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Prove: n(FuG) = (uF)n(uG)

  1. Aug 2, 2012 #1
    1. The problem statement, all variables and given/known data
    (Title is wrong)

    I was able to prove the similar [itex] \bigcup(F \cup G)=(\cup F)\bigcup(\cup G) [/itex] but I'm not to sure how to go about this one.

    Let F and G be nonempty families of sets. Prove
    [itex] \bigcap(F \cup G)=(\cap F)\bigcap(\cap G) [/itex]

    2. The attempt at a solution
    To prove [itex]\bigcap(F \cup G) \subset (\cap F)\bigcap(\cap G)[/itex],
    let [itex] x \subset \bigcap(F \cup G) [/itex] be arbitrary. Clearly, [itex] x \in S[/itex] for some set [itex] S \subset \bigcap F \cup G [/itex] containing all common elements in [itex] F \cup G [/itex]. We have [itex] S \in F \cup G.[/itex]

    Do I go by..

    We have two cases: (only one needs to be proven really.)
    Case 1: [itex] If S \in F [/itex], clearly [itex] S \subset \cup F [/itex]...
    (I'm stuck here)

    or Do I go by...

    Suppose [itex] S \in F [/itex] and [itex] S \in G [/itex]. Clearly [itex] S \subset \cup F [/itex] and [itex] S \subset \cup G [/itex]. (I'm stuck here)

    I'll try proving the converse and restart the proof since I misread the problem.
     
    Last edited: Aug 2, 2012
  2. jcsd
  3. Aug 3, 2012 #2
    By definition of the operation [itex]\bigcap[/itex], any x in [itex]\bigcap(F\cup G)[/itex] must be contained in EVERY SET of [itex]F\cup G[/itex]. Continue from here.
     
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