Math Help - Another Axiom Question

1. Another Axiom Question

2. Construct the set $Y = \{ x\in X | x\not \in x\}$. Notice that $Y\subseteq X$ therefore $Y\in \mathcal{P}(X)$. However, $Y\not \in X$, to show this, assume to contrary that $Y\in X$. Now what we have is essentially Russel's paradox, because if $Y\in X$ we must have either $Y\in Y$ or $Y\not \in Y$. If $Y\in Y$ then by construction $Y\not \in Y$, a contradiction. If $Y\not \in Y$ then by construction $Y\in Y$ for $Y\in X$ and $Y\not \in Y$, a contradiction. Thus, we must have $Y\not \in X$.