Help on interpreting subsequential limit proof

[shoeburg]

Homework Statement

My analysis study guide asks me to prove the following:
If a_n is a sequence of real numbers whose only subsequential limit in the extended reals is finite, then a_n is bounded.

Homework Equations

The Attempt at a Solution

Is it right to say that since it has only one subsequential limit, call it L, which is finite in the extended reals (meaning we were considering infinity, so it for sure does not diverge), then since a_n is a subsequence of itself, then it also converges to L? If this were true, I know how to show it is bounded. What makes me think this is wrong is that, below the problem, the study guide asks "additionally, show that a_n is convergent under the above conditions." But I was going to use its convergence to show its boundedness.. any help appreciated! got a big test coming up next week.

 

[Dick]

Homework Statement

My analysis study guide asks me to prove the following:
If a_n is a sequence of real numbers whose only subsequential limit in the extended reals is finite, then a_n is bounded.

Homework Equations

The Attempt at a Solution

Is it right to say that since it has only one subsequential limit, call it L, which is finite in the extended reals (meaning we were considering infinity, so it for sure does not diverge), then since a_n is a subsequence of itself, then it also converges to L? If this were true, I know how to show it is bounded. What makes me think this is wrong is that, below the problem, the study guide asks "additionally, show that a_n is convergent under the above conditions." But I was going to use its convergence to show its boundedness.. any help appreciated! got a big test coming up next week.

How did you show that the sequence is bounded? Having a limit point doesn't show a sequence converges. You have to give an argument.

 

[shoeburg]

How did you show that the sequence is bounded? Having a limit point doesn't show a sequence converges. You have to give an argument.

That is exactly my question. I'm asking if it's right to say, that since it has only one subsequential (finite) limit, call it L, then must the original sequence a_n also converge to that L? If this is true, I know how to proceed to show it is bounded. I'm just asking if this bridging step is correct.

The definition of subsequential limit I'm using is: if L is a subsequential limit, then there exists some subsequence that converges to that L.

 

[Dick]

That is exactly my question. I'm asking if it's right to say, that since it has only one subsequential (finite) limit, call it L, then must the original sequence a_n also converge to that L? If this is true, I know how to proceed to show it is bounded. I'm just asking if this bridging step is correct.

The definition of subsequential limit I'm using is: if L is a subsequential limit, then there exists some subsequence that converges to that L.

It is true that if there is one finite limit point and the sequence is bounded that the sequence converges. But you have to give an argument showing that. And first you have to show that it's bounded. You said you did it, but you can't do it by ASSUMING that the sequence converges. You know +infinity is not a limit point. What does that tell you?

 

[shoeburg]

Ohh, I think I see what you're saying. Here's what I have:

Let e=1. Let a_n be a sequence, let it have only one subsequential limit that is finite, call it L. Then there exists a subsequence b_n which converges to L.
Then there exists a natural number N such that for all n>N, abs(b_n - L) < 1.
Then b_N - 1 < b_n < b_N +1, so the set {b_n for n>N} is bounded.
The set {b_n for n <= N}, being a finite set, is also bounded.
The union of these two sets then is also bounded, so b_n is bounded.
Since b_n is bounded, then a_n is also bounded??

The bridge between talking about b_n and a_n still kinda confuses me. I appreciate your help.

 

[Dick]

Ohh, I think I see what you're saying. Here's what I have:

Let e=1. Let a_n be a sequence, let it have only one subsequential limit that is finite, call it L. Then there exists a subsequence b_n which converges to L.
Then there exists a natural number N such that for all n>N, abs(b_n - L) < 1.
Then b_N - 1 < b_n < b_N +1, so the set {b_n for n>N} is bounded.
The set {b_n for n <= N}, being a finite set, is also bounded.
The union of these two sets then is also bounded, so b_n is bounded.
Since b_n is bounded, then a_n is also bounded??

The bridge between talking about b_n and a_n still kinda confuses me. I appreciate your help.

You are getting a little tangled up with this whole subsequence thing. Just try to show it's bounded first. +infinity in the extended reals is not a limit point. What does that tell you?

 

[shoeburg]

I don't know why this is so hard for me to think about haha. +infinity not being a limit point means that the entire sequence is less than some number. Could I maybe start the proof with: Since the only subsequential limit of a_n is finite, and since a_n is a subsequence of itself, then I can find some B where a_n < B for all n.

 

[Dick]

I don't know why this is so hard for me to think about haha. +infinity not being a limit point means that the entire sequence is less than some number. Could I maybe start the proof with: Since the only subsequential limit of a_n is finite, and since a_n is a subsequence of itself, then I can find some B where a_n < B for all n.

You keep bringing up this subsequence thing and that's confusing you right now. If +infinity is not a limit point then there is some neighborhood of infinity [M,+infinity) that contains only a finite number of elements in your sequence. So?

 

[shoeburg]

Well the set of those finite elements has got to be bounded above by some number A. And because of your construction of the neighborhood, the rest of the sequence has got to be less than that same A (because they will all be less than M)? Hence the whole sequence is bounded?

 

[Dick]

Well the set of those finite elements has got to be bounded above by some number A. And because of your construction of the neighborhood, the rest of the sequence has got to be less than that same A (because they will all be less than M)? Hence the whole sequence is bounded?

Yes, they are all bounded above by A. Same thing for -infinity. So now the elements of your sequence are contained in [-R,R] for some real R. And you have exactly one limit point -R<=L<=R. Can you show the limit must be L? I think you need to use compactness of closed intervals.

 

[shoeburg]

The study guide says showing that it is convergent is a bonus, so for now I'll probably prioritize on the other things I need to study. Thank you for your patience and your help! This one had been bugging me.

 

[Dick]

The study guide says showing that it is convergent is a bonus, so for now I'll probably prioritize on the other things I need to study. Thank you for your patience and your help! This one had been bugging me.

Fair enough. Very welcome.