1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook