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Math Help - Questions on Proof for "Every bounded seq has a convergent subseq"

  1. #1
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    Questions on Proof for "Every bounded seq has a convergent subseq"

    Hello everyone,

    I have two questions with a proof for the theorem that "every bounded sequence has a convergent subsequence" in R.

    I have typed my questions and the proof below.

    Thank you very much. Please let me know if the whole proof should be shown.

    scherz0

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    Attached Thumbnails Attached Thumbnails Questions on Proof for "Every bounded seq has a convergent subseq"-p27thm118.jpg  
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  2. #2
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    By definition, a sequence of real numbers is a function  f:N \rightarrow R where we denote each term  f(n) = x_n. So, every sequence has infinitely many terms (although a term may be repeated infinitely many times). Since  (x_n) is bounded, it is entirely contained in the interval  [a,b] . Splitting the interval in half, we see that the sequence is entirely contained within  [a, \frac{1}{2}(a+b)] \cup [\frac{1}{2}(a+b), b] , since this union is the same as the original interval. Now, since the sequence has infinitely many terms (though not necessarily infinitely many distinct terms), one of those two halves of [a,b] has to contains  x_n for infinitely many of the ns, otherwise there would be terms outside the interval [a,b] (which we're assuming is not the case).

    Does that help?

    Edit: It doesn't say that the sub-interval contains every term of the sequence. It just says that it contains an infinite number of the terms.
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  3. #3
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    Have you left something out? Does the sequence contain infinitely many distinct terms?

    The usual way to prove this is a two step process.
    Theorem: Every sequence contains a monotone subsequence.

    Theorem: Every bounded monotone sequence converges.
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  4. #4
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    Thank you for your responses, Math Major and Plato.

    @MathMajor: I understand a bit more of my questions now. However, why can it be true that only one subinterval contains  x_n for infinitely many  n ?

    If you have infinitely many terms within  [a,b] , wouldn't each of the two subintervals still have infinitely many terms, after the partition?

    For instance, if I consider the real number on  [0,1] , then if I split this interval into  [0, 0.5] \cup [0.5, 1] , each subinterval also contains an infinite number of terms.

    Thank you again!
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  5. #5
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    It could certainly be true that both of them contain infinitely many of the terms, but it could also be the case that one of them doesn't. Consider the sequence  (x_n) defined by  x_n = 1/n . Then this sequence is bounded and clearly contained in the interval  [0, 2] . Splitting this into two sub intervals, we can conclude that  (x_n) is in at least one of  [0,1] or  [1,2] for infinitely many n. But notice that it is not in the second subinterval, [1,2] infinitely many times (actually, it's only in there once). All that's required for this proof though is that at least one (though possibly both) of the subintervals contain infinitely many of the terms.
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  6. #6
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    Ah ok! Now I understand!

    Thank you very much for your help, Math Major.
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