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Set theory Proof help

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose {Ai| i [tex]\in[/tex] I} is an indexed family of sets and I does
    equal an empty set. Prove that [tex]\bigcap[/tex] i [tex]\in[/tex] I Ai
    [tex]\in[/tex] [tex]\bigcap[/tex] i[tex]\in[/tex] I P(Ai ) and P(Ai) is the
    power set of Ai
    2. Relevant equations
    none


    3. The attempt at a solution
    Suppose x [tex]\in[/tex] {Ai| i [tex]\in[/tex] I}. Let i be an arbitrary element of
    I where x [tex]\in[/tex] Ai . Then let y be an arbitrary element of x. Since x
    is an element of Ai and y [tex]\in[/tex] x it follows that ...

    maybe i want to show that [tex]\bigcap[/tex] i [tex]\in[/tex] I Ai [tex]\subseteq[/tex] [tex]\bigcap[/tex] i [tex]\in[/tex] I Ai and then
    I could say that [tex]\bigcap[/tex] i [tex]\in[/tex] I Ai [tex]\in[/tex] [tex]\bigcap[/tex] i[tex]\in[/tex] I P(Ai )
     
  2. jcsd
  3. Apr 30, 2009 #2
    Let [tex]\left\{ A_{i} \right\}_{i \in I} [/tex] be your indexed set of family.

    Do you mean this [tex]\bigcap_{i=1} A_i = \left\{ x : \forall i \in I: x \in A_i \right\} [/tex]?
     
  4. Apr 30, 2009 #3
    Yes sry about the horrible looking symbols
     
  5. Apr 30, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Did you mean "does not equal and empty set"?
     
  6. Apr 30, 2009 #5
    I [tex]\neq[/tex] [tex]\oslash[/tex] is what I meant
     
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