1. ## Sets proof

I need some help with this proof.It says

prove that A = B if and only if P(A) = P(B). (where A and B are sets)

I don't know how to go about it.

2. Suppose that $\displaystyle A = B$
Then $\displaystyle \begin{array}{lcr} {C \subseteq A} & \Leftrightarrow & {C \subseteq B} \\ {P(A)} & = & {P(B)} \\ \end{array}$.

Now suppose that $\displaystyle P(A) = P(B)$.
$\displaystyle \begin{array}{rcl} {\forall x \in A} & \Leftrightarrow & {\left\{ x \right\} \subseteq A} \\ {} & \Leftrightarrow & {\left\{ x \right\} \in P(A)} \\ {} & \Leftrightarrow & {\left\{ x \right\} \in P(B)} \\ {} & \Leftrightarrow & {x \in B} \\ \end{array}$

3. Now suppose that $\displaystyle P(A) = P(B)$.
$\displaystyle \begin{array}{rcl} {\forall x \in A} & \Leftrightarrow & {\left\{ x \right\} \subseteq A} \\ {} & \Leftrightarrow & {\left\{ x \right\} \in P(A)} \\ {} & \Leftrightarrow & {\left\{ x \right\} \in P(B)} \\ {} & \Leftrightarrow & {x \in B} \\ \end{array}$[/QUOTE]

Thanks for helping me out.I still have a question.I don't understand this part.What's ure logic behind it.Can you be more explicit because I'm new with sets.

Thanks

4. Originally Posted by NidhiS
Originally Posted by Plato
Now suppose that $\displaystyle P(A) = P(B)$.
$\displaystyle \begin{array}{rcl} {\forall x \in A} & \Leftrightarrow & {\left\{ x \right\} \subseteq A} \\ {} & \Leftrightarrow & {\left\{ x \right\} \in P(A)} \\ {} & \Leftrightarrow & {\left\{ x \right\} \in P(B)} \\ {} & \Leftrightarrow & {x \in B} \\ \end{array}$
Thanks for helping me out.I still have a question.I don't understand this part.What's ure logic behind it.Can you be more explicit because I'm new with sets.

Thanks
he showed that $\displaystyle x \in A \Longleftrightarrow x \in B$, which implies $\displaystyle A = B$.

so taking any $\displaystyle x \in A$, note that $\displaystyle \{ x \}$ is a subset of A and is hence in its power set. but that is the same as the power set of B, by our assumption, so that $\displaystyle \{ x \} \in \mathcal{P}(B)$. but that only happens if $\displaystyle x \in B$