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Prove U (Ai X Bi) [itex]\subseteq[/itex] (U Ai) X (U Bi)

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose {Ai l i [itex]\in[/itex]I} and {Bi l i [itex]\in[/itex]I} are indexed families of sets.

    Prove that U i [itex]\in[/itex]I(Ai X Bi) [itex]\subseteq[/itex] (Ui [itex]\in[/itex]IAi) X (Ui [itex]\in[/itex]IBi)


    2. Relevant\subseteq equations
    From How to Prove It, 2nd Edition, Sec. 4.1 #11a)


    3. The attempt at a solution

    Let (x, y) be arbitrary. Suppose (x, y) [itex]\in[/itex] [itex]\bigcup[/itex]i[itex]\in[/itex]I (Ai X Bi).

    Since (x, y) [itex]\in[/itex][itex]\bigcup[/itex]i[itex]\in[/itex]I(Ai X Bi), there exists an i[itex]\in[/itex]I with x[itex]\in[/itex]Ai and y[itex]\in[/itex]Bi.

    So x [itex]\in[/itex]{xl[itex]\exists[/itex]i[itex]\in[/itex]I(x[itex]\in[/itex]Ai)} and
    y[itex]\in[/itex]{yl[itex]\exists[/itex]i[itex]\in[/itex]I(y[itex]\in[/itex]Bi)}

    Therefore, x [itex]\in[/itex][itex]\bigcup[/itex]i[itex]\in[/itex]I Ai and y[itex]\in[/itex][itex]
    \bigcup[/itex]i[itex]\in[/itex]I Bi.

    This is equivalent to ([itex]\bigcup[/itex]i[itex]\in[/itex]I Ai) X ([itex]\bigcup[/itex]i[itex]\in[/itex]I Bi). Hence, [itex]\bigcup[/itex]i[itex]\in[/itex]I (Ai X Bi)[itex]\subseteq[/itex](Ui[itex]\in[/itex]I Ai) X ([itex]\bigcup[/itex]i[itex]\in[/itex]I Bi).
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 5, 2011 #2

    micromass

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    That looks very good!!!!

    It might be instructive to find a counterexample to the reverse inclusion...
     
  4. Oct 5, 2011 #3
    Thanks for commenting, I appreciate it. I have a very picky professor.
     
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