Exam IV Prob 45) Given that {xn} is a bounded, divergent, infinite sequence of real numbers, which of the following must be true?
(A) {xn} contains infinitely many convergent subsequences
(B) {xn} contains convergent subsequences with different limits.
(C) {yn = min xk } is convergent. k≤n
(D) All the above
(E) (A) and (C) only
The answer is D supposedly but i don't see why for example if one considers the sequence a_n=11/n if n is not prime and n if n is prime then what other limit could a subsequence that is not one? My answer was E.
Another REA Question

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Re: Another REA Question
Your example sequence is unbounded, so it doesn't satisfy the premises of the question. To see why B, holds, let {x_n} be such a sequence. Then by the BolzanoWeierstrass Theorem, it has a convergent subsequence, say with limit L. Since {x_n} isn't convergent, in particular it doesn't converge to L, which means for some epsilon > 0, we can find infinitely many terms in {x_n}, i.e. a subsequence {x_n_k}, such that all of these terms are a distance of at least epsilon away from L. By BolzanoWeierstrass again, this subsequence must have its own subsequence with limit L'. Since the terms of this new subsequence cannot get arbitrarily close to L, then we have L =/= L' and we are done.
Re: Another REA Question
Oh shoot i see! Thanks!