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Math Help - rationals and union of intervals that covers it

  1. #1
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    rationals and union of intervals that covers it

    I have a question, if \{q_{n}\}_{n=1}^{\infty} is a sequence of all rational numbers, let I_{n}=[q_{n}-\frac{1}{10^{n}},q_{n}+\frac{1}{10^{n}}]

    why do we have the following inclusion \mahbb{Q}\subset \displaystyle\cup_{n=1}^{\infty}I_{n}?
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  2. #2
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    Re: rationals and union of intervals that covers it

    Himm, if x ∈ A and y ∈ B, is it clear to you that {x, y} ⊆ A ∪ B?
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    Re: rationals and union of intervals that covers it

    yes it is, but I still don't understand the inclusion. Each of I_{n} is a set (closed interval) containing infinitely many rational numbers bounded by the end points of each interval. Will their union include all rationals?
    Last edited by rayman; October 9th 2013 at 10:39 AM.
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    Re: rationals and union of intervals that covers it

    Quote Originally Posted by rayman View Post
    Will their union include all rationals?
    Yes, it will. In my example, A ∪ B covers (i.e., is a superset of) the whole set {x, y} because each element of that set is covered by (i.e., belongs to) at least one of the sets in the union: either A or B. A completely parallel statement holds about ℚ and \bigcup_{n=1}^\infty I_n.
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    Re: rationals and union of intervals that covers it

    Ah yes of course, I got it thank you
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    Re: rationals and union of intervals that covers it

    Quote Originally Posted by rayman View Post
    I have a question, if \{q_{n}\}_{n=1}^{\infty} is a sequence of all rational numbers, let I_{n}=[q_{n}-\frac{1}{10^{n}},q_{n}+\frac{1}{10^{n}}]
    why do we have the following inclusion \mahbb{Q}\subset \displaystyle\cup_{n=1}^{\infty}I_{n}?
    Quote Originally Posted by rayman View Post
    yes it is, but I still don't understand the inclusion. Each of I_{n} is a set (closed interval) containing infinitely many rational numbers bounded by the end points of each interval. Will their union include all rationals?
    To prove that A\subset B is suffices to show x \in A \to x \in B.

    If t\in\mathbb{Q} then \exists j\in\mathbb{N}[q_j=t].
    But t=q_j\in I_{j}=[q_{j}-\frac{1}{10^{j}},q_{j}+\frac{1}{10^{j}}]

    Hence, every rational number is in the large union.
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