Question: This exercise presents Russell's paradox. Let S be the set that contains a set x if the set x does not belong to itself, so that S = {x | x /∈ x}.
a) Show the assumption that S is a member of S leads to a contradiction.
b) Show the assumption that S is not a member of S leads to a contradiction. By parts (a) and (b) it follows that the set S cannot be de- fined as it was. This paradox can be avoided by restricting the types of elements that sets can have.