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Index Family Subset Proof: Union subset Intersection

  1. May 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Let [itex] {B_j: j \in J} [/itex] be an indexed family of sets. Show that [itex] \bigcup_{i \in J} B_i \subseteq \bigcap_{j \in J} B_j [/itex] iff for all i, j, [itex] \in [/itex] J, Bi = Bj.


    2. Relevant equations



    3. The attempt at a solution

    First show that [itex] \bigcup_{i \in J} B_i \subseteq \bigcap_{j \in J} B_j \Rightarrow [/itex] for all i, j, [itex] \in [/itex] J, Bi = Bj.

    By contrapositive, [itex] B_i \neq B_j \Rightarrow \bigcup_{i \in J} B_i \not\subset \bigcap_{j \in J} B_j [/itex]

    Suppose [itex] B_i \neq B_j [/itex].

    Let [itex] x \in \bigcup_{i \in J} B_i [/itex]. So there exists an i in J such that x in Bi. But since Bi [itex]\neq[/itex] Bj, [itex]i \neq j[/itex] and there exists an [itex] i \in J \ni x \notin B_j. [/itex].



    I know that the definition of the index family of intersections is for all j in J, x in Ej. But I'm not sure how to say that this isn't the case in the above proof...


    Any guidance for a lost soul?
     
  2. jcsd
  3. May 2, 2013 #2

    LCKurtz

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    The contrapositive would be if there exist p and q such that ##B_p\neq B_q## then ##\bigcup_{i \in J} B_i \not\subset \bigcap_{j \in J} B_j##. Don't use i and j for the indices and also the particular values.

    Why don't you start with with x in one of ##B_p,B_q## and not the other? Then look at where it fits in the result.
     
    Last edited: May 2, 2013
  4. May 2, 2013 #3
    Maybe I entirely mistook your guidance, but this is where I ended up going with it...not sure it's logically consistent though?


    [itex] \exists p, q \in J \ni B_p \neq B_q \Rightarrow \bigcup_{i_J} B_i \not\subset \bigcap_{j \in J} B_j [/itex]

    Suppose [itex] \exists p, q \in J \ni B_p \neq B_q [/itex]. Let [itex] x \in \bigcup_{i \in J} B_i [/itex], then [itex] \exists i \in J \ni x \in B_i [/itex]. Then suppose [itex] x \in B_p [/itex] then [itex] x \notin B_q [/itex].

    Therefore, for all [itex] j \in J [/itex], namely j = q, [itex] x \notin B_j [/itex].

    So [itex] \bigcup_{i \in J} B_i \not\subset \bigcap_{j \in J} B_j [/itex].


    A bit of these steps seem dodgy to me. Your input is greatly appreciated! :)
     
  5. May 2, 2013 #4

    LCKurtz

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    No. Don't start with saying x is in the union. Without loss of generality you can say there is an x that is ##B_p## and not ##B_q##. Start with that x and show it is in the left side but not the right side.
     
  6. May 2, 2013 #5

    So, again my copious stupidity may be valiantly working against my attempts to understand this, but...

    By supposition [itex] B_p \neq B_q [/itex], [itex] x \in B_p [/itex] AND [itex] x \notin B_q [/itex]. Since there exists [itex] p \in J \ni x \in B_p [/itex], then [itex] x \in \bigcup_{i \in J} B_i. [/itex]

    Then, since there exists [itex] j \in J, [/itex] namely j = q, [itex] \ni x \notin B_j [/itex], then [itex] x \notin \bigcap_{j \in J} B_j [/itex].

    Therefore, [itex] \bigcup_{i \in J} B_i \not\subset \bigcap_{j \in J} B_j [/itex].
     
  7. May 2, 2013 #6

    LCKurtz

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    Good. You are now halfway done with the original problem. The other direction is even easier :smile:
     
  8. May 2, 2013 #7
    Thanks!!
     
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