Another REA Question

Forum for the GRE subject test in mathematics.
marco
Posts: 263
Joined: Mon Jun 11, 2012 2:27 am

Another REA Question

Postby marco » Wed Oct 10, 2012 12:49 pm

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=1-1/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.

saywhatagain
Posts: 3
Joined: Sat Sep 01, 2012 1:51 pm

Re: Another REA Question

Postby saywhatagain » Wed Oct 10, 2012 1:09 pm

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 Bolzano-Weierstrass 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 Bolzano-Weierstrass 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.

marco
Posts: 263
Joined: Mon Jun 11, 2012 2:27 am

Re: Another REA Question

Postby marco » Thu Oct 11, 2012 4:08 pm

Oh shoot i see! Thanks!




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