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Math Help - [SOLVED] Proof involing power sets

  1. #1
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    [SOLVED] Proof involing power sets

    Is it okay?

    For sets A,B:
    Prove that P(power set)(A) U P(B) is contained in P(AUB)

    I wrote:

    Let S be a subset and let S belong to P(AUB)
    Then S belongs to P(AUB) iff S is contained in (AUB)
    and S is contained in (AUB) iff S is contained in A and S is contained in B
    iff S belongs to the power set of A and the power set of B, which is equivalent to S belongs to P(A)U P(B)

    Is it okay?
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    For sets A,B:
    Prove that P(power set)(A) U P(B) is contained in P(AUB)
    Then S belongs to P(AUB) iff S is contained in (AUB)
    Almost all of what you wrote is not correct.
    This is an easy proof.
    First, C \subseteq D\; \Rightarrow \;P(C) \subseteq P(D)

    Second, for any set X it is true that A \subseteq A \cup X which means P\left( A \right) \subseteq P\left( {A \cup X} \right).
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  3. #3
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    Let me try this then, which seems too easy.

    Let X be a set and let X be contained in A and also let X be contained in B.

    So P(X) is contained in P(A) And in P(B) so P(X) is contained in P(A)UP(B).

    So X is contained in (AUB), then P(X) is contained in P(AUB)

    Seems wrong, but I don't know.
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  4. #4
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    Here is the complete proof.
    \begin{gathered}<br />
  A \subseteq A \cup B\; \Rightarrow \;P(A) \subseteq P(A \cup B) \hfill \\<br />
  B \subseteq A \cup B\; \Rightarrow \;P(B) \subseteq P(A \cup B) \hfill \\<br />
  P(A) \cup P(B) \subseteq P(A \cup B) \hfill \\ <br />
\end{gathered}
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  5. #5
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    Wow....that's so incredibly simple that I would never have thought of it. Thank you.

    If you have some time, would you mind checking the proof I have here at the last post as well:

    http://www.mathhelpforum.com/math-he...low-level.html
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