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Math Help - algebra of sets law

  1. #1
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    algebra of sets law

    Hi,

    it is determined that (\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c. Does (\bigcup_{n=1}^\infty A_n)^c = \bigcap_{n=1}^\infty (A_n)^c holds?
    Last edited by waytogo; June 3rd 2011 at 08:37 AM.
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  2. #2
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    Quote Originally Posted by waytogo View Post
    it is determined that (\bigcup_{n=1}^m A_n)^c = \bigcup_{n=1}^m (A_n)^c.
    That is not true.

    It is (\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c
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  3. #3
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    No. For example, let m = 2 and suppose A_1=A_2 = \{0\}, A_n=\{i\in\mathbb{N}\mid i>0\} for n>2. Then (A_1\cup A_2)^c=A_1^c=A_2^c=A_3=A_1^c\cup A_2^c. However, \big(\bigcup_{n=1}^\infty A_n\big)^c=\mathbb{N}^c=\emptyset, while \bigcup_{n=1}^\infty A_n^c=\mathbb{N}.
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  4. #4
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    you are right Plato, my mistake. made it correct now.
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  5. #5
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    Quote Originally Posted by waytogo View Post
    it is determined that (\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c. Does (\bigcup_{n=1}^\infty A_n)^c = \bigcap_{n=1}^\infty (A_n)^c holds?
    In the set \mathbb{N}=\{0,1,2,3,\cdots\} define A_n=\{n\}.
    For each m does (\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c hold ?
    Last edited by Plato; June 3rd 2011 at 09:06 AM.
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  6. #6
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    Quote Originally Posted by Plato View Post
    In the set \mathbb{N}=\{0,1,2,3,\cdots\} define A_n=\{n\}.
    For each m does (\bigcup_{n=1}^m A_n)^c = \bigcap_{n=1}^m (A_n)^c hold ?
    yes, for me it seems correct in this case.
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  7. #7
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    I have a bad feeling that we are missing some fine points here.
    Are you asking about a particular collection of sets with the same index set? Or are you asking about a collection with a finite index set that can be expanded into an infinite collect?

    It would be nice if you posted an exact, fully worded question.
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  8. #8
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    In what theory? In Z set theories, there is no absolute complement of a set. So suppose this is relative to some universe S:

    Suppose x in S\U[n = 1 to oo]An. Show that for all n, we have x in S\An. Well, x is in S. And if there were an n such that x in An, then x would be in U[1 = n to oo]An, but x is not in U[n = 1 to oo]An.

    Suppose that for all n, we have x in S\An. Show x in S\U[n = 1 to oo]An. Well, x is in S. And if x where in U[n = 1 to oo]An, then x would be in An for some n, but for no n is x in An.

    Indeed, in even the most general case, the relative complement of the union of B equals the intersection of the relative complements of the members of B.

    This is generalized De Morgan's law for sets.
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