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Proof Check: Closure of Union Contains Union of Closures

  1. Oct 5, 2013 #1
    I intend to show, for a set ##X## containing ##A_i## for all ##i##, $$\overline{\bigcup A_i}\supseteq \bigcup \overline{A_i}.$$
    //Proof: We proceed to prove that ##\forall x\in X,~x\in\bigcup\overline{A_i}\implies x\in\overline{\bigcup A_i}##. Equivalently, ##\forall x\in X,~x\not\in\overline{\bigcup A_i}\implies x\not\in\bigcup\overline{A_i}##. Thus, let ##x\not\in\overline{\bigcup A_i}##.

    From this, we conclude ##\exists N\ni x: N\cap(\bigcup A_i)=\emptyset\implies\bigcup (N\cap A_i)=\emptyset\implies\forall i,~N\cap A_i=\emptyset##.

    ##\therefore \forall i,~x\not\in\overline{A_i}##. Thus, ##x\not\in\bigcup\overline{A_i}##, and we conclude ##\overline{\bigcup A_i}\supseteq \bigcup \overline{A_i}##. Halmos.

    This proof is correct, right?
     
  2. jcsd
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