Questions from Enderton elements of Set-theory regarding Unions and Powersets I could not solve these two questions from second chapter of Enderton's set theory.
6.(b) Show that $A \subseteq P(\cup A)$, under what conditions does equality hold?
7.(b) Show that $P(A) \cup P(B) \subseteq P(A\cup B)$, under what conditions does equality hold?
In both of them, I have no problem to show the first part but in the second part (conditions for equality) I can't go any further than suggesting a set of examples for which the equality holds (for example, $\{\emptyset\}$ and $\{\emptyset, \{\emptyset\}\}$ for 6.(b)). I would appreciate any suggestions on how to solve them.
 A: For 6.b:
Notice that $\cup P(B)=B$ and so if $A=P(B)$ for some $B$ we have $P(\cup A)=P(\cup P(B))=P(B)=A$.
If $A\ne P(B)$ for all $B$ then in praticular we have $P(\cup A)\ne A$, otherwise $B=\cup A$ would satisfy $A=P(B)$.
So $P(\cup A)=A$ if and only if there exists $B$ such that $A=P(B)$.
For 7.b:
First notice that if $A\subseteq B$ than $C\subseteq A\implies C\subseteq B$ therefore $P(A)\subseteq P(B)$ hence $P(A)\cup P(B)=P(B)$, and $P(A\cup B)=P(B)$
If we don't have $A\subseteq B$ and $B\subseteq A$, then there exists $a\in A$ and $b\in B$ such that $a\notin B,b\notin A$ therefore $\{a,b\}\notin P(A)\cup P(B)$ but $\{a,b\}\in P(A \cup B)$.
So $P(A) \cup P(B) = P(A\cup B)$ if and only if $A\subseteq B$ or $B\subseteq A$
A: 6.b.  It’s always true that $\emptyset \in \mathscr P(A)$, so for $A=\mathscr P(\cup A)$ to hold, we must have $\cup A=\emptyset$.  But then $\cup A$ has no elements at all, whereas $\mathscr P(\cup A)$ has at least one element; namely, $\emptyset$, so equality doesn’t hold there either.  So equality never happens.
7.b.  Let $x \in \mathscr P(A \cup B) \setminus \mathscr P(A) \cup \mathscr P(B)$.  Then $x$ has elements in $A \setminus B$ and also in $B \setminus A$.  That’s possible unless $A \subseteq B$ or $B \subseteq A$, in which case you have equality.
