How is a set subset of its power set? This question is from S C Kleene's Introduction to Metamathematics, page 38: 

If we prescribe as admissible elements of sets (a) $\varnothing$ and (b) arbitrary sets whose members are admissible elements, so that sets have only sets as members, then when $M$ is the set of all sets, $ P(M)=M$.
  (Here $P(X)$ is power set of X.)

I can see why $P(M)\subset M$, but not $M\subset P(M)$.
 A: Since sets have only sets as members, every set is a set of sets.
A: A set $x$ is in $\mathcal{P}(M)$ if and only if all of the elements of $x$ are in $M$. Every set is in $M$, so for any $x$, all elements of $x$ are in $M$. Thus, every $x$ is in $\mathcal{P}(M)$, so $\mathcal{P}(M)$ is the set of all sets, $M$.
A: It may help to see what happens when we ignore Kleene's requirement that sets only contain admissible elements as members.
Suppose that sets can contain things that are not themselves sets (like, say, a banana). The set $\{\mathrm{banana}\}$, containing a single banana is an element of $M$, since it is a set, but $\mathrm{banana} \notin M$, since a banana is not a set. But then $\{\mathrm{banana}\} \nsubseteq M$, since $\mathrm{banana} \in \{\mathrm{banana}\}$, but $\mathrm{banana} \notin M$. This means that $\{\mathrm{banana}\} \notin \mathcal{P}(M)$, even though $\{\mathrm{banana}\} \in M$, showing that $M \nsubseteq \mathcal{P}(M)$.
But if only sets can ever be members of sets, then we can never do anything like the above, and indeed $M \subseteq \mathcal{P}(M)$. We can show this directly. Take any set $A \in M$, and we will show $A \in \mathcal{P}(M)$. This is the same as showing $A \subseteq M$, which just means that every member of $A$ is a member of $M$. Since every member of $A$ is a set by the requirement that sets contain only sets (and $A$ is a set, of course), every member of $A$ is a member of $M$, since $M$ contains all sets as members. This shows $A \subseteq M$, which is what was wanted.
