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Math Help - another question

  1. #1
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    another question

    Can you conclude that A = B if A and B are two sets with the same power set?
    I think it cannot conclude that A = B because the same power set does not mean that they have the same set or same order. Am i right?
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  2. #2
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    Given that \mathcal{P}(A)= \mathcal{P}(B) that means that C \subseteq A \Leftrightarrow C \subseteq B.
    Now just stop and think carefully about the implications of that statement.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Given that \mathcal{P}(A)= \mathcal{P}(B) that means that C \subseteq A \Leftrightarrow C \subseteq B.
    Now just stop and think carefully about the implications of that statement.
    A = {1,2,3,4,5,6,7,8}
    B = {0,2,4,6,8,10}
    C = {2,4,6}

    if C \subseteq A \Leftrightarrow C \subseteq B, \mathcal{P}(A)= \mathcal{P}(B) ?

    And if A and B has the same power set. For eg:
    A ={a1,a2}
    B = {b1,b2}
    How could A and B the same?
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  4. #4
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    Quote Originally Posted by zpwnchen View Post
    A = {1,2,3,4,5,6,7,8}
    B = {0,2,4,6,8,10}
    if C \subseteq A \Leftrightarrow C \subseteq B, \mathcal{P}(A)= \mathcal{P}(B) ?
    And if A and B has the same power set.
    How could A and B the same?
    I got to be honest with you: I donít think that you have caught on to any of this material.
    The above is no example at all.
    Using  A = \{1,2,3,4,5,6,7,8\}~\&~B = \{0,2,4,6,8,10\}, then it is clear to anyone who understands the definitions that \{1,3,5\}\in\mathcal{P}(A)\text{ BUT }\{1,3,5\} \notin\mathcal{P}(B) .
    Therefore, \mathcal{P}(A)\not= \mathcal{P}(B) which in contrary to the given.

    So start with \mathcal{P}(A)= \mathcal{P}(B) .
    We are trying to show that A \subseteq B\;\& \;B \subseteq A
    Letís do it.
    x\in A\Rightarrow \{x\}\in\mathcal{P}(A) \Rightarrow \{x\}\in \mathcal{P}(B) \Rightarrow x\in B . That shows that A \subseteq B.
    Similarly show that B \subseteq A.
    Together they mean A =B
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  5. #5
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    Why it's not the example at all?
    <br />
A = \{1,2,3,4,5,6,7,8\}~\&~B = \{0,2,4,6,8,10\}<br />
    <br />
\{2,4,6\}\in\mathcal{P}(A) \text   \&  \{2,4,6\} \in\mathcal{P}(B)<br />
    Therefore: <br />
\mathcal{P}(A) = \mathcal{P}(B)<br />
    What is wrong with it? Please.. the more I know, I more i'm confused. yeh i'm getting lost

    I understand the second part A=B. Thanks
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  6. #6
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    Quote Originally Posted by zpwnchen View Post
    <br />
A = \{1,2,3,4,5,6,7,8\}~\&~B = \{0,2,4,6,8,10\}<br />
    <br />
\{2,4,6\}\in\mathcal{P}(A) \text   \&  \{2,4,6\} \in\mathcal{P}(B)<br />
    Therefore: \color{red}<br />
\mathcal{P}(A) = \mathcal{P}(B)
    Explain why you think that is true.
    Because if you understand any of this, then you know it is completely false.
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  7. #7
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    It is obvious that it's completely false.

    But according to what you said here:

    Quote Originally Posted by Plato View Post
    Given that \mathcal{P}(A)= \mathcal{P}(B) that means that C \subseteq A \Leftrightarrow C \subseteq B.
    <br />
\{2,4,6\} \subseteq (A) \text \& \{2,4,6\} \subseteq (B)<br />
Therefore, \mathcal{P}(A)= \mathcal{P}(B) ?
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  8. #8
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    \mathcal{P}(A)= \mathcal{P}(B) that means that C \subseteq A \Leftrightarrow C \subseteq B
    That means any subset of A is a subset of B and any subset of B is a subset of A.
    C is a subset of A if and only if C is a subset of B
    That means that every subset of A is also a subset of B and visa versa.
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  9. #9
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    <br />
A = \{1,2,3,4,5,6,7,8\}~\&~B = \{0,2,4,6,8,10\}<br />

    Why cant we say that?
    <br />
\{2,4,6,8\} \subseteq (A) \Leftrightarrow \{2,4,6,8\} \subseteq (B)<br />
which means C is a subset of A if and only if C is a subset of B.
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