Question about sets

Discussion in 'Physics & Math' started by arfa brane, Feb 25, 2017.

  1. arfa brane call me arf Valued Senior Member

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    If {a,b} ∩ {b,c} = {b}, and {a,a} ∩ {b,c} = ∅, what is {a,a} ∩ {a,c}?

    1) {a,a}
    2) {a}
     
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  3. QuarkHead Remedial Math Student Valued Senior Member

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    \( \{a,a\}\) is not a set (no element can appear more than once in a well-formed set)
     
    rpenner likes this.
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  5. arfa brane call me arf Valued Senior Member

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    Ok. Nonetheless I recall being told by a professor that the set {a} is distinct from the set {a,a}. But apparently {a,a} isn't a set.
    If I accept that then I need to accept that {a,a} cannot be the union or intersection of other sets, nor can it be the complement of a set . . .?
     
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  7. rpenner Fully Wired Staff Member

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    According to the axiom of pairing, if X and Y are set-valued variables, then there is a set W such that Z ∈ W means Z = X OR Z = Y. In other words, W = { X, Y }.
    According to the axiom of extensionality, if X and Y are set valued variables then if for every Z, Z ∈ X was true exactly whenever Z ∈ Y, then X is the same set as Y.
    Therefore {a, a} = {a} and {a,a} ∩ {a,c} = {a}. To write {a, a} as a technically correct solution while ignoring {a} would be a flaw in your understanding of sets.


    https://en.wikipedia.org/wiki/Axiom_of_pairing
    https://en.wikipedia.org/wiki/Axiom_of_extensionality


    {a} and {a, a} are distinct in the mathematical realm of multisets, but that is not the topic you asked about.

    In mulitsets, {a,a} ∩ {a,c} = {a} because 1 = min(2,1) and 0 = min(0,1).

    https://en.wikipedia.org/wiki/Multiset

     
  8. arfa brane call me arf Valued Senior Member

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    Ok, thanks for the expl.

    All I remember being told is that {a,a} is not the same "kind of object" as {a}.

    Another thing I still haven't quite grasped is about Cartesian products of sets.
    Any set A × ∅ = ∅. But isn't the product a set of duples: {(a,∅), . . . }, how is that mathematically = ∅?
     
  9. rpenner Fully Wired Staff Member

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    Your example is misleading, because there is literally no element which can form the second part of that ordered pair.
     
  10. QuarkHead Remedial Math Student Valued Senior Member

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    Because in set theory, the emptyset symbol is often (always?) used to denote non-existence. For example writing \(A \cap B=\emptyset\) means "there does not exist any element in the intersection of A and B"

    So \(A \times \emptyset= \emptyset\) is perhaps technically correct, since, as rpenner says, there exists no ordered pair which can serve as an element in this Cartesian product.

    But it is not helpful. Far from it.
     
  11. arfa brane call me arf Valued Senior Member

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    I wonder what is helpful then, in trying to understand the equality A × ∅ = ∅.

    Since it is something you encounter, and the explanation or proof if you will, is that there are ordered pairs (a,b) such that a ∈ A and b ∈ ∅. But there is no b ∈ ∅. So there are no ordered pairs and the product is empty.

    Is the 'axiom of pairing' useful (or helpful) here?
     
  12. rpenner Fully Wired Staff Member

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    The axiom of pairing says that if A is a set and B is a set then U = {A, B} is a set. So V= { A }, W= { B }, X = { ∅ }, Y= { A, ∅ }, and Z = { B, ∅ } are also also sets by the axiom of pairing. But that means ∅ is an element of { ∅ }, so { ∅ } is not the empty set just like V≠A and W≠B.

    So I don't think it's useful, since the properties of ∅ come from it being empty.

    card ∅ = 0
    card X = 1
    card U = 2
    card ∅×∅ = 0
    card ∅×X = 0
    card ∅×U = 0
    card X×X = 1
    card X×U = 2
    card U×U = 4
    card A×B = (card A)×(card B) (when A and B are finite sets)
     

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