Here is Case 1 to my problem i need help with Case2 : Case 1: Assume s n is incr

Question

Here is Case 1 to my problem i need help with Case2: Case 1: Assume sn is increasing. By Hypothesis theset A ={sn|n a natural number} is bounded above. (M ||sn| M) By the completeness axiom: ther exists and L in the realnumbers with L= sup A ( L is the least upper bound). Let > 0 be given. know that L- is no longer an upper bound for A ( b/c Lwas the least upper bound) thus there exists an n where sn> L- =>sn > L- <=> > L-sn..... (1) Now for all n L > sn so L-sn> 0 and we have L - sn > -..... (2) Combining (1) and (2) : > L-sn >- or - < L-sn   < => |L-sn| < .... (*) so now we know there exists an n that is a natural number suchthat |sn-L| < Now we need to show that for all   k>n theinequality still holds b/c sn is increasing k > n =>sk > sn thus if   > |L-sn| >|L-sk| So this inequality holds for all k > n. CASE 2: Help me show wheresn is decreasing => L = inf A (L+ => and goes above) Here is Case 1 to my problem i need help with Case2: Case 1: Assume sn is increasing. By Hypothesis theset A ={sn|n a natural number} is bounded above. (M ||sn| M) By the completeness axiom: ther exists and L in the realnumbers with L= sup A ( L is the least upper bound). Let > 0 be given. know that L- is no longer an upper bound for A ( b/c Lwas the least upper bound) thus there exists an n where sn> L- =>sn > L- <=> > L-sn..... (1) Now for all n L > sn so L-sn> 0 and we have L - sn > -..... (2) Combining (1) and (2) : > L-sn >- or - < L-sn   < => |L-sn| < .... (*) so now we know there exists an n that is a natural number suchthat |sn-L| < Now we need to show that for all   k>n theinequality still holds b/c sn is increasing k > n =>sk > sn thus if   > |L-sn| >|L-sk| So this inequality holds for all k > n. CASE 2: Help me show wheresn is decreasing => L = inf A (L+ => and goes above)

Explanation / Answer

from your answer , i realized the question to be " amonotonic sequence bounded above is convergent andconsequently a monotonic function decreasing which is bounded belowis convergent". you want to prove the second part. if i am correct , then given {Sn} is decreasing. given {Sn} is bounded below. so, {Sn} is a non empty set of real numbers boundedbelow. suppose every lower bound of {Sn} is of the form x. let M = { x / x is a lower bound of {Sn} } clearly, every member of {Sn} is an upper bound of M. so, M is a non empty set of real numbers which isbounded above by every member of {Sn}. by completeness axiom , M has supremum. let it be L. this iswhat you gave. now, we wish to show that L is the infimum of {Sn} and {Sn}converges to L. since L is the supremum of M , every x < L is in Mand so a lower bound of {Sn}. further , any member > L is not a lower bound of{Sn}. L is the greatest lower bound of {Sn}. now, let be any smallest positive realnumber. then L + > L and so not a lower bound of {Sn}. since {Sn} is monotonically decreasing and L + is not alower bound of {Sn} , there exist atleast one y in {Sn} suchthat L < y < L+. any deleted - neighbourhood of L is non emptysubset of {Sn}. so, L is the limit point of {Sn}. further, if L1 is any other limit point of L , the n we caneasily show that L 1 = L. thus { Sn} converges to L. hence the theorem.

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