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 $\displaystyle d(s_n, t_n) < 1/n$ for all n. Prove that $\displaystyle \left( {t_n} \right) \to s$.

-------------------------------------------------------------------

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?

2. ## Re: Advanced Calculus Problem

Originally Posted by MissMousey
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 d(sn, tn) < 1/n for all n. Prove that {tn} converges to s.
Don't over-think the problem.
$\displaystyle d(S,t_n)\le d(S,s_n)+d(s_n,t_n)$.
Now make each of the last two terms $\displaystyle <\frac{\epsilon}{2}$

3. ## Re: Advanced Calculus Problem

Originally Posted by Plato
Don't over-think the problem.
$\displaystyle d(S,t_n)\le d(S,s_n)+d(s_n,t_n)$.
Now make each of the last two terms $\displaystyle <\frac{\epsilon}{2}$
Why $\displaystyle d(S, t_n)$?

4. ## Re: Advanced Calculus Problem

Originally Posted by MissMousey
Why $\displaystyle d(S, t_n)$?
Well, that is what you are asked to prove:
$\displaystyle \left( {t_n } \right) \to S$

5. ## Re: Advanced Calculus Problem

Originally Posted by Plato
Well, that is what you are asked to prove:
$\displaystyle \left( {t_n } \right) \to S$
But I'm asked to show that $\displaystyle \left( {t_n} \right) \to s$, not the set S.

6. ## Re: Advanced Calculus Problem

Originally Posted by MissMousey
But I'm asked to show that $\displaystyle \left( {t_n} \right) \to s$, not the set S.
That was just my mistake in using a upper case. Change it to lower case.

7. ## Re: Advanced Calculus Problem

Originally Posted by Plato
That was just my mistake in using a upper case. Change it to lower case.
Sorry for the late response but one more question. Why are you able to set $\displaystyle d(s_n,t_n)$ to be $\displaystyle <\frac{\epsilon}{2}$? Because the problem says that $\displaystyle d(s_n,t_n) < 1/n$, which I'm assuming is equal to $\displaystyle {\epsilon}$. I know that we're trying to show that $\displaystyle d(s,t_n) < {\epsilon}$, but with what they provided me in the problem, can I just disregard their $\displaystyle d(s_n,t_n) < 1/n$?

8. ## Re: Advanced Calculus Problem

Originally Posted by MissMousey
Sorry for the late response but one more question. Why are you able to set $\displaystyle d(s_n,t_n)$ to be $\displaystyle <\frac{\epsilon}{2}$? Because the problem says that $\displaystyle d(s_n,t_n) < 1/n$, which I'm assuming is equal to $\displaystyle {\epsilon}$. I know that we're trying to show that $\displaystyle d(s,t_n) < {\epsilon}$, but with what they provided me in the problem, can I just disregard their $\displaystyle d(s_n,t_n) < 1/n$?
The idea is we can make $\displaystyle d(s_n,s)<\tfrac{\epsilon}{2}$ as we can make $\displaystyle d(s_n,t_n)<\tfrac{1}{n}<\tfrac{\epsilon}{2}$

9. ## Re: Advanced Calculus Problem

Ah, ok. Great. Thank you very much for your help.