# Prove/Disprove that if two sets have the same power set then they are the same set

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.

• What do you mean by "same"? Commented Sep 19, 2011 at 2:37
• I think he means "same" in the sense of the axiom of extensionality. $(\forall x)(x \in A \Leftrightarrow x \in B) \Rightarrow (A = B)$ Commented Sep 19, 2011 at 2:41
• "the same" isn't the same, depending on the context! Commented Sep 19, 2011 at 2:59

Suppose $$A \neq B$$. Without loss of generality, there exists an $$x \in A$$ such that $$x \notin B$$. Then $$\{x\} \in \mathscr{P}(A)$$ whereas $$\{x\} \notin \mathscr{P}(B)$$. Thus $$\mathscr{P}(A) \neq \mathscr{P}(B)$$.

Conversely, if $$\mathscr{P}(A) = \mathscr{P}(B)$$, then all their singletons are the same. Thus $$A = B$$.

$$A = B$$ if and only if $$\mathscr{P}(A) = \mathscr{P}(B)$$.

• +1 Is the statement that $\{ x \} \notin \mathscr P(B)$ evident or does it need proof? (It seems "clearly" true, but I do not know what to say if someone asks me to justify it.) Commented Sep 19, 2011 at 2:44
• I would just use the definition of subset. $D \subset E$ if and only if $(\forall n)(n \in D \Rightarrow n \in E)$. So by assumption $x \notin B$ and $x \in A$. So we have there exists $n$ (in particular that $x$) such that $n \in \{x\}$ and $n \notin B$. Thus I have proved $(\exists n)(n \in \{x\} \wedge n \notin B) = \neg((\forall n)(n \in \{x\} \Rightarrow n \in B))$. Thus $\neg(\{x\} \subset B)$ and hence $\{x\} \notin \mathscr{P}(B)$. Commented Sep 19, 2011 at 2:49
• "Conversely"...but you have shown the same direction twice. Commented Sep 20, 2016 at 21:56

To add on William's answer with a positive proof, first one has to note the following observation:

$$A=\bigcup\{B\mid B\subseteq A\}$$

To prove this, the inclusion $A\subseteq\bigcup\{B\mid B\subseteq A\}$ is trivial since $A\subseteq A$, so we take $A$ into the union. In the other direction, since every $B$ in the union is a subset of $A$ the union is a subset of $A$.

Now we can proceed. The above identity can be written in terms of the power set as $A=\bigcup\mathcal P(A)$.

Assume $\mathcal P(A)=\mathcal P(B)$, therefore $\bigcup\mathcal P(A)=\bigcup\mathcal P(B)$, therefore $A=B$.

• @anon: $\bigcup_{B\subseteq A} B$ can be written as $\bigcup\{B\mid B\subseteq A\}$. Commented Sep 19, 2011 at 5:35
• Hmm. Never seen the notation, guess I'll keep it in mind.
– anon
Commented Sep 19, 2011 at 5:45

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:

$$\begin{array}{ll} & \mathcal{P}(A) = \mathcal{P}(B) \;\equiv\; A = B \\ \equiv & \;\;\;\text{"double inclusion, twice"} \\ & \mathcal{P}(A) \subseteq \mathcal{P}(B) \land \mathcal{P}(B) \subseteq \mathcal{P}(A) \;\equiv\; A \subseteq B \land B \subseteq A \\ \Leftarrow & \;\;\;\text{"logic"} \\ & (\mathcal{P}(A) \subseteq \mathcal{P}(B) \;\equiv\; A \subseteq B) \;\land\; (\mathcal{P}(B) \subseteq \mathcal{P}(A) \;\equiv\; B \subseteq A) \\ \equiv & \;\;\;\text{"the other theorem, twice"} \\ & \text{true} \\ \end{array}$$

• Instead of posting the same answer twice, you can post it once and point out that the question is the same. Commented Mar 16, 2013 at 20:36
• @AsafKaragila I'm sorry, I did not intend to post the same answer twice. Which two answers are you referring to? Commented Mar 16, 2013 at 20:43
• While not word for word, it is the same question and the same answer: math.stackexchange.com/a/332186/622 Commented Mar 16, 2013 at 20:45
• The questions are actually slightly different: this one is about $\mathcal{P}(A) = \mathcal{P}(B) \Rightarrow A = B$, while the other is about $\mathcal{P}(A) \subseteq \mathcal{P}(B) \equiv A \subseteq B$. But you are of course correct that these two theorems are very much related: (a stronger version of) the former follows directly from the latter using extensionality. Commented Mar 16, 2013 at 20:54
• @MarnixKlooster: +1. I'm very grateful that you uncloak all the detail and steps. I made two ancillary edits for layout but please feel free to revert them. Just a question: How is the update "a stronger version of this theorem"? In this answer, you're only proving in two ways the same result: $A = B \iff P(A) = P(B)$?
– user53259
Commented Jan 7, 2014 at 6:58