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This question, in general, has been already asked here.
My question is whether following approach can be applied too.

For any $A$ and $\mathcal P(A)$ we know that $A \in \mathcal P(A)$
If we know that some given set is a power set, seems we are able to choose the "main" element from the power set.
In other words we can define a function $f(\mathcal P(A))=A$
So if $f(\mathcal P(A))= f(\mathcal P(B))$, then $A=B$

Is it a valid approach?

Update

I will try to explain my question better.
Of course it's not a problem to fetch $A$ from $\mathcal P(A)$ by $\cup \mathcal P(A)$. And it's not a problem to prove the statement in many other ways.
But my question is intended to achieve a better understanding of functions. And the specific question about P(A)=P(B) - is no more just an example. So...
Usually when we define some function we describe the algorithm of this function (or assume it's obvious).
But what about defining function without known algorithm?
Instead of telling the function "Return me, please, a union of all elements of $\mathcal P$" I want to tell the function "I know that $\mathcal P$ is a power set and it contains the 'main' element [the set that this $\mathcal P$ was created from]. So, please, dear function, go and fetch this element for me".

I don't see here any contradiction to logic or to common sense. But I wonder is there any contradiction to axioms of sets theory.
Hope, my question is clear now.

Update 2

I think I understood what is my mistake. I still haven't figured out if there is any meaning in my question in principle. But I realized that the example I gave is really bad.
If we have some function that is not one to one it's clear that there is no way do define some sort of inverse function [e.g. if we have $f:\mathbb R\to \mathbb R, f(x)=x^2$ it's not legitimate to define function $g$ (upgraded $f^{-1}$) that will in some way reveal secrets - whether appropriate 4 was product of $f(2)$ or of $f(-2)$]
In our question we know that $P(A)=P(B)$ but we don't know that it's kind of one to one (e.g. may be $A\ne B$), so it's impossible do define an inverse function that will fetch the 'main' element.
Sorry for all the chatter.

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Shimon S
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  • How do you define the "main" element though? – Wojowu Aug 03 '19 at 21:24
  • @Wojowu the "main" element is a set, that the power set was created from. – Shimon S Aug 03 '19 at 21:27
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    So, how about the union of all the elements of $\mathcal P(A)$ – GEdgar Aug 03 '19 at 21:28
  • The nature of the question isn't "Prove $P(A) = P(B) \to A = B$". That is so obvious a claim that it goes without saying. This is the kind of question where they are really asking you to "Prove system X is sufficiently strong enough to show that $P(A) = P(B) \to A = B$" (sometimes disguised in the language of "Show that you know how to use system X to..."). – DanielV Aug 03 '19 at 21:32
  • @GEdgar Of course, you are right. But on this question, what interests me most is the following point: Can we choose the element we need from the set? – Shimon S Aug 03 '19 at 21:35
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    "the "main" element is a set, that the power set was created from." How do you know there is only one possible set? In a way you seem to be thinking of $P$ as a function from class of sets to sets of power sets and you need to prove it is injective. This can be done, but in doing so you will probably need to prove you consjecture as a step toward this. – fleablood Aug 03 '19 at 22:06

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Picking out a "main" element of the power set is possible (though you won't have a full proof until you describe how).

However, it is easier to simply take the union of all the elements of the power set and be done with that: $$ \bigcup \mathcal P(A) = A $$ Do this on both sides of $\mathcal P(A)=\mathcal(B)$, and your're done.

hmakholm left over Monica
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  • On this question, what interests me most is the following point: Can we choose the element we need from the set? – Shimon S Aug 03 '19 at 21:34
  • @ShimonS: Yes -- it is the unique element that is the union of all the elements. – hmakholm left over Monica Aug 03 '19 at 21:35
  • OK. But can I understand from your words, that if it wasn't a union of all elements, we can not choose it? – Shimon S Aug 03 '19 at 21:42
  • @ShimonS: Then you shouldn't choose it because the set you're starting with can't be its power set. – hmakholm left over Monica Aug 03 '19 at 22:07
  • @ShimonS If you want a way to choose the element, as a choice function on the (proper) class of all power sets, then the function $f:P\mapsto \bigcup P$ itself is this choice function. Of course you have to show that $\bigcup P\in P$, which is quite trivial, since $P$ is a power set. – Vsotvep Aug 04 '19 at 07:18