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Math Help - Union of a set

  1. #1
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    Union of a set

    This is from a proof in my book:

    If  A = \{ x : f(x) > 0 \} , define  A_n = \{ x : f(x) > \frac {1}{n} \} , then  A = \bigcup ^ \infty _ 1 A_n

    Why is this true?
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    This is from a proof in my book:

    If  A = \{ x : f(x) > 0 \} , define  A_n = \{ x : f(x) > \frac {1}{n} \} , then  A = \bigcup^\infty_1 A_n

    Why is this true?
    To show set equality we just need to show inclusion, right? One inclusion is obvious:

    Clearly A_n\subseteq A for each n so
     A \supseteq \bigcup^\infty_1 A_n .

    The other inclusion is not much harder. Let x\in A. Then f(x)>0 so by the Archimedean principle there exists an integer M such that Mf(x)>1, and so f(x)>\dfrac{1}{M} thus x\in A_M\subseteq \bigcup^\infty_1 A_n. Therefore A \subseteq \bigcup^\infty_1 A_n and so A = \bigcup^\infty_1 A_n !
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  3. #3
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    Quote Originally Posted by tttcomrader View Post
    This is from a proof in my book:

    If  A = \{ x : f(x) > 0 \} , define  A_n = \{ x : f(x) > \frac {1}{n} \} , then  A = \bigcup ^ \infty _ 1 A_n

    Why is this true?
    One way is completely obvious.
    a \in A\, \Rightarrow \,f(a) > 0\, \Rightarrow \,\left( {\exists n} \right)\left[ {\frac{1}{n} < f(a)} \right]\, \Rightarrow \,a \in A_n
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