Let (S, d) be a metric space. Suppose {sn belongs to S} be a sequence converging to s, where s belongs to S. Suppose {tn belongs to S} is a sequence of points so that for all n. Prove that .
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I know this may come as a silly question but should I prove that tn is bounded?
Because I'm stuck on finishing this. Here's what I have so far.
We know that {sn} is a converging sequence and that it converges to some s in S so we write for each epsilon > 0, there exists an N belonging to the set of natural numbers such that n > N implies that |sn - s| < epsilon (or lim sn = s).
From Kenneth Ross, Definition 13.2: A sequence (sn) in a metric space (S,d) converges to s in S if the limit (n -> infinity) d(sn, s) = 0. A sequence (sn) in S is a Cauchy sequence if for each epsilon > 0, there exists an N such that m,n > N implies d(sn,s) < epsilon.
We know that {tn} is a sequence in S and that the d(sn,tn) < 1/n for all n in the set of natural numbers. So we know that at some point s, both {sn} and {tn} meet but the sequence {sn} stops at s so...this where I get a little stuck.
I know that the lim d(sn, tn) = 0 but I don't think that's suffice to show that they converge (or am I wrong)? What's getting me stuck is whether or not I have to prove that tn is bounded.
Can you lend a hint?