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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.

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What do you mean by "same"? –  Qiaochu Yuan Sep 19 '11 at 2:37
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I think he means "same" in the sense of the axiom of extensionality. $(\forall x)(x \in A \Leftrightarrow x \in B) \Rightarrow (A = B)$ –  William Sep 19 '11 at 2:41
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"the same" isn't the same, depending on the context! –  The Chaz 2.0 Sep 19 '11 at 2:59

3 Answers 3

up vote 14 down vote accepted

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

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

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

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+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.) –  Srivatsan Sep 19 '11 at 2:44
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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 $\{x\} \notin \mathscr{P}(B)$. –  William Sep 19 '11 at 2:49
    
Thanks, I appreciate your answer and the follow up. –  ae0709 Sep 19 '11 at 3:00
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+1 for "without loss of geneality" ;) –  Chris Taylor Sep 19 '11 at 8:41

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 $A=B$.

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@anon: $\bigcup_{B\subseteq A} B$ can be written as $\bigcup\{B\mid B\subseteq A\}$. –  Asaf Karagila Sep 19 '11 at 5:35
    
Hmm. Never seen the notation, guess I'll keep it in mind. –  anon Sep 19 '11 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 (http://math.stackexchange.com/a/332186/11994). In fact, we can directly prove the stronger version of this question's theorem from that one:

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

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Instead of posting the same answer twice, you can post it once and point out that the question is the same. –  Asaf Karagila Mar 16 '13 at 20:36
    
@AsafKaragila I'm sorry, I did not intend to post the same answer twice. Which two answers are you referring to? –  Marnix Klooster Mar 16 '13 at 20:43
    
While not word for word, it is the same question and the same answer: math.stackexchange.com/a/332186/622 –  Asaf Karagila Mar 16 '13 at 20:45
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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. –  Marnix Klooster Mar 16 '13 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: $A = B \iff P(A) = P(B)$? –  LePressentiment Jan 7 at 6:58

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