# If $\mathcal{P}(A)=\mathcal{P}(B)$, then $A=B$? [duplicate]

Prove, disprove, or give a counterexample:

If $\mathcal{P}(A)=\mathcal{P}(B)$, then $A=B$.

Assume $\mathcal{P}(A)=\mathcal{P}(B)$. Since we know $A \subseteq A$, we know $A \in \mathcal{P}(A)$.

Since $\mathcal{P}(A)=\mathcal{P}(B)$, we know that $A \in \mathcal{P}(B)$.

Therefore, $A \subseteq B$ and $A=B$.

Is this proof okay?

Edit: I should note this isn't probability, $\mathcal{P}$ is the power set.

• It's missing the $B \subseteq A$ bit, but the idea is right. Commented Mar 15, 2014 at 19:52

You can prove: $$\bigcup \mathcal{P}(A)=A$$$$\bigcup\mathcal{P}(B)=B$$ and by hypothesis you have $\mathcal{P}(A)=\mathcal{P}(B)$ therefore $$B=\bigcup\mathcal{P}(B)=\bigcup\mathcal{P}(A)=A$$

Everything in your answer is correct until the implication "Therefore $A \subseteq B$ and $A=B$". The second implication (i.e. that $A=B$ is not yet justified). So keep only the first, i.e. that $$A \subseteq B \tag 1$$ Reason in exactly the same way to deduce that $$B \subseteq A \tag2$$ Now combining $(1)$ and $(2)$ yields the result $$A=B$$

The proof is almost okay. You need to argue why $B\subseteq A$. But it follows from the same argument.

• I should use the same logic that showed A$\subseteq$B to show B$\subseteq$A? And thus, A=B? Commented Mar 15, 2014 at 20:03
• @CharlesNosbig Just say something like, "By symmetry, $B \subseteq A$ as well." Commented Mar 15, 2014 at 20:11
• @Charles: Yes, you should. Commented Mar 15, 2014 at 20:11

$A\in \mathcal P (A)\rightarrow A \in \mathcal P (B)\rightarrow A\subset B$

$B\in \mathcal P (B)\rightarrow B \in \mathcal P(A)\rightarrow B\subset A$

Using the conclusions of each line $A=B$

If we know $A$ we can determine unequivocally $\mathcal{P}(A)$ and if we know $\mathcal{P}(A)$ we can determine unequivocally A, as element of $\mathcal{P}(A)$ which includes any subset , so if $A=B$ then $\mathcal{P}(A)=\mathcal{P}(B)$ and if $A\neq B$ then $\mathcal{P}(A)\neq\mathcal{P}(B)$ what completes the proof of your theorem.

Another way of proving: $$(\mathcal{P}(A)=\mathcal{P}(B))\implies [(A\in\mathcal{P}(B)) \wedge (B\in\mathcal{P}(A))]\implies [(A \subseteq B) \wedge (B \subseteq A)]\\ [(A \subseteq B) \wedge (B \subseteq A)] \implies [(A\cup B=B) \wedge (B\cup A=A)]\implies (A=B)$$

PS. -1 as rating of my answer is more against reputation of math.stackexchange.com , than mine.

• This construction only works for finite sets. We should take the element that is maximal under inclusion. Commented Mar 15, 2014 at 20:18
• Injection, not bijection. Commented Mar 15, 2014 at 20:25
• ... and exactly the assertation that $A\mapsto P(A)$ (not $P(A)$ which is a set, not a function) is an injection is what needs to be proved here. Commented Mar 15, 2014 at 20:26
• @Darius You were using the term “bijection”. Commented Mar 15, 2014 at 21:20
• Whoa calm down there. Nobody said anything about persecuting or sanctioning. Commented Mar 16, 2014 at 16:50