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Math Help - Proof with sets

  1. #1
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    Proof with sets

    "Let I be the set of real numbers that are greater than 0. For each x \inI, let A_x be the open interval (0,x). Prove that \cap _{x\in I} A_x = \emptyset, \cup _{x\in I} A_x = I. For each x \in I, let B_x be the closed interval [o, x]. Prove that \cap _{x\in I} B_x = {0}, \cup _{x\in I} B_x = I \cup{0}."
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  2. #2
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    Re: Proof with sets

    Quote Originally Posted by Borkborkmath View Post
    "Let I be the set of real numbers that are greater than 0. For each x \inI, let A_x be the open interval (0,x). Prove that \cap _{x\in I} A_x = \emptyset, \cup _{x\in I} A_x = I. For each x \in I, let B_x be the closed interval [o, x]. Prove that \cap _{x\in I} B_x = {0}, \cup _{x\in I} B_x = I \cup{0}."
    All of these follow from two basic facts:
    \left( {\frac{1}{n}} \right) \to 0\;\& \,(n) \to \infty .

    That means that if x\in\mathcal{I} then \left( {\exists k \in \mathbb{Z}^ +  } \right)\left[ {\frac{1}{k} < x} \right].
    So x \notin \left( {0,\frac{1}{k}} \right). That is the first.

    Moreover, for the second: \left( {\exists K \in \mathbb{Z}^ +  } \right)\left[ {K > x} \right] so x\in (0,K).
    Last edited by Plato; September 30th 2011 at 01:02 PM.
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  3. #3
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    Re: Proof with sets

    I might be wording this wrong, but the idea is similar to finite subcovers?
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    Re: Proof with sets

    Quote Originally Posted by Borkborkmath View Post
    I might be wording this wrong, but the idea is similar to finite subcovers?
    What? This question is about generalized intersections and unions.
    What do covers have to do with that?
    You are asked to show \bigcap\limits_{x \in I} {A_x }  = \emptyset \;\& \;\bigcup\limits_{x \in I} {A_x }=\mathcal{I}
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