# Math Help - Doubt on the Axioma of Separation (Russel's Paradox)

1. ## Doubt on the Axioma of Separation (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

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.