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Math Help - Proof - sets

  1. #1
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    Proof - sets

    Hey guys, after spending a few hours on this question and not getting anywhere, I think I need some help.

    Question

    \bigcap_{n \in N} A_n denotes the intersection of the sets A_n and  n \in N and  \bigcup_{n \in N} A_n denotes the union of the sets A_n and  n \in N . ( N is the set of Natural Numbers)

    Prove that  (0, \frac{4}{3} ] = \bigcup_{n \in N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]

    and prove that  \{1\} = \bigcap_{n \in N} \left[1, 1+ \frac{1}{n} \right]

    __________________________________________________ _______________

    We have been learning about the Archimedean property, so I've been trying to mess around with that. I mean the second one is fairly obvious. 1 will be included in the set for any n. And therefore it is the only number than is always in the set of natural numbers and the set [1, 1+ 1/n]...but I don't know how to write this formally.

    Regarding the second one, again, I can see that 1/(n+2) tends to 0 as n tends to infinity. Perhaps I need to prove this via the archimedean property? But I don't think that'd be enough. 4/3 is the highest that 1+ 1/n+2 (n=1) could possibly be if n belongs to the set of natural numbers...but I'm just stating the obvious here.

    Please help!
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  2. #2
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    Because I do not know the level of rigor demanded in your course, I can only give general directions.

    For #1. We know that \frac{4}{3} \in U_1  = \left[ {\frac{1}{3},\frac{4}{3}} \right] and \forall n\left[ {1 + \frac{1}{{n + 2}} \le \frac{4}{3}} \right]<br />
so that takes care of the right endpoint being included. Next suppose that 0 < t < 1 then \left( {\exists J} \right)\left[ {\frac{1}{{J + 2}} < t} \right]\quad  \Rightarrow t \in \left[ {\frac{1}{{J + 2}},1 + \frac{1}{{J + 2}}} \right]. That should suffice for the left endpoint. You may need to point out that \forall n\left[ {0 \notin \left[ {\frac{1}{{n + 2}},1 + \frac{1}{{n + 2}}} \right]} \right].
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  3. #3
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    Thanks for the reply, Plato!

    Quote Originally Posted by Plato
    Because I do not know the level of rigor demanded in your course, I can only give general directions.

    For #1. We know that \frac{4}{3} \in U_1  = \left[ {\frac{1}{3},\frac{4}{3}} \right] and \forall n\left[ {1 + \frac{1}{{n + 2}} \le \frac{4}{3}} \right]<br />
    Ah, clever - I understand this bit.

    Quote Originally Posted by Plato
    Next suppose that 0 < t < 1 then \left( {\exists J} \right)\left[ {\frac{1}{{J + 2}} < t} \right]\quad  \Rightarrow t \in \left[ {\frac{1}{{J + 2}},1 + \frac{1}{{J + 2}}} \right]. That should suffice for the left endpoint.
    I'm probably being very dull here, but why does that suffice? I don't understand much of this. I see why t is in that interval, but what does this mean about 0? Why does this show that 0 is the 'lowest' possible point? If it's not too much trouble, can you please explain this a bit more?

    Quote Originally Posted by Plato
    You may need to point out that \forall n\left[ {0 \notin \left[ {\frac{1}{{n + 2}},1 + \frac{1}{{n + 2}}} \right]} \right].
    Not a problem, I think I should be able to prove this without any difficulty.

    __________________________________________________ __________

    Using the 'trick' you used for the first bit of this question, I will attempt a formal proof of #2 and post it. Thanks.
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  4. #4
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    Quote Originally Posted by Joel24 View Post
    I'm probably being very dull here, but why does that suffice? I don't understand much of this. I see why t is in that interval, but what does this mean about 0? Why does this show that 0 is the 'lowest' possible point? If it's not too much trouble, can you please explain this a bit more?
    That shows that for any positive number less that 1 belongs to one of those intervals.
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  5. #5
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    Quote Originally Posted by Plato View Post
    That shows that for any positive number less that 1 belongs to one of those intervals.
    *D'oh* Of course!

    I think I've completed the second proof relatively ok, so this thread can be closed.

    Thanks for your help, Plato.
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  6. #6
    Senior Member JaneBennet's Avatar
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    To show that two sets A and B are equal, you must show that A\subseteq B and B\subseteq A. So to show that \left(0,\frac{4}{3}\right]=\bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right], show both (i) and (ii) below:

    \color{white}.\quad. (i) \left(0,\frac{4}{3}\right]\ \subseteq\ \bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]


    Let x\in\left(0,\frac{4}{3}\right]. First, note that \forall\,n\in\mathbb{N}, 1+\frac{1}{n+2}\leq1+\frac{1}{3}=\frac{4}{3}

    If 0<x\leq1, then choose an N\in\mathbb{N} such that \frac{1}{N+2}\leq x. (This always exists, by the Archimedean property of \mathbb{N}.) Then \frac{1}{N+2}\leq x\leq1\leq1+\frac{1}{N+2}, i.e. x\in\left[\frac{1}{N+2},1+\frac{1}{N+2}\right].

    If 1<x\leq\frac{4}{3}, then \frac{1}{3}<1<x\leq\frac{4}{3}, i.e. x\in\left[\frac{1}{1+2},1+\frac{1}{1+2}\right].

    So x\in\left[\frac{1}{n+2},1+\frac{1}{n+2}\right] for some n\in\mathbb{N}, i.e. x\in\bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]. This proves (i).


    \color{white}.\quad. (ii) \bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]\subseteq\left(0,\frac{4}{3}\right].

    Let x\in\bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]. Thus x\in\left[\frac{1}{n+2},1+\frac{1}{n+2}\right] for some n\in\mathbb{N}

    \Rightarrow\ 0<\frac{1}{n+2}\leq x\leq1+\frac{1}{n+2}\leq\frac{4}{3}

    \Rightarrow\ x\in\left(0,\frac{4}{3}\right]. This proves (ii).


    (i) & (ii) \Rightarrow\ \left(0,\frac{4}{3}\right]=\bigcup_{n\,\in\,N} \left[ \frac{1}{n+2} , 1 + \frac{1}{n+2} \right]
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