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.

Recall the definitions for a set to be:
 * ordinary it is not an element of itself
 * extraordinary it is an element of itself.

Thus, $\map P S = S \in S$ is a property on which we can use the to build the set $T$ of all extraordinary:


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

which is the set of all sets which contain themselves.

Or we can apply the to build the set $T$ of all ordinary sets:


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

($R$ 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.