set include itself? If it does, it doesn't, and if it doesn't, it does.

Russell's solution to this paradox is simply to deny that sets can be members of themselves. There are cats, trees, and sets, but sets are a different type of thing from cats, and you shouldn't talk about both the same way. Russell developed a whole theory, called the Theory of Types, to account for this.

How does Russell'$ solution to this puzzle about sets apply to the Barber Paradox? Can you come up with a better solution to either problem?

Bibliography

Russell, Bertrand, Introduction to Mathematical Philosophy, 1919.