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Set Theory Question: a ∩ b ⊆ a

  1. Jan 20, 2016 #1
    1. The problem statement, all variables and given/known data

    I'm reviewing my powerpoints from class and see the formula A ∩ B ⊆ A. Is this a correct formula? I interpret this as all elements of set A intersected with set B is a subset of set A. I don't think this is a true statement, is it? Sorry it's been a while since I have studied set theory, probably back in high school days or so. I don't see how it could be true because elements of B are not necessarily elements of just A alone.

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Jan 20, 2016 #2

    phinds

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    What does A-intersect-B MEAN to you?
     
  4. Jan 20, 2016 #3
    To me A intersect B is something like this
    155px-Venn_A_subset_B.svg.png
    visual representation where A intersect B is the whole entire circle with all space in A and B included

    That's why I don't understand, if I use this meaning and visual aid then certainly the space in B is not a subset of A. I don't understand why the equation would be true using this visual aid.
     
  5. Jan 20, 2016 #4

    SammyS

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    No. It looks like you're thinking of union, not intersection.
     
  6. Jan 20, 2016 #5
    I see this
    http://www.wolframalpha.com/input/?i=(a+intersection+b)+subset+a
    I understand in the vendiagram that it is true. However in the case of the picture posted previously

    were B is considered the all the space inside B excluding the space in A (a circle with a hole in it)
    A is considered it's on space independent of the space in B
    then A intersect B is all the space in B plus the hole filled in as a solid circle

    Ahhh this picture is a bad representation I think it makes since now A intersect B is were the space in A "overlaps the space in B" making the equation true. That was never to clear to me. Sorry for tangent question.

    I think its better thought of as truth densities were A intersect B is the truth density of A and B occurring at the same time making the truth density of A a subs set of the truth density of both A and B occurring.

    I always wondered, why is Boolean logic similar to set theory representation? I remember studying De Morgan's law in both digital logic class and some linear algebra class.

    Thanks for the help!
     
  7. Jan 20, 2016 #6

    HallsofIvy

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    No! "A intersect B" is, by definition, the set of all points that are both set A and set B. In this case, that is exactly set A, not set B.
    You can prove that [itex]A\cap B\subseteq A[/itex] by "Let x be a point in [itex]A\cap B[/itex]. Then x is in both A and B. In particular x is in A. Since x can be any element of [itex]A\cap B[/itex] any member of [itex]A\cap B[/itex] is a member of A, by definition of "subset", [itex]A\cap B\subseteq A[/itex].
    I have no idea what "truth densities" are. Perhaps it is a translation problem. Boolean logic is "similar" to set theory because Boolean logic has 2 values, "true", and "false" while a point can be in or not in a given set.
     
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