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Math Help - Showing a family is actually an algebra

  1. #1
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    Showing a family is actually an algebra

    Hello!

    I am struggling with just the very last stage of the solution and any suggestions would be much appriciated!! Thanks!

    I am trying to show that given a set X and a collection of subsets U of X, and given that
    (i) X \in U and \emptyset \in U
    (ii) E \in U implies that the complement  X\setminus E \in U
    ( U is closed under complements)
    (iii) U is closed under finite (or, case 2: countable) intersections, (ie, E_1, E_2, ..., E_N \in U \Rightarrow \bigcap_{n=1}^{N}E_n \in U} )

    I am trying to show that U is an algebra (or, case 2: a sigma algebra)

    Handling case 1, closed under finite intersections, I think I have come up with a proof using some set theory but I was hoping someone might be able to verify.

    So, firstly, I think all we need to show is that U is closed under finite unions to show that it is an algebra.

    To do this, take E_1, E_2, ..., E_N \in U. Then since U is closed under complements and also under finite intersections, \bigcap_{n=1}^{N}X\setminus E_n \in U .
    Then by Demorgans Law, \bigcap_{n=1}^{N}X\setminus E_n = X\setminus {\bigcup_{n=1}^{N}X\setminus E_n. Since the left-hand side is in U, the right-hand side will be as well. Also, Since, we are closed under complements, X\setminus {\bigcup_{n=1}^{N}X\setminus E_n \in U \Rightarrow {\bigcup_{n=1}^{N}X\setminus E_n \in U.

    Here is where I am stuck. Can I show that {\bigcup_{n=1}^{N}X\setminus E_n \in U \Rightarrow {\bigcup_{n=1}^{N}E_n \in U ? Because then I would be done (we would have shown it is closed under finite unions). And I believe a similar argument would work for countable unions and sigma algebras?

    Again, any help appriciated!! Thank you!
    Last edited by matt.qmar; January 12th 2012 at 10:44 AM.
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  2. #2
    MHF Contributor

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    Re: Showing a family is actually an algebra

    You are using the wrong LaTeX tags.
    [TEX]X\setminus E[/TEX] gives X\setminus E
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