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Thread: Union of a set

  1. #1
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    Union of a set

    This is from a proof in my book:

    If $\displaystyle A = \{ x : f(x) > 0 \} $, define $\displaystyle A_n = \{ x : f(x) > \frac {1}{n} \} $, then $\displaystyle 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 $\displaystyle A = \{ x : f(x) > 0 \} $, define $\displaystyle A_n = \{ x : f(x) > \frac {1}{n} \} $, then $\displaystyle 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 $\displaystyle A_n\subseteq A$ for each $\displaystyle n$ so
    $\displaystyle A \supseteq \bigcup^\infty_1 A_n $.

    The other inclusion is not much harder. Let $\displaystyle x\in A$. Then $\displaystyle f(x)>0$ so by the Archimedean principle there exists an integer $\displaystyle M$ such that $\displaystyle Mf(x)>1$, and so $\displaystyle f(x)>\dfrac{1}{M}$ thus $\displaystyle x\in A_M\subseteq \bigcup^\infty_1 A_n$. Therefore $\displaystyle A \subseteq \bigcup^\infty_1 A_n $ and so $\displaystyle 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 $\displaystyle A = \{ x : f(x) > 0 \} $, define $\displaystyle A_n = \{ x : f(x) > \frac {1}{n} \} $, then $\displaystyle A = \bigcup ^ \infty _ 1 A_n $

    Why is this true?
    One way is completely obvious.
    $\displaystyle 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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