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Math Help - intro analysis help

  1. #1
    Senior Member Danneedshelp's Avatar
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    intro analysis help

    Q1: Consider the statement of the Nested Intervals Property. As in the proof, consider the
    set of all left ends A = <br />
\{a_{n} : n\in\mathbb{ N}\}. Show that every right end b_{n} is an upper bound of A.

    Q2: Consider the set A = <br />
\{ \frac{1}{n} : n\in\mathbb{ N} \}. Prove that inf(A) = 0.

    A1: By the axiom of completeness, A has a greatest upper bound x=sup(A). Thus, for \forall\\n\in\mathbb{N}, a_{n}\leq\\x and for any \epsilon>0 there exists a b_{0}\in\\B such that x<b_{0}<x+\epsilon. Therefore, for \forall\\n\in\mathbb{N}, b_{n} is an upper buond for A.

    A2: Suppose there exists a lower bound l such that inf(A)=0<l. Then, l-0>0, so we can consider \epsilon=l-0>0. By lemma 1.3.7, there exists an a\in\\A such that, for any \epsilon>0, a<l+\epsilon. But, if we consider \epsilon=l-0, then a<2l. Therefore, \frac{a}{2}<l. This contradicts l being a lower bound for A. Hence, inf(A)=0.

    How do this proofs look? I feel I am not showing enough in the first one.

    thanks
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  2. #2
    Senior Member Sampras's Avatar
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    For Q1,  A \subset \mathbb{R} . But to show  \sup A exists, you have to also show that  A is bounded above to invoke the least upper bound property. So I think you are using what you are trying to prove.
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  3. #3
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    Quote Originally Posted by Danneedshelp View Post
    Q1: Consider the statement of the Nested Intervals Property. As in the proof, consider the
    set of all left ends A = <br />
\{a_{n} : n\in\mathbb{ N}\}. Show that every right end b_{n} is an upper bound of A.
    Let I_j = [a_j,b_j] be your intervals for j\geq 1. They are nested, so, I_i\subseteq I_j if j>i. Assume, by contradiction, not every right end was an upper bound of \{ a_n\}. Thus, it means there exists, a left end a_m so that a_m which was not an upper bound for A. That means, there exists b_n so that a_m>b_n. Now a_1\leq a_2\leq ... \leq a_n\leq b_n, therefore if a_m > b_n then it must mean that m>n. But we know that I_m \subseteq I_n. Now a_m\in I_m but a_m\not \in I_n because a_m > b_n. This is a contradiction by the nested interval property.
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  4. #4
    Senior Member Danneedshelp's Avatar
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    Can I simply break Q1 into cases?

    Let I_{n}=[a_{n},b_{n}]. Since I_{n+1}\subseteq\\I_{n}, we can see a_{1}\leq\\b_{1}, a_{2}\leq\\b_{2},.... Thus, there are three cases:

    Case 1: If i=j, then it is a given that a_{i}\leq\\b_{j}.
    Case 2: If i<j, note that a_{i}\leq\\a_{j}\leq\\b_{j}.
    Case 3: If i>j, then a_{i}\leq\\b_{i}\leq\\b_{j}.

    Therefore, b_{n} is an upper bound for A, for every n\in\mathbb{N}.

    Do I need to add any thing more to Q2?

    Thanks for your help!
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