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Math Help - Sets

  1. #1
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    Sets

    Let P(S) be denoted as the power set of S and,

    Let S \oplus T be denoted as the symmetric diference of the sets S and T and,

    Let |S| be denoted as the # of elements in set S.

    Then, given |A| = 10 and |B| = 3,

    a.) Determine the largest value possible of |P(A \oplus B)|

    b.) Determine the smallest value possible of |P(A \oplus B)|

    c.) Determine the largest value possible of |P(A) \oplus P(B)|

    d.) Determine the smallest value possible of |P(A) \oplus P(B)|
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  2. #2
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    In the first two you should consider:
    \begin{array}{l}<br />
 a)\;A \cap B = \emptyset  \\ <br />
 b)\;B \subseteq A \\ <br />
 \end{array}<br />
.
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  3. #3
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    Quote Originally Posted by Plato View Post
    In the first two you should consider:
    \begin{array}{l}<br />
 a)\;A \cap B = \emptyset  \\ <br />
 b)\;B \subseteq A \\ <br />
 \end{array}<br />
.
    Hmm, well initially the problem was n and m ( n > m > 0) but then it was changed to 10 and 3, respectively for n and m to make it easier. Not sure how |B| would be a subset |A|. This whole prob is confusing w/ power sets/symmetric stuff.
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  4. #4
    Senior Member JaneBennet's Avatar
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    (a) The symmetric difference between two sets is largest when both sets are disjoint. Then |\mathcal{P}(A \oplus B)|=|\mathcal{P}(A\dot{\cup}B)|=2^{|A\dot{\cup}B|}  =2^{|A|+|B|}. ( \dot{\cup} indicates disjoint union.)

    (b) The symmetric difference between two sets is when one set is a subset of the other. Here A canít be a subset of B as A has more elements than B; so B must be a subset of A. Then |\mathcal{P}(A \oplus B)|=|\mathcal{P}(A\dot{\backslash}B)|=2^{|A\dot{\b  ackslash}B|}=2^{|A|-|B|}. ( \dot{\backslash} indicates subset exclusion.)

    (c) If A\cap B=\O, then their power sets have nothing in common other than the empty set. In this case, |\mathcal{P}(A) \oplus \mathcal{P}(B)| =|\mathcal{P}(A) \cup \mathcal{P}(B)|-|\mathcal{P}(A) \cap \mathcal{P}(B)|=|\mathcal{P}(A)|+|\mathcal{P}(B)|-2|\mathcal{P}(A)\cap\mathcal{P}(B)|=2^{|A|}+2^{|B|  }-2.

    (d) If B\subseteq A, then every subset of B is also a subset of A. In this case, |\mathcal{P}(A) \oplus \mathcal{P}(B)|=|\mathcal{P}(A)\dot{\backslash}\ma  thcal{P}(B)|=|\mathcal{P}(A)|-|\mathcal{P}(B)|=2^{|A|}-2^{|B|}.
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  5. #5
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    Thanks a lot! I would have never been able to come up with that...

    Thanks again.
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  6. #6
    Senior Member JaneBennet's Avatar
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    Youíre welcome. Do you understand what Iíve written though? Thatís much more important.
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  7. #7
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    Quote Originally Posted by JaneBennet View Post
    Youíre welcome. Do you understand what Iíve written though? Thatís much more important.
    Yes, well everything but the "disjoint union"
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