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Math Help - Sets proof

  1. #1
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    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.
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  2. #2
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    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}<br />
   {\forall x \in A} &  \Leftrightarrow  & {\left\{ x \right\} \subseteq A}  \\<br />
   {} &  \Leftrightarrow  & {\left\{ x \right\} \in P(A)}  \\<br />
   {} &  \Leftrightarrow  & {\left\{ x \right\} \in P(B)}  \\<br />
   {} &  \Leftrightarrow  & {x \in B}  \\ \end{array}
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  3. #3
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    Now suppose that P(A) = P(B).
    \begin{array}{rcl}<br />
   {\forall x \in A} &  \Leftrightarrow  & {\left\{ x \right\} \subseteq A}  \\<br />
   {} &  \Leftrightarrow  & {\left\{ x \right\} \in P(A)}  \\<br />
   {} &  \Leftrightarrow  & {\left\{ x \right\} \in P(B)}  \\<br />
   {} &  \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
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NidhiS View Post
    Quote Originally Posted by Plato
    Now suppose that P(A) = P(B).
    \begin{array}{rcl}<br />
   {\forall x \in A} &  \Leftrightarrow  & {\left\{ x \right\} \subseteq A}  \\<br />
   {} &  \Leftrightarrow  & {\left\{ x \right\} \in P(A)}  \\<br />
   {} &  \Leftrightarrow  & {\left\{ x \right\} \in P(B)}  \\<br />
   {} &  \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
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