Commutativity of $\bigcup$ and $\wp$ In his Naive Set Theory, Halmos in Section 5, Complements and Powers, asks the following.

Show that $E$ is always equal to $\bigcup_{X\in\wp (E)} X$ (that is $E=\bigcup\wp (E)$), but that the result of applying $\wp$ and $\bigcup$ to $E$ in the other order is a set that includes $E$ as a subset, typically a proper subset.

I’ve been able to show the first part. For the second part, I don’t think that $E$ can be a subset of $\wp\bigl(\bigcup E\bigr)$.

$\wp (X)$ means power set of $X$.

A simple illustration of the second fact: let $E=\{\{\emptyset\}\}$, then $\bigcup E = \{\emptyset\}$ (all elements of elements of $E$ together) and $\mathscr{P}\left(\bigcup E)\right) = \{\emptyset, \{\emptyset\}\}$ which indeed properly contains $E$ as a subset.
That $E \subseteq \mathscr{P}(\bigcup E)$ is clear: 
suppose that $x \in E$. Then $x \subseteq \bigcup E$ (because every $y \in x$ will be in the union $\bigcup E$ by definition) and so $x \in \mathscr{P}\left(\bigcup E\right)$ (a subset of the union is an element of the powerset of the union).
As Henning Makholm pointed out, iff $E$ is a powerset of some set, we will have $E= \mathscr{P}\left(\bigcup E\right)$. 
