- #1

Bashyboy

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## Homework Statement

Given that ##\{x_n\}## is a bounded, divergent sequence of real numbers, which of the following must be true?

(A) ##(x_n)## contains infinitely many convergent subsequences

(B) ##(x_n)## contains convergent subsequences with different limits

(C) The sequence whose ##n##-th term is ##y_n = \min \{ x_k ~|~ k \le n \}## converges

(D) All of the above

(E) Only options (A) and (C)

## Homework Equations

## The Attempt at a Solution

Here is my justification for why (D) is the answer. Since the sequence is bounded, it has a convergent subsequence by the BW theorem. Clearly this convergent sequence has an infinite number of convergent subsequences (just remove the single ##n##-th term for each ##n##). Since these sub-subsequences are themselves subsequence of ##(x_n)##, we see that ##(x_n)## has infinitely many convergent subsequences. Since ##\{ x_k ~|~ k \le n \} \subseteq \{ x_k ~|~ k \le n+1 \}##, we get ##\min \{ x_k ~|~ k \le n+1 \} \le \min \{ x_k ~|~ k \le n \}## or ##y_{n+1} \le y_n##, implying that ##(y_n)## is a decreasing, bounded sequence and therefore must converge. Similarly, the sequence ##z_n = \max \{ x_k ~|~ k \le n \}## is increasing and bounded, so it must converge.

Here is where things get fun. Intuitively I can see that the ##y_n## and ##x_n## have different limits, but I could use some help in justifying this. Initially I thought that ##y_n## would converge to ##inf \{x_n ~|~ n \in \Bbb{N} \} = \inf (S)## and ##z_n## would converge to ##\sup (S)##, and if it were the case that they had the same limit, ##\inf (S) = \sup (S)## would imply ##S## is a singleton, which is a contradiction. But I know recognize that this isn't necessarily the case . What

*is true*is that ##y_n## will converge to the infimum of ##S_m = \{y_n ~|~ n \in \Bbb{N} \} \subseteq S##, and ##z_n## will converge to the supremum of ##S_M = \{z_n ~|~ n \in \Bbb{N} \}##, but this doesn't seem to pose any issue if I assume that ##y_n## and ##z_n## converge to the same value.

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