Axiom of Regularity needed to show no set has it's power set as a subset? In Jech's book, one of the very first exercises at the end of the first chapter is to show that there is no set $X$ that
$\mathscr{P}(X) \subseteq X$
With the axiom of regularity it's extremely easy, since $X \in \mathscr{P}(X)$ and therefore $X \in X$.
But it's so early he's hardly mentioned that axiom at this point, saving it for a later chapter. So I wonder if I'm missing something and that there's another reason this can't happen?
If there were some such set $X$, then it would have to contain $\varnothing$ and itself, and then also {$\varnothing, X$}, and then {$\varnothing, X,$ {$\varnothing, X$}}, and more in the same way.
It seems pretty clear no finite set could satisfy this, and I have no clue how to show a set like {$\varnothing, X,$ {$\varnothing, X$}$,$ $...$} exists, but I can't see how to show it doesn't exist either (without Regularity).
 A: This seems to call for Cantor's diagonal argument: There is no surjective map $X\to \mathscr P(X)$, but there is a shortcut not even using the concept of map.
Using the Axiom Schema of Separation, we define
$$S=\{\,x\in X\mid x\notin x\,\}.$$
In other words, we have
$$ \forall x\colon(x\in S\leftrightarrow (x\in X\land x\notin x)),$$
hence specifically 
$$ S\in S\leftrightarrow (S\in X\land S\notin S).$$
By mere first order logic (namely: $p\leftrightarrow (q\land \neg p)$ is equivalent to $\neg p\land\neg q$, hence implies $\neg q$) we infer $S\notin X$. Since $S\in\mathscr P(X)$ by definition, this shows $\mathscr P(X)\not\subseteq X$.
A: First of all, you are right it isn't easy to show that the set $S_X= \{\varnothing, X, \{\varnothing, X\}, ...\}$ exists or doesn't exist. In fact taking $X=\{\varnothing\}$ the existence of $S_X$ is the "Axiom of Infinity" and is known to be independent from all other axioms of ZFC. However, the existence of such a set, doesn't give you a counterexample.  Now to answer your question: If $\mathscr{P}(X) \subseteq X$, then there is a surjection $f:X \to \mathscr{P}(X)$ defined by $x \to x$ if $x \in \mathscr{P}(X)$ and $x \to \varnothing$ otherwise. Let $A=\{x\in X: x \notin f(x)\}$. Since $A \in \mathscr{P}(X)$, $f(A) = A$. By definition of $A$ we have $A \in A$ if and only if  $A \notin A$ giving the desired contradiction.
