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I need help proving that if ${A}\sim{B}$ then ${P(A)}\sim{P(B)}$

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You asked several questions in the past day and a half, not specifying their origin I would have to assume either homework or preparing for some exam. I should tell you something about mathematics in general and set theory in particular. It is very important to try and tackle things on your own and not come and ask for help as soon as you get stuck, sure it's convenient and easy to try once or twice and then ask here. You don't really learn to chew problems if you haven't spent long hours chewing them on your own first. The one good advice I can give you in attempting problems (...) –  Asaf Karagila Apr 11 '12 at 10:06
    
(...) is to have the definitions of everything open in front of you. What is the definition of $A\sim B$, what is the definition of $P(A)$, etc. from the definitions everything follows in mathematics, and in intro-level it is usually straightforward and does not require complicated tricks and all that. –  Asaf Karagila Apr 11 '12 at 10:08
    
@Asaf thank you for your advice. Personally I do believe in what you say. –  JanosAudron Apr 11 '12 at 11:16
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Let $f:A\to B$ be a bijection. Define $F:\wp(A)\to\wp(B):S\mapsto\{f(s):s\in S\}$. Prove that $F$ is a bijection.

Added: It may be useful to suggest how you might come up with this idea. You could look at some small examples. For instance, $A=\{1,2,3\}\sim\{a,b,c\}=B$, and you could write down a bijection from $A$ to $B$, say

$$\begin{align*} &1\mapsto a\\ &2\mapsto b\\ &3\mapsto c\;. \end{align*}$$

Now look at a ‘typical’ member of $\wp(A)$, perhaps $\{1,3\}$. You want to match it up with a subset of $B$. It seems to me that there’s just one natural candidate, namely $\{a,c\}$:

$$\begin{align*} &\color{red}{1\mapsto a}\\ &2\mapsto b\\ &\color{red}{3\mapsto c}\;. \end{align*}$$

Explore this idea just a little, and you should quickly see that it works in general.

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