Are Sequence Definitions for Bounded Sequences Adequate for Proving Boundedness?

In summary, we are given a bounded sequence {x[n]} in Reals and we define {y[k]} and {z[k]} as y[k]=sup{x[n]: n \geq k} and z[k]=inf{x[n]: n \geq k}. The claim states that (i) both y[k] and z[k] are bounded sequences, (ii) {y[k]} is a decreasing sequence, and (iii) {z[k]} is an increasing sequence. To prove these claims, we first assume that y[k] and z[k] are not bounded, which leads to a contradiction since it would imply that x[n] is unbounded. Therefore, we can conclude that both y[k] and z
  • #1
christianrhiley
4
0
Definitions: Let {x[n]} be a bounded sequence in Reals.
We define {y[k]} and {z[k]} by
y[k]=sup{x[n]: n [tex]\geq[/tex] k}, z[k]=inf{x[n]: n [tex]\geq[/tex] k}

Claim: (i) Both y[k] and z[k] are bounded sequences
(ii){y[k]} is a decreasing sequence
(iii){z[k]} is an increasing sequence

Proof: (i) suppose y[k] and z[k] are not bounded. this implies x[n] is unbounded, a contradiction. therefore, we conclude that both y[k] and z[k] are bounded.
(ii)Let S[k] = {x[n]: n [tex]\geq[/tex] k} and S[k+1] = {x[n]: n [tex]\geq[/tex] k + 1}
S[k] is a [tex]\subset[/tex] S[k+1], and if sup(S[k]) [tex]\leq[/tex] sup(S[k+1]), it follows that S[k] [tex]\leq[/tex] S[k+1]. We conclude that {y[k]} is decreasing.
(iii)similar to part (ii) except inf(S[k+1]) [tex]\leq[/tex] inf(S[k])

Note: inf(A) [tex]\leq[/tex] inf(B) and sup(B) [tex]\leq[/tex] sup(A) have already been proven in an earlier exercise.

Where I Need Help: I need input regarding all three parts. I have made, at best, an informal sketch of a proof, and I would like some input on how to turn it into a rigorous proof.
 
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  • #2
If you mean that B is a subset of A implies that sup(B)<=sup(A) and inf(B)>=inf(A), I really don't see the need for more 'rigor'. I think you have it.
 

FAQ: Are Sequence Definitions for Bounded Sequences Adequate for Proving Boundedness?

What does it mean for a sequence to be bounded?

A sequence is considered bounded if there exists a number M such that all the terms in the sequence have an absolute value less than or equal to M. This means that the sequence does not grow infinitely large or infinitely small.

How do you prove that a sequence is bounded?

To prove that a sequence is bounded, you need to show that the absolute value of each term in the sequence is less than or equal to a specific number, M. This can be done using mathematical induction or by finding a specific value of M that satisfies the condition for all terms in the sequence.

Why is it important for a sequence to be bounded?

A bounded sequence is important because it allows us to make certain conclusions about the behavior of the sequence. For example, a bounded sequence must have a limit, and the limit can be determined using the properties of bounded sequences.

Can a sequence be bounded from above but not from below?

Yes, it is possible for a sequence to be bounded from above but not from below. This means that there exists a number M such that all the terms in the sequence are less than or equal to M, but there is no number that is less than or equal to all the terms in the sequence. In this case, the sequence is considered bounded above but unbounded below.

How does the boundedness of a sequence relate to its convergence?

The boundedness of a sequence is closely related to its convergence. If a sequence is bounded, then it must have a limit, and the limit can be determined using the properties of bounded sequences. On the other hand, if a sequence is unbounded, then it may not have a limit or its limit may be infinite.

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