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Set theory question

  1. Aug 9, 2009 #1
    Letting A, B_1, B_2, ..., B_n subsets of X, then show

    A\cap\bigcup_{n}^{i=1}B_{i} = \bigcup_{n}^{i=1}\left(A \cap\right B_{i})


    Is it sufficient to say...

    By De Morgan law, we have

    \left(A\cap \right B_{i})\cup\left(A \right \cap\ B_{2})\cup\ --- \cup\left(A\cap \right\ B_{n}) = \bigcup_{i=1}^{n}\left(A\cap \right B_{i})

    Is that sufficient, or is there a better/more complete way to do it?
  2. jcsd
  3. Aug 9, 2009 #2
    Well, that looks fine to me. But the other way to do it, if you prefer to avoid [tex]...[/tex] type notation, is to choose

    [tex]x\in A\cap\bigcup_{n}^{i=1}B_{i}[/tex],

    and say that therefore [tex]x\in A[/tex] and [tex]x\in B_i[/tex] for some [tex]i\in\mathbb{N}[/tex]. Therefore

    [tex]x\in A\cap B_i\subseteq\bigcup_{n}^{i=1}\left(A \cap\right B_{i})[/tex]

    [tex]\Rightarrow[/tex] [tex]x\in\bigcup_{n}^{i=1}\left(A \cap\right B_{i})[/tex].

    Then choose [tex]y\in\bigcup_{n}^{i=1}\left(A \cap\right B_{i})[/tex]

    [tex]\Rightarrow[/tex] [tex]y\in A\cap B_i\subseteq A\cap\bigcup_{n}^{i=1}B_{i}[/tex]

    [tex]\Rightarrow[/tex] [tex]y\in A\cap\bigcup_{n}^{i=1}B_{i}[/tex].

    So [tex]A\cap\bigcup_{n}^{i=1}B_{i}= A\cap\bigcup_{n}^{i=1}B_{i}[/tex]. [tex]\blacksquare[/tex]

    Obviously, though, that way requires a bit more writing. It's a matter of preference which you wish to use; both are valid.
  4. Aug 13, 2009 #3
    For the last line, don't you mean

    A\cap\bigcup_{n}^{i=1}B_{i}=\bigcup_{n}^{i=1}\left(A \cap\right B_{i})[/tex]

    But A and B_1, B_2, ... , B_n are subsets of X, so does the Demorgan law really apply here (since A and B_i are in the same family)? I thought they had to be distinct.
    Last edited: Aug 13, 2009
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