Sets proof

NidhiS · #1 $Old$ November 16th, 2008, 04:42 PM

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.

Plato · #2 $Old$ November 16th, 2008, 05:39 PM

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

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

NidhiS · #3 $Old$ November 16th, 2008, 05:58 PM

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

Jhevon · #4 $Old$ November 16th, 2008, 06:03 PM

Quote:

Originally Posted by NidhiS $View Post$

Quote:

Originally Posted by Plato

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

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 $x \in A \Longleftrightarrow x \in B$ , which implies $A = B$ .

so taking any $x \in A$ , note that $\{ x \}$ 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 $\{ x \} \in \mathcal{P}(B)$ . but that only happens if $x \in B$