- #1

Eclair_de_XII

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## 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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