Russell's Paradox/Proof 1
Theorem
The Axiom of Abstraction leads to a contradiction.
Proof
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 if and only if it is not an element of itself
- extraordinary if and only if it is an element of itself.
Thus, $\map P S = S \in S$ is a property on which we can use the Axiom of Abstraction 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 Axiom of Abstraction to build the set $T$ of all ordinary sets:
- $R = \set {S: S \notin S}$
($R$ for Russell, 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 Axiom of Abstraction.
$\blacksquare$
Sources
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