Showing P(A) not a subset of A, for all A Given only the following axioms:
-Separation (if A is set, and B a partial class of A, then B is a set)
-Pairing (the class {a,b} is a set)
-Union (∀ A a set, ∪A is a set)
-Infinity ({x | x ∈ ℕ} is a set)
Prove that ∀ A a set, P(A) = {B | B ⊆ A} ⊈ A
The question hints to use Russell’s paradox
To try by contradiction means:
∃ A a set such that P(A) ⊆ A 
then A ∈ A
which doesn’t lead to contradictions using those axioms.
Thanks for the help,

Suppose that P(A) is a subset of A. Then every subset of A is an element of A.
Let B be the set of elements of A which do not contain themselves. Since $B$ is a subset of $A$, it is an element of $A$, by assumption. Then $B$ contains itself if and only if it does not contain itself; contradiction.
