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Math Help - Computing Powers of Sets

  1. #1
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    Computing Powers of Sets

    I have the following question:

    Let A = {a,b,c,d},
    Let B = {c,d,e}.

    Compute P(A) and P(A (intersection) B).

    Here's my answer:

    P(A) = {{a,b,c,d}}
    A(intersection)B = {c,d}
    P(A(intersection)B) = {{c,d}}

    Is my answer correct?
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  2. #2
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    Re: Computing Powers of Sets

    no. the power set of a set, contains ALL subsets of the original set.

    for example, if C = {a,b}, then P(C) = {,{a},{b},{a}U{b}} = {,{a},{b},{a,b}} = {, {a},{b},C}

    the power set is "bigger" than just {C}.
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  3. #3
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    Re: Computing Powers of Sets

    Quote Originally Posted by maclunian View Post
    I have the following question:
    Let A = {a,b,c,d},
    Let B = {c,d,e}.
    Compute P(A) and P(A (intersection) B).
    Here's my answer:
    P(A) = {{a,b,c,d}}
    A(intersection)B = {c,d}
    P(A(intersection)B) = {{c,d}}
    Is my answer correct?
    No, not even close to being correct.

    \mathcal{P}(A) is a set of all subsets of A.
    It is true that A\in\mathcal{P}(A) and \emptyset\in\mathcal{P}(A).
    But there are fourteen more sets in \mathcal{P}(A).
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    Re: Computing Powers of Sets

    Quote Originally Posted by Plato View Post
    No, not even close to being correct.

    \mathcal{P}(A) is a set of all subsets of A.
    It is true that A\in\mathcal{P}(A) and \emptyset\in\mathcal{P}(A).
    But there are fourteen more sets in \mathcal{P}(A).
    How do you know that there are fourteen more sets in \mathcal{P}(A)?
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  5. #5
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    Re: Computing Powers of Sets

    Quote Originally Posted by maclunian View Post
    How do you know that there are fourteen more sets in \mathcal{P}(A)?
    If a set, A contains n elements the \mathcal{P}(A) contains 2^n elements.
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  6. #6
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    Re: Computing Powers of Sets

    Quote Originally Posted by Deveno View Post
    no. the power set of a set, contains ALL subsets of the original set.

    for example, if C = {a,b}, then P(C) = {,{a},{b},{a}U{b}} = {,{a},{b},{a,b}} = {, {a},{b},C}

    the power set is "bigger" than just {C}.
    So, if A = {a,b,c,d}, then P(A) = {,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},{a,b,c,d}} = {,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},A}

    Is this correct?
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  7. #7
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    Re: Computing Powers of Sets

    Quote Originally Posted by maclunian View Post
    So, if A = {a,b,c,d}, then P(A) = {,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},{a,b,c,d}} = {,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},A}
    Is this correct?
    This much is correct: P(A) = {,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},{a,b,c,d}}
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    Re: Computing Powers of Sets

    What about the second bit?

    P(A) = {,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},A}
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  9. #9
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    Re: Computing Powers of Sets

    Quote Originally Posted by maclunian View Post
    What about the second bit?
    = {,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},A}
    They are identical. You just renamed one set.
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  10. #10
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    Re: Computing Powers of Sets

    one way to understand "why" |P(A)| = 2^{|A|} is to think of all possible functions:

    f:A→{0,1}.

    for example, one such function is:

    f(a) = 1
    f(b) = 1
    f(c) = 0
    f(d) = 1

    which corresponds to the set {a,b,d}. see what i did? i just picked a subset of A, in this case, {a,b,d}, and defined f(x) = 1, if x is in {a,b,d}, f(x) = 0, otherwise.

    there are 2^{|A|} such functions (since we have 2 choices for every element f(x), and |A| elements).

    if you like, you can think of this as a truth-table for the proposition: "is x in this subset X of A?" a value of 1, means "yes"(true), and a value of 0 means "no"(false).

    if you have a set of 3 elements, its power set will have 8 elements. and there are precisely 8 functions from {x,y,z} to {0,1}.
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