Russell's Paradox/Proof 1

Proof
Sets have elements.

Some of those elements may themselves be sets.

So, given two sets $S$ and $T$, we can ask the question: Is $S$ an element of $T$? The answer will either be yes or no.

In particular, given any set $S$, we can ask the question: Is $S$ an element of $S$? Again, the answer will either be yes or no.

Thus, $\map P S = S \in S$ is a property on which we can use the comprehension principle to build this set:


 * $T = \set {S: S \in S}$

which is the set of all sets which contain themselves.

Or we can apply the comprehension principle to build this set:


 * $R = \set {S: S \notin S}$

($R$ for, of course.)

We ask the question: Is $R$ itself an element of $R$?

There are two possible answers: yes or no.

If $R \in R$, then $R$ must satisfy the property that $R \notin R$, so from that contradiction we know that $R \in R$ does not hold.

So the only other answer, $R \notin R$, must hold instead. But now we see that $R$ satisfies the conditions of the property that $R \in R$, so we can see that $R \notin R$ does not hold either.

Thus we have generated a contradiction from the comprehension principle.