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Math Help - Accumlation Points

  1. #1
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    Accumlation Points

    Find all accumulation points of the following set in R^2 with the usual metric.

    { (1/n+m , 1/n + 1/m) : n, m ∈ N }



    If anyone helps me, I will be thankful.
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  2. #2
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    Here is a starting point.
    Is (0,0) an accumulation point of that set? Why or why not?

    Is (0, \frac{1}{2}) an accumulation point of that set? Why or why not?

    Is (1,0) an accumulation point of that set? Why or why not?
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Is the answer (in white) all the numbers in [0,2]X[0,2] since this is the set of all rationals in (0,2]X(0,2]and the rationals are dense in the reals?
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    Is the answer (in white) all the numbers in [0,2]X[0,2] since this is the set of all rationals in (0,2]X(0,2]and the rationals are dense in the reals?
    Find a pair of positive integers n,m such that (1/n)+(1/m) is in (3/2,2)

    CB
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  5. #5
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    I think the answer should be [0,1/2] X [0,2] . Am I right?

    Since when n=m=1, (1/2, 2)

    when n=m=2, (1/4, 1)

    when n and m go to infinity, (0,0)
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  6. #6
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    Quote Originally Posted by selinunan View Post
    I think the answer should be [0,1/2]X [0,2] . Am I right?
    I think that both poster and Mathstud28 need to consider the makeup of the set
    S = \left\{ {\left( {\frac{1}{{n + m}},\frac{1}{n} + \frac{1}{m}} \right):\left\{ {n,m} \right\} \subseteq \mathbb{Z}^ +  } \right\}.
    Note that the set S is in the quarter closed disk, centered at (0,0) with radius <br />
\frac{{\sqrt {17} }}{2}.
    The point furthest from the origin is \left( {\frac{1}{2},2} \right) happens when n = 1\,\& \,m = 1.
    If we let n = 1\,\& \,m = k then we get the points \left( {\frac{1}{1+k},\frac{k+1}{k}} \right).

    Now, we must remind ourselves of the definition of accumulation point.
    If p is an accumulation point of S then p must be the limit of a sequence of distinct points in S.

    Again think about letting n = 1\,\& \,m = k then we get the points \left( {\frac{1}{1+k},\frac{k+1}{k}} \right) as k \to \infty .

    Does that help you to think about this set?
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  7. #7
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    Then, the only accumulation point is (0,0), isn't it?

    Because as n and m goes to infinity, the set goes to (0,0) and D( (0,0), r) contains at least one point of the set.

    Is it true now?
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  8. #8
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    Quote Originally Posted by selinunan View Post
    the only accumulation point is (0,0), isn't it? NO
    Because as n and m goes to infinity,
    If you hold n fixed, n = J\,\& \,m = k then we get the points \left( {\frac{1}{J+k},\frac{k+J}{Jk}} \right) as k \to \infty
    you get the limit \left( 0 ,\frac{1}{J} \right) .
    Do you see that?
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  9. #9
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    But it will change if you hold m constant.

    If you hold m fixed,then the limit will be (0, 1/k).

    Then is it possible to say that the accumulation points are (0,1/m) and (0,1/n)?

    Sorry for asking too much but I've got to learn that.
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  10. #10
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    Quote Originally Posted by selinunan View Post
    But it will change if you hold m constant. NO!
    Take a good look at the definition of S.
    The sums are symmetric. So holding n constant and varying m yields the same limit as holding m constant and varying n for a given constant.
    The set of accumulation points of S is \left\{ {\left( {0,0} \right)} \right\} \cup \left\{ {\left( {0,\frac{1}{J}} \right):J \in \mathbb{Z}^ +  } \right\}.
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