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Set theory proof

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  1. Apr 2, 2016 #1
    1. The problem statement, all variables and given/known data
    Attached is the problem

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    3. The attempt at a solution
    So I have to show that each side is a subset of the other side

    Assume x∈ A ∪ (∩Bi)
    so x∈A or x∈∩Bi

    case 1 x∈ ∩ Bi

    so x∈ (B1∩B2∩B3.......∩Bn)
    which implies x∈B1 and x∈B2 ......... and x∈Bn
    so x∈B1∪A and x∈B2∪A.......... and x∈Bn∪A
    so x∈∩(A∪Bi)

    My teacher told me that this approach doesn't work because the sets can be infinite and told me to use "words". I don't what he meant with that.
     

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  3. Apr 2, 2016 #2

    haruspex

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    You don't need to use words instead of symbols, though you could. What you do need to do is avoid writing it as a finite (or even countably infinite) intersection. It is not hard to convert your proof to one that works.
     
  4. Apr 2, 2016 #3
    can I say n->infinity?
     
  5. Apr 2, 2016 #4
    what about if I split (∩Bi) into two finite sets call it S and B so it equals SnB
     
  6. Apr 2, 2016 #5

    andrewkirk

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    The problem is that by writing x∈ (B1∩B2∩B3.......∩Bn) you have assumed that the index set ##I## has only ##n## elements. But in fact it may be infinite, or worse, uncountable. That assumption cannot be made because it may be false. Consider for instance if the index set ##I## is the set of all real numbers in the interval [0,1) and ##B_j## for ##j\in I## is the set of all positive real numbers whose fractional part is ##j##. Then ##I## is uncountably infinite.

    To avoid making that invalid assumption, use the quantifier ##\forall##, which means 'for all'.

    So you have ##x\in\bigcap_{j\in I} B_j## and instead of writing

    x∈B1 and x∈B2 ......... and x∈Bn

    you write ##\forall j\in I:\ x\in B_j##.

    Similarly for the rest of your proof: wherever you find yourself using ##n## or an ellipsis (that's the '.....' you've written in the middle of lists), get rid of them by re-writing using ##\forall##.

    When you come to do similar proofs for unions of indexed sets you will need to use the other quantifier ##\exists##, which means 'there exists' ('there is at least one').
     
  7. Apr 2, 2016 #6
    thanks!!!!!!
     
  8. Apr 2, 2016 #7
    by the way I am theMathNoob. can you ask the administrator to unban my account?. I promise I won't post silly ps4 questions.
     
  9. Apr 3, 2016 #8

    berkeman

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