TITLE: Prove/Disprove that if two sets have the same power set then they are the same set
QUESTION [27 upvotes]: I am really sure that if two sets have the same power set, then they are the same set. I just am wondering how does one exactly go about proving/showing this?
I'm usually wrong, so if anyone can show me an example where this fails, I'd like that too.
The homework just asks for true/false, but I'm wanting to show it if possible. My thoughts are that since the power set is by definition the set of all subsets of a set, if each of the two power sets are identical, we have an identity map between each set, thus it's indistinguishable which power set is a given set's power set. I hope that wasn't verbose. Since a set has only one power set, we can conclude they are in fact the same set.

REPLY [8 votes]: An alternative way to answer this old question: for all sets A and B,
$$
\begin{array}{ll}
& \mathcal{P}(A) = \mathcal{P}(B) \\
\equiv & \;\;\;\text{"extensionality"} \\
& \langle \forall V :: V \in \mathcal{P}(A) \equiv V \in \mathcal{P}(B) \rangle \\
\equiv & \;\;\;\text{"definition of $\mathcal{P}$, twice"} \\
& \langle \forall V :: V \subseteq A \equiv V \subseteq B \rangle \\
\Rightarrow & \;\;\;\text{"choose $V:=A$, and choose $V:=B$"} \\
& (A \subseteq A \equiv A \subseteq B) \;\land\; (B \subseteq A \equiv B \subseteq B) \\
\equiv & \;\;\;\text{"$\subseteq$ is reflexive, so $A \subseteq A$ and $B \subseteq B$"} \\
& A \subseteq B \land B \subseteq A \\
\equiv & \;\;\;\text{"definition of set equality"} \\
& A = B \\
\end{array}
$$
Update: As a comment rightly points out, the above proof is very similar to my answer to another question.  In fact, we can directly prove the stronger version of this question's theorem from that one:
$$
\begin{array}{ll}
& \mathcal{P}(A) = \mathcal{P}(B) \;\equiv\; A = B \\
\equiv & \;\;\;\text{"double inclusion, twice"} \\
& \mathcal{P}(A) \subseteq \mathcal{P}(B) \land \mathcal{P}(B) \subseteq \mathcal{P}(A) \;\equiv\; A \subseteq B \land B \subseteq A \\
\Leftarrow & \;\;\;\text{"logic"} \\
& (\mathcal{P}(A) \subseteq \mathcal{P}(B) \;\equiv\; A \subseteq B) \;\land\; (\mathcal{P}(B) \subseteq \mathcal{P}(A) \;\equiv\; B \subseteq A) \\
\equiv & \;\;\;\text{"the other theorem, twice"} \\
& \text{true} \\
\end{array}
$$