# Limits of Subsequences

by soul
Tags: limits, subsequences
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,553 Limits of Subsequences Specifically, you can do this: suppose the subsequence of {an} has subsequence {an}i which converges to P and subsequence {an}j which converges to Q. Assume that {an} converges to L, and take $\epsilon$= (1/2)|P- Q|. For any N, there will be n1> N such that an1, in the first subsequence, is arbitratily close to P and n2> N such that an2, in the second subsequence is arbitrarily close to Q. If they are not within $2\epsilon$ of each other, they cannot both be within $\epsilon$ of L, a contradiction.
 P: 47 Suppose $$\lim a_n=L$$. Let $$a_{n_i}$$ be an arbitrary subsequence. We wish to prove that $$\lim a_{n_i}=L$$. We know that $$\forall \epsilon>0$$ $$\exists N$$ such that $$n\geq N$$ implies $$|a_n-L|<\epsilon$$. We need to show that $$\forall \epsilon>0$$ $$\exists I$$ such that $$i\geq I$$ implies $$|a_{n_i}-L|<\epsilon$$. But if we choose $$I$$ such that $$i\geq I$$ implies that $$n_i\geq N$$, then $$i\geq I$$ implies $$|a_{n_i}-L|<\epsilon$$. The existence of such an $$I$$ is guaranteed by the definition of a subsequence (something to check). So what this proof says is that if a sequence is convergent, then all it's subsequences converge to the same limit. Now take the contra-positive of this statement and compare it with your question. I hope that helped.  Just noticed the date of the OP's post... oh well.