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Understanding Distributive Law Using Arbitrary Union

  1. Apr 24, 2014 #1
    1. The problem statement, all variables and given/known data
    In the textbook I'm reading it tells me that [itex]A \cup \bigcap B = \bigcap \left\{ A \cup X | X \in B \right\}[/itex] for B not equal to ø


    2. Relevant equations



    3. The attempt at a solution
    I don't understand how this would work, the left side of the equation creates a set where all of the elements of [itex]A[/itex] are included but the right side of the equation takes the arbitrary intersection of [itex]A \cup X[/itex], so wouldn't this mean that some of the elements of [itex]A[/itex] aren't necessarily included in this newly constructed set as the elements of [itex]A[/itex] aren't by necessity also elements of [itex]X[/itex]?

    How can the left side of the equation be equal to the right?
     
  2. jcsd
  3. Apr 24, 2014 #2

    Ray Vickson

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    The only way this makes sense is if ##B = \{ X_{\gamma}, \gamma \in \Gamma \}## is a class class of sets, so when you write ##X \in B## you have ##X## itself is a set in the class ##B##. In other words, I think it is saying that
    [tex] A \cup \bigcap_{\gamma \in \Gamma} X_{\gamma}
    = \bigcap_{\gamma \in \Gamma} A \cup X_{\gamma}[/tex]
     
  4. Apr 24, 2014 #3

    HallsofIvy

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    No, the union of A with any set, by definition, includes all elements of A. You are confusing the union with the intersection, [itex]A\cap X[/itex].

     
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