Prove statement on a sequence of real numbers

In summary: In this case, it would be more clear to say "the limit does not exist", instead of saying "the limit is not equal to infinity". Thanks for the clarification!In summary, the conversation discusses how to prove that a sequence does not have any convergent subsequences if and only if the limit of the sequence is equal to positive infinity. One person presents their attempt at a solution, but struggles with proving the other implication. Another person suggests proving the contrapositive and carefully negating the definition of the limit. The conversation continues with a discussion about why saying the limit is not equal to infinity does not necessarily imply that the limit exists.
  • #1
mahler1
222
0
Homework Statement .

Prove that ##\{x_n\}_{n \in \mathbb N} \subset \mathbb R## doesn't have any convergent subsequence iff ##lim_{n \to \infty} |x_n|=+\infty##.

The attempt at a solution.

I think I could correctly prove the implication ##lim_{n \to \infty} |x_n|=+\infty \implies## it doesn't exist any convergent subsequence:

Suppose there exists ##\{x_{n_k}\}_{k \in \mathbb N}## convergent and call the limit ##A##. Then, for ##ε=1##, there exists ##n_0: \forall n≥n_0, |x_{n_k}-A|< 1##.

This means, ##\forall n≥n_0, |x_{n_k}|-|A|\leq |x_{n_k}-A|< 1 \implies |x_{n_k}|<1+|A| ##.

On the other hand, ##lim_{n \to \infty} |x_n|=+\infty##, so, for ##M=1+|A|##, there exists ##n_1 : \forall n≥n_1, |x_n|>M##.

Take ##N=max\{n_0,n_1\}##, then for all ##n_k≥N, \space M<|x_{n_k}|<M##, which is clearly absurd. This proves that with the given hypothesis, it can't exist any convergent subsequence of the original sequence.

I need help to prove the other implication: If there is no convergent subsequence then ##lim_{n \to \infty} |x_n|=+\infty##.
 
Physics news on Phys.org
  • #2
mahler1 said:
Homework Statement .

Prove that ##\{x_n\}_{n \in \mathbb N} \subset \mathbb R## doesn't have any convergent subsequence iff ##lim_{n \to \infty} |x_n|=+\infty##.

The attempt at a solution.

I think I could correctly prove the implication ##lim_{n \to \infty} |x_n|=+\infty \implies## it doesn't exist any convergent subsequence:

Suppose there exists ##\{x_{n_k}\}_{k \in \mathbb N}## convergent and call the limit ##A##. Then, for ##ε=1##, there exists ##n_0: \forall n≥n_0, |x_{n_k}-A|< 1##.

This means, ##\forall n≥n_0, |x_{n_k}|-|A|\leq |x_{n_k}-A|< 1 \implies |x_{n_k}|<1+|A| ##.

On the other hand, ##lim_{n \to \infty} |x_n|=+\infty##, so, for ##M=1+|A|##, there exists ##n_1 : \forall n≥n_1, |x_n|>M##.

Take ##N=max\{n_0,n_1\}##, then for all ##n_k≥N, \space M<|x_{n_k}|<M##, which is clearly absurd. This proves that with the given hypothesis, it can't exist any convergent subsequence of the original sequence.

I need help to prove the other implication: If there is no convergent subsequence then ##lim_{n \to \infty} |x_n|=+\infty##.

Try and think about proving the contrapositive. I.e. if it's not true that ##lim_{n \to \infty} |x_n|=+\infty## then ##x_n## has a convergent subsequence. You'll have to think carefully about how to negate the definition of the limit.
 
Last edited:
  • Like
Likes 1 person
  • #3
Dick said:
Try and think about proving the contrapositive. I.e. if it's not true that ##lim_{n \to \infty} |x_n|=+\infty## then ##x_n## has a convergent subsequence. You'll have to think carefully about how to negate the definition of the limit.

Suppose it's not true that ##lim_{n \to \infty} |x_n|=+\infty##. Then, there exists ##M>0## ,such that for every natural number ##j##, there is ##n_j>j## : ##|a_{n_j}|\leq M##. This means the subsequence ##\{a_{n_j}\}_{j \in \mathbb N}## is bounded. By the Bolzano Weierstrass theorem, there exits ##\{x_{n_{j_k}}\}_{k \in \mathbb N}## convergente subsequence. But ##\{x_{n_{j_k}}\}_{k \in \mathbb N}##is also a subsequence of the sequence ##\{x_n\}_{n \in \mathbb N}##. From here it follows the contraposivite is true, so if ##\{x_n\}_{n \in \mathbb N}## doesn't have any convergent subsequence, then ##lim_{n \to \infty} |x_n|=+\infty##.

Thanks, Dick!
 
Last edited:
  • #4
mahler1 said:
Suppose ##lim_{n \to \infty} |x_n|≠+\infty##. Pick ##M>0## arbitrary, then, for every natural number ##j##, there exists ##n_j>j## such that ##|a_{n_j}|\leq M##. This means the subsequence ##\{a_{n_j}\}_{j \in \mathbb N}## is bounded. By the Bolzano Weierstrass theorem, there exits ##\{x_{n_{j_k}}\}_{k \in \mathbb N}## convergente subsequence. But ##\{x_{n_{j_k}}\}_{k \in \mathbb N}##is also a subsequence of the sequence ##\{x_n\}_{n \in \mathbb N}##. From here it follows the contraposivite is true, so if ##\{x_n\}_{n \in \mathbb N}## doesn't have any convergent subsequence, then ##lim_{n \to \infty} |x_n|=+\infty##.

Thanks, Dick!

That's roughly the idea, but I'd stay away from writing things like ##lim_{n \to \infty} |x_n|≠+\infty##, it sort of implies that ##|x_n|## has a limit, but it's just not infinity. ##|x_n|## may have no limit. More seriously, you can't pick ##M>0## arbitrary, some M's may not work. Think about the negation of the limit definition again.
 
  • Like
Likes 1 person
  • #5
Dick said:
That's roughly the idea, but I'd stay away from writing things like ##lim_{n \to \infty} |x_n|≠+\infty##, it sort of implies that ##|x_n|## has a limit, but it's just not infinity. ##|x_n|## may have no limit. More seriously, you can't pick ##M>0## arbitrary, some M's may not work. Think about the negation of the limit definition again.

You're right, ##M## is not arbitrary. I've corrected it. However, I don't understand why saying it's not true ##lim_{n \to \infty} |x_n|=+\infty## is not equivalent to say ##lim_{n \to \infty} |x_n|≠+\infty##. If it is not infinity, then, it is different to infinity, I agree that the limit may not exist (the sequence could oscillate) but why saying ##lim_{n \to \infty} |x_n|≠+\infty## would imply ##|x_n|## has a limit?
 
  • #6
mahler1 said:
You're right, ##M## is not arbitrary. I've corrected it. However, I don't understand why saying it's not true ##lim_{n \to \infty} |x_n|=+\infty## is not equivalent to say ##lim_{n \to \infty} |x_n|≠+\infty##. If it is not infinity, then, it is different to infinity, I agree that the limit may not exist (the sequence could oscillate) but why saying ##lim_{n \to \infty} |x_n|≠+\infty## would imply ##|x_n|## has a limit?

Because if the limit doesn't exist, then you are saying that something that doesn't exist is not equal to infinity. Doesn't that sound funny to you? It does to me.
 
  • #7
Dick said:
Because if the limit doesn't exist, then you are saying that something that doesn't exist is not equal to infinity. Doesn't that sound funny to you? It does to me.

Yes, now I see it.
 

What is a sequence of real numbers?

A sequence of real numbers is a list of numbers that are ordered according to their position in the list. The numbers in a sequence can be positive, negative, or zero, and they can be whole numbers or decimals. For example, the sequence 1, 3, 5, 7, 9 is a sequence of odd numbers in ascending order.

How do you prove a statement on a sequence of real numbers?

To prove a statement on a sequence of real numbers, you need to use mathematical reasoning and logical arguments. This can involve using algebraic manipulation, properties of real numbers, and mathematical induction. It is important to clearly state your assumptions and provide evidence for each step of your proof.

What is the difference between a convergent and a divergent sequence?

A convergent sequence is one in which the numbers in the sequence get closer and closer to a specific value, known as the limit. In contrast, a divergent sequence is one in which the numbers in the sequence do not approach a specific value and instead either increase or decrease infinitely. In other words, a divergent sequence does not have a limit.

Can a sequence of real numbers have more than one limit?

No, a sequence of real numbers can only have one limit. This is because the limit of a sequence is a unique value that the numbers in the sequence get closer and closer to. If a sequence had more than one limit, it would mean that the numbers are getting closer to multiple values, which is not possible.

What is the importance of proving statements on sequences of real numbers?

Proving statements on sequences of real numbers is important because it allows us to understand and make predictions about the behavior of real numbers. It also helps us to establish the validity of mathematical theories and to solve problems related to real-life situations that involve sequences of numbers. Furthermore, proving statements on sequences of real numbers is an essential skill for students studying mathematics and other scientific fields.

Similar threads

  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
915
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
Replies
1
Views
519
  • Calculus and Beyond Homework Help
Replies
4
Views
834
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
964
  • Calculus and Beyond Homework Help
Replies
2
Views
640
Back
Top