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.
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} $
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
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 $\displaystyle x \in B$