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 $L$?


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.

  • $\begingroup$ You chose $\frac{\epsilon}{2}$ because we have two sequence that are converging to $L$ which are $x_n$ and $L_n$, correct? $\endgroup$ – Q.matin Apr 22 '13 at 5:34
  • $\begingroup$ @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 $L_{n'}$. $\endgroup$ – P.. Apr 22 '13 at 5:39
  • $\begingroup$ Opps, I meant to say that $x_n$ converges to $L_n$, but thanks for clearing that up! $\endgroup$ – Q.matin Apr 22 '13 at 5:46
  • $\begingroup$ 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 $k$...) $\endgroup$ – Lior B-S Apr 22 '13 at 7:10
  • 1
    $\begingroup$ 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 $x_{k}\in (L-\epsilon,L+\epsilon)$. $\endgroup$ – Lior B-S Apr 22 '13 at 11:26

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