Does a Bounded, Divergent Sequence Always Have Multiple Convergent Subsequences?

In summary: 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 EquationsThe Attempt at a SolutionHere is my justification for why (D) is the answer. Since the sequence is bounded, it
  • #1
Bashyboy
1,421
5

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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  • #2
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.
 
  • #3
FactChecker said:
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.
 
  • #4
Are you sure that you have the inequality of the statement of C) correct? It seems like yn = min{ xk, k≥n } would tell you more. The way you have it, if x1 = -999 and all other xis > 0, then yn = x1 = -999 doesn't tell you much.
 
  • #5
FactChecker said:
Are you sure that you have the inequality of the statement of C) correct? It seems like yn = min{ xk, k≥n } would tell you more. The way you have it, if x1 = -999 and all other xis > 0, then yn = 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.
 
  • #6
Bashyboy said:

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.
 
  • #7
Bashyboy said:
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.
 
  • #8
Ray Vickson said:
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?
 
  • #9
Bashyboy said:
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.
 

Related to Does a Bounded, Divergent Sequence Always Have Multiple Convergent Subsequences?

1. What is a bounded sequence?

A bounded sequence is a sequence of numbers that has both an upper bound and a lower bound. In other words, all the terms in the sequence fall within a certain range of values.

2. What is a divergent sequence?

A divergent sequence is a sequence of numbers that does not have a finite limit. This means that the terms in the sequence do not approach a specific value as the sequence goes on, but instead, they continue to increase or decrease without end.

3. How do you determine if a sequence is bounded or divergent?

To determine if a sequence is bounded, you need to find both the upper and lower bounds of the sequence. If the terms in the sequence do not exceed these bounds, then the sequence is bounded. To determine if a sequence is divergent, you need to calculate the limit of the sequence. If the limit does not exist or is infinite, then the sequence is divergent.

4. Can a sequence be both bounded and divergent?

No, a sequence cannot be both bounded and divergent. A bounded sequence has a finite limit, while a divergent sequence does not have a finite limit. Therefore, a sequence cannot have both characteristics at the same time.

5. What are some real-life examples of bounded and divergent sequences?

An example of a bounded sequence could be the population of a city, as it typically has an upper and lower limit. An example of a divergent sequence could be the value of a stock over time, as it can continue to increase or decrease without a specific limit.

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