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Set Theory: P(P{1}) and its cardinality

  1. Sep 3, 2012 #1
    The notation has me a bit confused...

    Heres my logic

    for the P({1}) on the inside

    {EmptySet, {{1}}}
    reason being, you always include the empty set, {1} is a part of the set. The cardinality is two

    You have the set: {EmptySet, {{1}}}, and now you have to consider the outer "P"
    the possibilities are:{EmptySet, {{1}}},EmptySet, {{1}}
    I feel like there should be one more as its cardinality is three, when it should be four, and I dont think order matters
     
  2. jcsd
  3. Sep 3, 2012 #2
    You're fine except that you're mistaken about thinking the empty set is an element of every set. What's true is that the empty set is a subset of every set. So the empty set is an element of the power set of every set.

    ∅ is the empty set.

    {∅} is the set containing the empty set. It's the set we call '1' in the von Neumann ordinals. In other words,

    {∅} = 1.

    Then {∅, 1} = {∅, {∅}} = 2 and so forth. Note that the number of elements of each set is equal to the name of the set. 1 has 1 element, 2 has 2 elements.

    Remember the distinction between an element of a set and a subset of a set. The empty set has no elements, but it does have one subset, namely the empty set. The empty set is a subset of every set; but the empty set may or may not be an element of some other given set.
     
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