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Prove Sequence is Bounded 
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#1
Oct1007, 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. 


#2
Oct1007, 11:11 PM

Sci Advisor
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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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