How can this evade Russel's Paradox?
If p is a property, for any X, there exists a set Y = {x E X;P(x)}
If P(x) = x ~E x (a set x cannot contain itself) ^ Y = x
Then
Y E X ^ Y ~E Y
A paradox:
Y E X and Y ~E Y cannot coexist
How can this evade Russel's Paradox?
If p is a property, for any X, there exists a set Y = {x E X;P(x)}
If P(x) = x ~E x (a set x cannot contain itself) ^ Y = x
Then
Y E X ^ Y ~E Y
A paradox:
Y E X and Y ~E Y cannot coexist
If $P(x) \stackrel{\text{def}}{=} x \not\in x$, then since this holds for EVERY set $x$, we have that $Y = \{x \in X: P(x)\} = X$.
Therefore, we have $Y \not\in X$, since this is the same as saying: $X \not\in X$, which is true. There is no paradox, here. $Y$ is a subset of $X$, not an element.