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Prove Sequence is Bounded

by christianrhiley
Tags: bounded, prove, sequence
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christianrhiley
#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.
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Dick
#2
Oct10-07, 11:11 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
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.


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