Russell's Paradox is a discovery by Bertrand Russell that the set theory of Gottlob Frege implies a contradiction.

The problem begins with the distinction between sets that are members of themselves, and sets that are not members of themselves. For example, we may have the set S = { 1, 2, 3 }, such that S∉S, or the set T = { 1, 2, 3, T }, in which case T∈T.

Now, consider the case of absolute complements, where the complement of set S, ∁(S), is the set of all things that are not S. Let's assume that the domain is all sets, and that S is the set of all sets that are not members of themselves.

Therefore, ∁(S) is the set of all sets that are not members of themselves. The paradox arises when you ask whether ∁(S) is a member of S or ∁(S). If we count ∁(S) as a member of itself, then in order to be a member of itself it must not be a member of itself.

However, if we count ∁(S) as not a member of itself, then it must be a member of the set S, and therefore it must be a member of itself, since S is the set of all sets that are members of themselves.

Frege's logic carries within it a Comprehension axiom, which states that for every predicate Px, where x is a free variable, there is a set of all things which satisfy the predicate Px. For example, if P is “is a philosopher”, then there is a set of all things that are philosophers.

Based on the Comprehension axiom, we can suppose that there is a set of all things that are not members of themselves. This set would be defined as: S = {x|xx}, in the domain of all sets, which simply states that S is the set of every set x such that x is not a member of itself.

Consider, then, is S∈S (is S a member of itself)?

Assume that yes, S∈S. Then, S does not satisfy the condition xx, and therefore, S∉S. Therefore (S∈S → S∉S)

Assume that no, S∉S. Then, then S does indeed satisfy the condition xx, and therefore S∈S. Therefore (S∉S → S∈S)

So, as a result we now have (S∈S ↔ S∉S).

Note that S∉S is equivalent to ¬(S∈S). The law of the excluded therefore middle gives us (S∈S ∨ S∉S).

Finally, we can conclude (S∈S ∧ S∉S), which is a contradiction.