Prove Sequence is Bounded


by christianrhiley
Tags: bounded, prove, sequence
christianrhiley
christianrhiley is offline
#1
Oct10-07, 10:17 PM
P: 4
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.
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
Dick
Dick is offline
#2
Oct10-07, 11:11 PM
Sci Advisor
HW Helper
Thanks
P: 25,174
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.


Register to reply

Related Discussions
Every sequence of bounded functions that is uniformly converent is uniformly bounded Calculus & Beyond Homework 4
prove sequence is bounded Calculus & Beyond Homework 3
Bounded Sequence, Cauchy Calculus & Beyond Homework 2
bounded sequence as convergent General Math 7
Prove D U D' is bounded Calculus 1