Russell's paradox: a set cannot contain its own powerset I've been trying to challenge myself but with no luck. Maybe one of you will have a better idea.
How can it be proven that there can't be a set which contains its own powerset, using only russell's paradox?
I've managed to prove it with other principles or axioms, such as that a set can not belong to itself, or that a powerset cannot belong to itself. 
 A: It is more complicated than necessary.
Let $A$ be a set with $P(A)\subseteq A$. Then, let $B = \{ S \in A \mid S\notin S \}$. That is in fact a set by axiom schema of specification.
Now, notice that from $B \subseteq A$ follows $B\in P(A) \subseteq A$.
If $B\in B$, then we have $B\notin B$. If $B\notin B$, then we have $B\in B$. Just like in the Russell's Paradox.
Note:
For sake of completeness, here is the contradiction to regularity:
$$ A\in P(A) \subseteq A. $$
A: By definition a power set is generated by subsets of the given set. So you can make a cardinality argument.
A: (EDIT: Borrowing from Noah)
Suppose to the contrary that for set $A$, we have $P(A)\subset A$. 
Now, by an axiom of set theory, there exists a subset $B\subset A$ such that $\forall a:[a\in B\iff a\in A \land a\notin a]$. 
Since $B\subset A$ and $P(a) \subset A$, we must have $B\in A$.
Along the lines of RP, we obtain the contradiction $B\in B\iff B\notin B$, thus negating our initial assumption.
See my machine-verified formal proof (updated).

Perhaps more intuitively satisfying  (without $a\notin a$ and such) would be:
If   $X \subset Y$ then there exists the obvious injection $f:X \to Y$, i.e. $f(x)=x$. And, for any set $S$, there cannot exist an injection $f :P(S)\to S$. So, we cannot have $P(S)\subset S$.
