Understanding Boundedness in Real Sequences

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Homework Help Overview

The discussion revolves around the concept of boundedness in real sequences, specifically within the context of the set of bounded real sequences denoted as \ell_\infty \mathbb{R}. Participants are examining various sequences and questioning their boundedness properties.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are assessing whether specific sequences belong to \ell_\infty \mathbb{R} based on their boundedness. They raise questions about the sequences (n), (2n^2+1), (1/n), and (4-1/n), discussing their upper and lower bounds.

Discussion Status

Some participants have confirmed the boundedness of certain sequences while others express uncertainty regarding the sequence (4-1/n), leading to further exploration of its properties. There is an ongoing examination of the definitions and implications of boundedness.

Contextual Notes

Participants are navigating potential typos and clarifications in the problem statements, particularly concerning the sequence (4-1/n) and its boundedness. The discussion reflects varying interpretations of what constitutes boundedness in this context.

bugatti79
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Homework Statement



Let \ell_\infty \mathbb({R}) be the set of bounded real sequences with k > 0 such that \left | x_n \right |\le k

a) (n)=(1,2,3...) \notin \ell_\infty \mathbb({R}). This is not bounded 'above'?

b) (2n^2+1) \notin \ell_\infty \mathbb({R}) Same answer as above?

c) (1/n)=(1,1/2,1/3,1/4...) \in \ell_\infty \mathbb({R}) Is bounded above?

d) (4-1/n) \notin \ell_\infty \mathbb({R}) Why is this not bounded? Is it because the value wll not go below 0?
 
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a) yes.

b) yes.

c) bounded above AND below: 0 < 1/n ≤ 1

d) i think there's a typo here
 


bugatti79 said:

Homework Statement



Let \ell_\infty \mathbb({R}) be the set of bounded real sequences with k > 0 such that \left | x_n \right |\le k

a) (n)=(1,2,3...) \notin \ell_\infty \mathbb({R}). This is not bounded 'above'?
Correct. No matter how large an M you pick, for some n, an > M.
bugatti79 said:
b) (2n^2+1) \notin \ell_\infty \mathbb({R}) Same answer as above?
Yes.
bugatti79 said:
c) (1/n)=(1,1/2,1/3,1/4...) \in \ell_\infty \mathbb({R}) Is bounded above?
Yes, by 1.
bugatti79 said:
d) (4-1/n) \notin \ell_\infty \mathbb({R}) Why is this not bounded? Is it because the value wll not go below 0?
Looks bounded to me. Every number in the sequence is less than 4. Why do you think it's not bounded?
 


Mark44 said:
Looks bounded to me. Every number in the sequence is less than 4. Why do you think it's not bounded?

Thanks guys,

Is d) bounded above AND below...because 0&lt;(4-1/n) \le 4...?
 


For d, you have 3 <= 4 - 1/n < 4, with n being a positive integer.
 


Thanks Mark.
 

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