We are tasked with using the Bolzano-Weierstrass theorem to prove that every bounded sequence has a monotone subsequence. The Bolzano-Weierstrass theorem states that every bounded sequence has a convergent subsequence. We'll leverage this fact to find a monotone subsequence within a bounded sequence.

Proof Outline:

1. Given: A bounded sequence $\{x_n\}_{n \in \mathbb{N}}$, meaning there exists some $M > 0$ such that $|x_n| \leq M$ for all $n \in \mathbb{N}$.

2. By the Bolzano-Weierstrass theorem: There exists a convergent subsequence $\{x_{n_k}\}_{k \in \mathbb{N}}$, converging to some limit $L \in \mathbb{R}$. Let: $L = \lim_{k \to \infty} x_{n_k}$

3. Now, distinguish two cases based on how the terms of $\{x_{n_k}\}_{k \in \mathbb{N}}$ relate to the limit $L$.

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Case 1: $x_{n_k} \geq L$ for infinitely many $k$

In this case, there exists an infinite subsequence of $\{x_{n_k}\}$ such that $x_{n_k} \geq L$ for all sufficiently large $k$. Let's define this subsequence as $\{x_{m_k}\}_{k \in \mathbb{N}}$, which is such that: $x_{m_k} \geq L \quad \text{for all } k \in \mathbb{N}.$ Since $\{x_{m_k}\}$ converges to $L$ (as it's a subsequence of $\{x_{n_k}\}$), and all its terms are greater than or equal to $L$, the subsequence must be monotonically decreasing as $k$ increases. Therefore, $\{x_{m_k}\}$ is a monotone decreasing subsequence of the original sequence $\{x_n\}$.

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Case 2: $x_{n_k} < L$ for all but finitely many $k$

In this case, there exists an infinite subsequence of $\{x_{n_k}\}$ such that $x_{n_k} \leq L$ for all but finitely many $k$. Let this infinite subsequence be $\{x_{p_k}\}_{k \in \mathbb{N}}$, which is such that: $x_{p_k} \leq L \quad \text{for all } k \in \mathbb{N}.$ Since $\{x_{p_k}\}$ converges to $L$ and all terms are less than or equal to $L$, it follows that $\{x_{p_k}\}$ is a monotonically increasing subsequence of the original sequence $\{x_n\}$.

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Conclusion:

In both cases, we have constructed a monotone subsequence:

• In Case 1, a monotonically decreasing subsequence exists.
• In Case 2, a monotonically increasing subsequence exists.

Since every bounded sequence has a convergent subsequence (by Bolzano-Weierstrass), and every convergent subsequence can be analyzed as above, every bounded sequence must have a monotone subsequence.

This completes the proof.

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Would you like more details about any of the steps? Here are 5 follow-up questions that can deepen your understanding:

1. Why is the Bolzano-Weierstrass theorem important for the existence of a convergent subsequence?
2. How does the boundedness of a sequence ensure that a convergent subsequence exists?
3. What happens if a sequence is not bounded? Can we still find a monotone subsequence?
4. How is the definition of a limit used to guarantee that the subsequences behave monotonically?
5. Can this result be extended to sequences in metric spaces?

Tip: When working with limits and subsequences, always think about the "tail behavior" of the sequence — what happens to terms as you move further along the sequence — this often guides you to construct useful subsequences.