Bounded, Divergent Sequence

[Bashyboy]

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.

 

[FactChecker]

Your statement of B) seems incomplete. Is B) the reason you are trying to prove that the sequences of mins converge to a different limit than the sequence of maxes? Otherwise, I don't see why you are trying to prove that.

 

[Bashyboy]

Your statement of B) seems incomplete. Is B) the reason you are trying to prove that the sequences of mins converge to a different limit than the sequence of maxes? Otherwise, I don't see why you are trying to prove that.

Sorry about that! I believe everything is in order now.

 

[FactChecker]

Are you sure that you have the inequality of the statement of C) correct? It seems like y_n = min{ x_k, k≥n } would tell you more. The way you have it, if x₁ = -999 and all other x_is > 0, then y_n = x₁ = -999 doesn't tell you much.

 

[Bashyboy]

Are you sure that you have the inequality of the statement of C) correct? It seems like y_n = min{ x_k, k≥n } would tell you more. The way you have it, if x₁ = -999 and all other x_is > 0, then y_n = x₁ = -999 doesn't tell you much.

No. I correctly transcribed part (C). Besides, if the inequality were the other way around, there is no guarantee the sequence would exist, since the minimum of an infinite set does not always exist.

 

[Ray Vickson]

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.

You are right that (D) is the correct answer, but I don't think looking at your sequences $(y_n)$ and $(z_n)$ is very useful. However, there are similar sequences that will do the job.

Hint: look at your next thread on Lim sup, etc.

 

[FactChecker]

No. I correctly transcribed part (C). Besides, if the inequality were the other way around, there is no guarantee the sequence would exist, since the minimum of an infinite set does not always exist.

Even if it is bounded? Perhaps infimum is a better word.

 

[Bashyboy]

You are right that (D) is the correct answer, but I don't think looking at your sequences $(y_n)$ and $(z_n)$ is very useful. However, there are similar sequences that will do the job.

Hint: look at your next thread on Lim sup, etc.

But the sequences with terms $s_n := \sup \{a_k ~|~ k \ge n \}$ and $i_n := \inf \{ a_k ~|~ k \ge n \}$, though they converge, won't necessarily be subsequences of the original sequence...right?

 

[Ray Vickson]

But the sequences with terms $s_n := \sup \{a_k ~|~ k \ge n \}$ and $i_n := \inf \{ a_k ~|~ k \ge n \}$, though they converge, won't necessarily be subsequences of the original sequence...right?

Right, but that does not matter. That is exactly why I suggested you look at your other thread about Lim sup.