Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $(x_n)$ is a sequence of real numbers and suppose that the $L_n$ are real numbers such that $L_n\to L$ as $n\to \infty$. If for each $k\ge 1$ there is a subsequence of $(x_n)$ converging to $L_n$, show that some subsequence converges to $L$.

My attempt:

Let $\lim_{n\to\infty}L_n =L$, then by definition, there are infinitely many $L_n$ such that $$L-\epsilon <L_n<L+\epsilon.$$ Which implies that $\exists$ $L_{n_k}$ such that $L_{n_k}\in \{a-\epsilon,a+\epsilon\}= a-\epsilon <L_n<a+\epsilon$ ...

How can I get this subsequence to converge to $L$?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

We assume that the sequence $(L_n)_{n\in\mathbb N}$ is not finally constant (i.e. we cannot find an $n_0\in\mathbb N$ such that $L_n=L, \ \forall n\geq n_0$).

It's enough to show that for any $\epsilon>0$ there is a $k\in\mathbb N$ s.t. $x_k\in(L-\epsilon,L+\epsilon)$.
So, let $\epsilon>0$. Since $L_n\to L$ there is an $n'\in\mathbb N$ s.t. $L_{n'}\in(L-\frac\epsilon2,L+\frac\epsilon2)$ and $L_{n'}\neq L$.
Find an element of the sequence $(x_n)_{n\in\mathbb N}$ which is $\min\{\frac\epsilon2,|L_{n'}-L|\}$ far from $L_{n'}$ and you are done (why?).

The case where $(L_n)_{n\in\mathbb N}$ is finally constant is left as an exercise.

share|improve this answer
    
You chose $\frac{\epsilon}{2}$ because we have two sequence that are converging to $L$ which are $x_n$ and $L_n$, correct? –  Q.matin Apr 22 '13 at 5:34
    
@Q.matin: The sequence $x_n$ doesn't converge to $L$. I choose $\frac\epsilon2$ because I know I can choose $L_n$ as close to $L$ as I want and I can choose $x_k$ as close to $L_n$ as I want. Therefore to find an $x_n$ which is $\epsilon$ far from $L$ I first find an $L_{n'}$ which is $\frac\epsilon2$ far from $L$ and then an $x_k$ which is $\frac\epsilon2$ far from $L_{n'}$. –  P.. Apr 22 '13 at 5:39
    
Opps, I meant to say that $x_n$ converges to $L_n$, but thanks for clearing that up! –  Q.matin Apr 22 '13 at 5:46
    
There's a minor detail missing: You need that the set of $k$'s chosen to be infinite. (In the event of $L_n=L$ and $x_k=L$, you proof may give only one $k$...) –  Lior B-S Apr 22 '13 at 7:10
1  
P.. wants to generate a subsequence of $x_n$ that tends to $L$. A criterion for that is that for every $\epsilon>0$ there is an infinite number of indices $k$ such that $x_{k}\in (L-\epsilon,L+\epsilon)$. –  Lior B-S Apr 22 '13 at 11:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.