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Homework Help: Easy convergent subsequence question.

  1. Oct 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider the sequence {x_k} = {(arctan(k^2+1),sink)} in R^2. Is there a convergent
    subsequence? Justify your answer.

    2. Relevant equations

    Every bounded sequence in R^n has a convergent subsequence.

    3. The attempt at a solution
    To show {x_k} is bounded: The range of arctan(k^2+1) is (-pi/2, pi/2) and the range of sin(k) is [-1,1].

    A sequence is bded from above and below if there exist an M and an m both belong to R, such that a_k<=M and a_k>=m for all a_k in the sequence. Here, I want to state let M=1 and m=-1. because they are max and min y coordinate value for {x_k}, is this correct?
  2. jcsd
  3. Oct 3, 2008 #2


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    No, it's not. But you're close. What you've stated is not the correct definition of a bounded sequence in R^2.
  4. Oct 3, 2008 #3


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    Actually, I think that depends on the precise definition of "bounded" in R2 in your text and, in turn, upon the norm used.

    Morphism's point (I think) is that you have shown that the x and y of (x,y) are each bounded. For R2 you must show that |(x,y)|= [itex]\sqrt{x^2+ y^2}[/itex] has an upper bound. That, of course, is easy.

    It is also possible to define |(x,y)| as max |x|,|y| as a norm on R2 in which case you are done! Of course, that is not the standard norm.
  5. Oct 3, 2008 #4
    we just have 2 similar concepts on the text, one is

    every bded sequence in R has a convergent subsequence

    the second is what I stated before..

    every bded sequence in R^n has a convergent subsequence

    or should I use the definition of upper and lower bd?
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