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Subsequencial limit set of a product sequence SiTi

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Let {si} and {ti} be bounded sequences of real numbers, let E, F, G be the sets of all subsequential limit points of {si}, {ti}, {siti} respectively. Prove that [tex]G\subseteq EF= \left\ ef|e\in E,f\in F \right\[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I have trouble understanding the behavior of the sequence siti. For this question, it seems that the right thing to do is to choose any x in G, show that x is in EF. Here's what I've done:
    There is a subsequence [tex]s_{\alpha _{i}}t_{\alpha_{i}}[/tex] that converges to x. Now, consider the sequences [tex]{s_{\alpha _{i}}}[/tex] and [tex]{t_{\alpha _{i}}}[/tex]. Being a sequence in the compact set (i.e. [-M,M] in R), some subsequence of [tex]{s_{\alpha _{i}}}[/tex] converges to a point a, let A be the set of these subscripts. And some subsequence of [tex]{t_{\alpha _{i}}}[/tex] converges to a point b, Let B be the set of these subscripts.
    If [tex]A\cap B[/tex] has infinitely many elements, then [tex]x=ab\in EF[/tex]. But what if [tex]A\cap B[/tex] is empty? I've no idea why this cannot happen. Any hint would be greatly appreciated! Is there any different way to prove it?
     
    Last edited: Nov 15, 2008
  2. jcsd
  3. Nov 15, 2008 #2

    Office_Shredder

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    Don't work with both sequences at the same time.

    [tex]s_{a_i}t_{a_i}[/tex] converges to x. Then [tex]s_{a_i}[/tex] has a convergent subsequence [tex]s_{a_{i_j}}[/tex]. What does the sub-subsequence [tex]t_{a_{i_j}}[/tex] do? Remember, you know [tex]s_{a_{i_j}}t_{a_{i_j}}[/tex] converges to x also
     
  4. Nov 16, 2008 #3
    yea, I think I get it now.
    if s_i (denotes the sub-subsequence...) ->a. a=0 is a trivial case. If a!=0, then the corresponding t_i converges to b such that ab=x, since
    |a*t_i-ab|<=|a*t_i-s_i*t_i|+|s_i*t_i-x|<=t_i*|s_i-a|+|s_i*t_i-x|<=Me+e
    Thanks a lot!
     
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