Equal powers sets in ZF theory

• Jun 29th 2012, 08:56 AM
cpcook
Equal powers sets in ZF theory
Is it possible to prove that equal power sets imply equal sets using the ZF axioms?
• Jun 29th 2012, 09:05 AM
Plato
Re: Equal powers sets in ZF theory
Quote:

Originally Posted by cpcook
Is it possible to prove that equal power sets imply equal sets using the ZF axioms?

Suppose that $\mathcal{P}(A)=\mathcal{P}(B)$.
If $\exists x\in A\setminus B$ then $\{x\}\subseteq A$ so $\{x\}\in\mathcal{P}(A)$.
What is wrong with that?
• Jun 29th 2012, 10:51 AM
cpcook
Re: Equal powers sets in ZF theory
I cannot use the subset notation. We can only use the "element of". That is our only operator in the universe we are using.
• Jun 29th 2012, 11:09 AM
Plato
Re: Equal powers sets in ZF theory
Quote:

Originally Posted by cpcook
I cannot use the subset notation. We can only use the "element of". That is our only operator in the universe we are using.

I think that you better tell us more about the course you are taking.
Maybe a textbook or a list of axioms and/or definitions?
• Jun 29th 2012, 02:28 PM
emakarov
Re: Equal powers sets in ZF theory
Quote:

Originally Posted by cpcook
I cannot use the subset notation. We can only use the "element of". That is our only operator in the universe we are using.

Well, yes, Zermelo–Fraenkel set theory has only two predicates: equality and membership. However, $A\subseteq B$ is easily expressible as $\forall x.\,x\in A\to x\in B$. Subset relation is used in the axiom of power set.
• Jun 29th 2012, 03:45 PM
Deveno
Re: Equal powers sets in ZF theory
in this case, i would be tempted to prove the contrapositive: distinct sets give rise to distinct power sets. you can use the same argument as in Plato's post:

if x is in A\B, then {x} is in P(A)\P(B).