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Subsequence of Non-Bounded Sequence, and fun Limit action

  1. Nov 4, 2006 #1
    Hello, here is the exercise from the book:
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    Let [itex](a_{n})_{n=0}^{\infty}[/itex] be a sequence which is not bounded. Show that there exists a subsequence [itex](b_{n})_{n=0}^{\infty}[/itex] of [itex](a_{n})_{n=0}^{\infty}[/itex] such that the limit [itex]\lim_{n\rightarrow \infty} 1/b_{n} = 0[/itex]. (Hint: for each natural number j, introduce the quantity [itex]n_{j} := min\{n \in \mathbb{N} : |a_{n}| \geq j \}[/itex], first by explaining why the set [itex]\{n \in \mathbb{N} : |a_{n}| \geq j \}[/itex] is non-empty. Then set [itex]b_{j} := a_{n_{j}}[/itex].)

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    First the hint seems incorrect to me. What I mean is that we could have a situation with b1 = b2 = b3 = a1. And have only one element in the sequence with the value of a1. Consider the following: An = -4,-3,-2,-1,0,1,2,3,4,.... This sequence is not bounded (a sequence {An} is bounded by a real number M iff |An| <= M for all n.) so it meets the requirement. But when we look at n0, n1, n2, we get the same value, 0. Thus our sequence bn would start out: -4,-4,-4,-4, then the rest, but this is not necessarily a subsequence of An. Am I missing something here, or is the hint incorrect?

    Regardless of that I think I have the idea behind a (the) solution. First let us pick b1 as the first element in An that is larger than 1. Then we will pick b2 as the next element in An that is larger than b1, which we can do since An is not bounded, and then continue. This will give us a sequence that is strictly increasing. Then when we look at 1/bn we will have a sequence that is strictly decreasing, but more importantly we will have [itex]0 \leq 1/b_{n} \leq 1/n[/itex] so that when we take the limit we can use the squeeze test to get [itex]\lim_{n\rightarrow \infty} 1/b_{n} = 0[/itex]. My problem with this is that I do not know how to formalize this, which is what I think our professor wants, and also I would like to know how to do such a thing. What I mean by formalize is give a more explicit definition of what bn is. Any ideas?

    Thanks!
     
    Last edited: Nov 4, 2006
  2. jcsd
  3. Nov 5, 2006 #2

    0rthodontist

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    Yeah, the hint is flawed, unless "subsequence" in this context allows repeated elements.

    You may have the right general idea, but you need to refine it.
    There might be no such element. An unbounded sequence can be bounded from above so long as it's not also bounded from below. You want b2 to be the next element in the sequence whose _absolute value_ is larger than b1, and larger by at least 1.
     
  4. Nov 5, 2006 #3

    AKG

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    There must be an unbounded subsequence which strictly increases in absolute value, otherwise the original, supposedly unbounded sequence would be bounded within [-max{|a1|, |a2|}, max{|a1|, |a2|}].
     
  5. Nov 5, 2006 #4

    0rthodontist

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    That doesn't follow. Also note that you don't just need it to increase in absolute value, because there are bounded sequences with that property.
     
  6. Nov 5, 2006 #5

    AKG

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    Sure it follows. And not only does it increase in absolute value, but since it's an unbounded subsequence, it increases without bound in absolute value, which is what you want. In fact, this is exactly what the hint says, the only difference is it gives a more explicit construction of this sequence.
     
  7. Nov 5, 2006 #6

    0rthodontist

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    The sequence {ak} where a1 = a2 = an = 0 for n > 3, but a3 = 2, has no unbounded subsequence that is increasing in absolute value, but it is not bounded by max(|a1|, |a2|). Even if what you said about a1 and a2 were true, you were trying to show by contradiction that "There must be an unbounded subsequence which strictly increases in absolute value," which is not sufficient for the problem. As mattmns pointed out, the hint is not correct since nj as defined in the hint can equal nk for j not equal to k, so b as defined in the hint need not be a subsequence of a.
     
    Last edited: Nov 5, 2006
  8. Nov 5, 2006 #7

    AKG

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    But your sequence (an) isn't unbounded. I didn't mean to say that any sequence that isn't bounded by max(|a1|, |a2|) has an unbounded subsequence that increases strictly in absolute value.
    If |bn| increases strictly without bound, then for any e > 0, there is a natural N such that for all n > N, |1/bn| < e. So (|1/bn|) converges to 0, hence so does (1\bn). As mattmns pointed out, the hint is not correct since nj as defined in the hint can equal nk for j not equal to k, so b as defined in the hint need not be a subsequence of a.[/QUOTE]That's right. But there's an obvious way to rework the hint. The general idea works though.
     
  9. Nov 5, 2006 #8

    0rthodontist

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    Well, just what did you mean then? The only interpretation I can see is,
    "if a sequence {an} does not contain an unbounded subsequence of strictly increasing absolute value, then {an} is bounded within [-max{|a1|, |a2|}, max{|a1|, |a2|}]." Did you have something else in mind?
    You're right--I misread you in that case, somehow I skipped over the word "unbounded" for b, misreading it as infinite length.
     
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