Subsequence of Non-Bounded Sequence, and fun Limit action

In summary: This is still not sufficient. You need to show that the sequence has a subsequence which increases without bound in absolute value.
  • #1
mattmns
1,128
6
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!
 
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  • #2
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.
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.
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.
 
  • #3
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|}].
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
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  • #7
0rthodontist said:
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|).
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.
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.
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.
 
  • #8
AKG said:
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.
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?
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).
You're right--I misread you in that case, somehow I skipped over the word "unbounded" for b, misreading it as infinite length.
 

What is a subsequence of a non-bounded sequence?

A subsequence of a non-bounded sequence is a sequence that is obtained by selecting elements from the original sequence in a specific order without changing their relative positions. Unlike a bounded sequence, a non-bounded sequence does not have a finite limit or bound, so its subsequence can include an infinite number of terms.

How is a subsequence of a non-bounded sequence different from a bounded sequence?

A subsequence of a non-bounded sequence is different from a bounded sequence in that it does not have a finite limit or bound. This means that its terms can increase or decrease without limit, and its subsequence can include an infinite number of terms. In contrast, a bounded sequence has a finite limit or bound, and its subsequence can only include a finite number of terms.

What is the significance of studying subsequence of non-bounded sequences?

Studying subsequence of non-bounded sequences is important because it allows us to understand the behavior of a sequence without a finite limit or bound. This can help us make predictions about the trend or pattern of the sequence, and also has applications in fields such as mathematics, physics, and computer science.

Can a subsequence of a non-bounded sequence have a limit?

Yes, a subsequence of a non-bounded sequence can have a limit. This limit may or may not be the same as the limit of the original non-bounded sequence. The limit of a subsequence is determined by the behavior of its terms, which can be different from the behavior of the terms in the original sequence.

What is "limit action" in relation to subsequence of non-bounded sequences?

"Limit action" refers to the behavior or trend of a sequence as its terms approach a limit. In the case of subsequence of non-bounded sequences, "limit action" refers to the behavior of the terms in the subsequence as they approach a limit, which may or may not be the same as the limit of the original sequence.

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