Prove: A bounded sequence contains a convergent subsequence.

In summary: I was told to prove that every bounded sequence with a finite range and an infinite range contains a convergent subsequence. I was using the least-upper-bound and greatest-lower-bound approach when I should have used the Bolzano-Weierstrass approach.In summary, the task was to prove that every bounded sequence with a finite range and an infinite range contains a convergent subsequence. This can be done using the Bolzano-Weierstrass approach rather than the least-upper-bound and greatest-lower-bound approach. The counterexample given in Post #5 shows that looking for a subsequence that converges to the least-upper-bound or greatest-lower-bound of the sequence is not sufficient.
  • #1
Eclair_de_XII
1,083
91

Homework Statement


"Let ##\{a_n\}_{n=1}^\infty## be a bounded, non-monotonic sequence of real numbers. Prove that it contains a convergent subsequence."

Homework Equations


Monotone: "A sequence ##\{\alpha_n\}_{n=1}^\infty## is monotone if it is increasing or decreasing. In other words, if a sequence is increasing, then ##\alpha_k \leq \alpha_{k+1}## for all ##k\in ℕ##. Similarly for if a sequence is decreasing."

The Attempt at a Solution


Note: I was asked to prove that the sequence contains both a convergent finite subsequence and a convergent infinite subsequence. I'm starting with the former:

"Assuming that ##\{a_n\}_{n=1}^\infty## is non-empty, then there exists a least-upper-bound and greatest-lower-bound. Then there exist ##m,M\in ℝ## such that if ##m=inf\{a_n\}_{n=1}^\infty##, and ##M=sup\{a_n\}_{n=1}^\infty##, then ##m \leq a_n \leq M## for all ##n\in ℕ##. Moreover, if ##\{a_n\}_{n=1}^\infty## is not monotonic, then if the sequence is increasing, then there exists a ##u\in ℕ## such that ##a_u > a_{u+1}##.

Then we define a subsequence ##\{a_n\}_{n=1}^u##. Then we have a monotonic, finite subsequence of ##\{a_n\}_{n=1}^\infty## that converges to ##inf\{a_n\}##."

I'm honestly not confident in this proof, though. I'm pretty sure what I wrote wasn't anywhere near the definition of convergence; please disregard the second paragraph...

But what happens if my sequence is ##\{(-1)^n,n\in ℕ\}##? I want to define a convergent subsequence for a sequence that fails to be monotone for more than one term in the sequence. Should I use the sequences from my previous topic in here? Not completely sure what to do here, right now.
 
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  • #2
I don't know what a 'convergent finite subsequence' would be. A finite sequence could just be an ordered finite set of numbers, but how could we make any sense of a statement that it is 'convergent' when the definition of convergence refers to sequence elements with unlimited indices? Indeed, if we apply the usual definition of convergence literally then any finite sequence is convergent to its last element.

I suggest you go back to the person that asked you to include the finite bit and ask them what they mean by a 'finite convergent subsequence'.
 
  • #3
I'll ask for elaboration from my professor. Just give me until morning to give you an answer.
 
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  • #4
Try to imagine what it is about. You have a real interval of a certain length with infinitely many points in it. Now try to place those points and keep them apart. You will simply run out of space.

My suspicion about the "finite subsequence" is, that you misunderstood the wording. Consider this interval and cut it in half. Then not both halves can contain only finitely many ##a_n##. There must be at least one half with infinitely many elements. Do you see, where this leads to?
 
  • #5
Eclair_de_XII said:
Then we define a subsequence ##\{a_n\}_{n=1}^u##. Then we have a monotonic, finite subsequence of ##\{a_n\}_{n=1}^\infty## that converges to ##inf\{a_n\}##."

I'm honestly not confident in this proof, though. I'm pretty sure what I wrote wasn't anywhere near the definition of convergence; please disregard the second paragraph...

To help you a bit. You might think that you can always find a subsequence that converges to the ##sup## or ##inf##. But, consider the sequence:

##0, 2, 1, 1, 1 \dots##

That counterexample to your "proof" shows that looking for a subsequence that converges to ##sup## or ##inf## is not going to work.

I know it's not easy, but you need to start finding these (simple) counterexamples for yourself.
 
  • #6
Eclair_de_XII said:

Homework Statement


"Let ##\{a_n\}_{n=1}^\infty## be a bounded, non-monotonic sequence of real numbers. Prove that it contains a convergent subsequence."

Homework Equations


Monotone: "A sequence ##\{\alpha_n\}_{n=1}^\infty## is monotone if it is increasing or decreasing. In other words, if a sequence is increasing, then ##\alpha_k \leq \alpha_{k+1}## for all ##k\in ℕ##. Similarly for if a sequence is decreasing."

The Attempt at a Solution


Note: I was asked to prove that the sequence contains both a convergent finite subsequence and a convergent infinite subsequence. I'm starting with the former:

"Assuming that ##\{a_n\}_{n=1}^\infty## is non-empty, then there exists a least-upper-bound and greatest-lower-bound. Then there exist ##m,M\in ℝ## such that if ##m=inf\{a_n\}_{n=1}^\infty##, and ##M=sup\{a_n\}_{n=1}^\infty##, then ##m \leq a_n \leq M## for all ##n\in ℕ##. Moreover, if ##\{a_n\}_{n=1}^\infty## is not monotonic, then if the sequence is increasing, then there exists a ##u\in ℕ## such that ##a_u > a_{u+1}##.

Then we define a subsequence ##\{a_n\}_{n=1}^u##. Then we have a monotonic, finite subsequence of ##\{a_n\}_{n=1}^\infty## that converges to ##inf\{a_n\}##."

I'm honestly not confident in this proof, though. I'm pretty sure what I wrote wasn't anywhere near the definition of convergence; please disregard the second paragraph...

But what happens if my sequence is ##\{(-1)^n,n\in ℕ\}##? I want to define a convergent subsequence for a sequence that fails to be monotone for more than one term in the sequence. Should I use the sequences from my previous topic in here? Not completely sure what to do here, right now.

The insight provided by the counterexample of Perok in Post #5 shows why your argument won't work. However, you can replace ##\inf \{a_n\}## and ##\sup \{ a_n\}## by ##\liminf \{a_n\}## and ##\limsup \{a_n\}## and then proceed.
 
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  • #7
fresh_42 said:
My suspicion about the "finite subsequence" is, that you misunderstood the wording. Consider this interval and cut it in half. Then not both halves can contain only finitely many ##a_n##. There must be at least one half with infinitely many elements. Do you see, where this leads to?

It leads to the Bolzano-Weierstrauss Theorem. In any case, I greatly misunderstood what he said. He said to prove that every bounded sequence with a finite range and an infinite range contains a convergent subsequence. It greatly simplified what I was asked to do. In any case, I no longer need help with this problem.

PeroK said:
That counterexample to your "proof" shows that looking for a subsequence that converges to ##\text{sup}## or ##\text{inf}## is not going to work.

Duly noted.
 

Related to Prove: A bounded sequence contains a convergent subsequence.

1. What is a bounded sequence?

A bounded sequence is a sequence of numbers that is limited or confined within a certain range. This means that the values in the sequence do not exceed a certain upper bound and do not fall below a certain lower bound.

2. What does it mean for a sequence to be convergent?

A convergent sequence is one in which the terms of the sequence approach a certain fixed value as the index of the sequence increases. In other words, as the sequence progresses, the numbers get closer and closer to a specific value.

3. How do you prove that a bounded sequence contains a convergent subsequence?

To prove that a bounded sequence contains a convergent subsequence, you must show that there is a subsequence within the original sequence that has a limit. This can be done by finding a subsequence that is decreasing or increasing and using the Monotone Convergence Theorem.

4. Can a bounded sequence contain more than one convergent subsequence?

Yes, a bounded sequence can contain multiple convergent subsequences. This is because a sequence can have multiple subsequences that approach different limits as the index increases.

5. Why is it important to prove that a bounded sequence contains a convergent subsequence?

Proving that a bounded sequence contains a convergent subsequence is important because it is a fundamental concept in real analysis and is used to prove many theorems and properties. Additionally, understanding the convergence of sequences is crucial in many areas of mathematics and science, such as calculus and statistics.

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