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Math Help - Sequence

  1. #1
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    Sequence

    I need help with the following problem in analysis:

    Find a sequence of closed intervals I_1 I_2 ... I_n ... whose end points are rational numbers and such that I_n = {e}.
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  2. #2
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    Quote Originally Posted by tarheelborn View Post
    I need help with the following problem in analysis:

    Find a sequence of closed intervals I_1 I_2 ... I_n ... whose end points are rational numbers and such that I_n = {e}.
    The rational numbers are dense in the reals.
    \left( {\exists p_1  \in \mathbb{Q}} \right)\left[ {e - 1 < p_1  < e} \right]\;\& \;\left( {\exists q_1  \in \mathbb{Q}} \right)\left[ {e < q_1  < e + 1} \right].

    Let \delta _1  = \max \left\{ {p_1 ,e - \frac{1}{2}} \right\}\;\& \,\varepsilon _1  = \min \left\{ {q_1 ,e + \frac{1}{2}} \right\}.
    \left( {\exists p_2  \in \mathbb{Q}} \right)\left[ {\delta _1  < p_2  < e} \right]\;\& \;\left( {\exists q_2  \in \mathbb{Q}} \right)\left[ {e < q_2  < \varepsilon _1 } \right]<br />

    If n>1
    Let \delta _n  = \max \left\{ {p_n ,e - \frac{1}{n+1}} \right\}\;\& \,\varepsilon _n  = \min \left\{ {q_n ,e + \frac{1}{n+1}} \right\}.
    \left( {\exists p_{n+1}  \in \mathbb{Q}} \right)\left[ {\delta _n  < p_{n+1}  < e} \right]\;\& \;\left( {\exists q_{n+1}  \in \mathbb{Q}} \right)\left[ {e < q_{n+1}  < \varepsilon _n } \right]

    See what you can do with the sequence I_n=\left[p_n,q_n\right].
    Last edited by Plato; October 18th 2009 at 03:47 PM.
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  3. #3
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    I feel certain that {e} is a set with one element, though the problem doesn't specifically state that is the case. However, our text does use the normal symbol as the empty set, so I imagine with the context of this section added, we are talking about a set with one element. Thank you so much!
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    Quote Originally Posted by tarheelborn View Post
    I feel certain that {e} is a set with one element, though the problem doesn't specifically state that is the case. However, our text does use the normal symbol as the empty set, so I imagine with the context of this section added, we are talking about a set with one element. Thank you so much!
    Please read my reply to your post.
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  5. #5
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    Now I am really confused... I think that e (the exponential function) is supposed to be contained in each of the I_n's, and it is the only element that is contained in all of the I_n's. But I am not sure how to define a set of closed intervals such that the intersection of those intervals is e. Thank you.
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  6. #6
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    So would this work? Let I_1 = {e}, I_2 = {e, 1}. I_3 = {e, 1, 2}, ..., I_n= I_n-1 U {n-1}? Thanks.
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  7. #7
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    Quote Originally Posted by tarheelborn View Post
    Now I am really confused... I think that e (the exponential function) is supposed to be contained in each of the I_n's, and it is the only element that is contained in all of the I_n's. But I am not sure how to define a set of closed intervals such that the intersection of those intervals is e. Thank you.
    Well I have given you a sequence of nested closed intervals with rational endpoints.
    The intersection of these intervals is the set \{e\}.
    But using what I gave you, you still must prove it all.
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  8. #8
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    Quote Originally Posted by tarheelborn View Post
    So would this work? Let I_1 = {e}, I_2 = {e, 1}. I_3 = {e, 1, 2}, ..., I_n= I_n-1 U {n-1}? Thanks.
    Of course NOT.
    For one, I_1=\{e\} is not a closed interval, much less with rational endpoints.
    The whole point of this question is this: e is not a rational number.
    BUT e is the limit of an increasing sequence of rational numbers and the limit of an decreasing sequence of rational numbers.
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  9. #9
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    I really appreciate your help and I can almost understand your example, but I believe it is several levels above that of my class. Thanks, though!
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  10. #10
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    OK, now something is sinking in. I apologize for my denseness and I do appreciate you.
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  11. #11
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    Well I hope so.
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