Prove: A bounded sequence contains a convergent subsequence.

[Eclair_de_XII]

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.

 

[andrewkirk]

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'.

 

[Eclair_de_XII]

I'll ask for elaboration from my professor. Just give me until morning to give you an answer.

 

[fresh_42]

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?

 

[PeroK]

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.

 

[Ray Vickson]

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.

 

[Eclair_de_XII]

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.

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.