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Homework Help: Finding The Power Set Of A Given Set

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    My answer is {∅, {{∅}}, {∅, {∅}}}
    but the actual answer is: {∅,{∅},{{∅}},{∅,{∅}}}
    I don't understand how the second element, {∅}, appears...
  2. jcsd
  3. Sep 28, 2012 #2


    Staff: Mentor

    your set { a, b } so the power set has all proper subsets including the null set and the set itself:

    { nullset, {a}, {b}, {a,b} }

    and in your example then the { a } corresponds to the { nullset } and the { b } corresponds to { { nullset } }
  4. Sep 28, 2012 #3
    Okay, thank you.
  5. Sep 28, 2012 #4
    Actually, I have another question concerning power sets. My book says, "If a set has n elements, then its power set has 2^n elements." Why does this formula work?
  6. Sep 28, 2012 #5


    Staff: Mentor

    okay the best way to thing of this is using a binary string of digits where each digit of the string represents an element of the set so I can encode the various subsets as strings of binary digis as follows:

    000000 represents the nullset { }
    100000 represents the first element {a}
    010000 the 2nd { b }
    001000 the 3rd... { c }
    111000 a subset containing {a,b,c}
    111111 the set itself { a,b,c,d,e,f }

    so in creating a subset then

    a is or isnt in the subset: 2 choices
    b is or isnt in the subset : x 2 =4 choices
    f is or isn't in the subset x 2 = 64 choices

    or more succinctly:

    a x b x c x d x e x f
    2 x 2 x 2 x 2 x 2 x 2 = 2^6 = 64 possible subsets

    and therefore the powerset contains 64 elements.
  7. Sep 28, 2012 #6


    User Avatar
    Science Advisor

    Another proof, by induction. If set A has no elements, it is empty. The only subset of the empty set is the empty set itself so its power set contains only the empty set and so has 1= 20 member.

    Suppose that, for some number k, any set with k member has 2k members. Let A be a set containing k+1 members. Since A is not empty, it contains at least one member. Pick a specific member and call it "x". Every subset of A contains x or not.

    1) If subset B of A does not contain x, it is a subset of A\{x}. Since A contains k+1 members, A\{x} contains k and so has 2k subsets. That is, there are 2k subsets of A that do not include x.

    2) If subset B contains x, then B\x does not and so is a subset of A\{x}. That is, eery subset containing x is just a subset that doesn't contain a, union {a}. Since there are 2k such sets, there are also 2k subsets of A that do contain x.

    Therefore, there are 2k+ 2k= 2(2k)= 2k+1 subsets of A and we are done.
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