# Subsequence and sequences

Suppose $(x_n)_{n=1}^\infty$ is a sequence in $\mathbb R$, and that $L_k$ are real numbers with $\lim_{k\to\infty}L_k=L$. If for each $k\geq 1$, there is a subsequence of $(x_n)_{n=1}^\infty$ converging to $L_k$, show that some subsequence converges to $L$. HINT: Find an increasing sequences $n_k$ such that $|x_{n_k}-L|<1/k$.

Can someone tell me what $L_k$ actually are? Is that a sequence or is it something else? I thought it was a sequence first, but the following sentence suggests it isn't (you can't converges to a sequence)

-
I would find it confusing that the $k$ in $L_k$ and the $k$ in $n_k$ have nothing to do with eachother. The author of your question should've written "Hint: find an increasing sequence $n_i$ such that $|x_{n_i} - L| < 1/i$", swapping $k$ for $i$. –  Arthur Oct 7 '12 at 8:21
add comment

## 4 Answers

$L_k$ seem to be terms of a sequence $\left(L_k\right)_{k=1}^\infty$ with limit $L$. The question is basically asking you to prove that if the sequence $\left(x_n\right)_{n=1}^\infty$ has convergent subsequences to each term $L_k$ then it also has a subsequence which converges to $L$.

-
What does the hint mean? Same confusion "Find an increasing sequence $n_k$ such that $|x_{n_{k}} - L| < 1/k$". Do they mean $x_{n}$, not $n_k$? –  sidht Oct 7 '12 at 7:44
$n_k$ is (yet) another sequence. This time it is a sequence of indices. What they mean is to find a subsequence of $(x_n)$ indexed by the sequence $(n_k)$. For example, if we let $(n_k)$ start as $1,\ 4,\ 5,\ 8,\ \cdots$ then the subsequence of $(x_n)$ we choose would be $x_1,\ x_4,\ x_5,\ x_8,\ \cdots$. –  EuYu Oct 7 '12 at 7:47
Lietrally, the hint just say that you should find a subsequence with sepcific convergence rate. As such the hint is bogus in my opinion. But the $k$ occuring there might indicate that you find the $k$th term of your subsequence via a $L_k$ and a subsequence converging to it. –  Hagen von Eitzen Oct 7 '12 at 9:18
I thank you all for shredding some confusion over this question. But as it stands now, I am still confused –  sidht Oct 8 '12 at 4:03
add comment

"...$L_k$ are real numbers with $\lim_{k\to\infty}L_k=L$."

So for each $k$, $L_k\in\mathbb R$. $L_k$ is the $k$th term of the sequence $(L_k)_{k=1}^\infty$.

-
add comment

$(L_k)_{k=1}^\infty$ is a sequence that converges to the limit $L$. So the question supposes that for each $k$ there is a corresponding subsequence of $(x_n)_{n=1}^\infty$ which has limit each $L_k$.

-
add comment

If you are uncomfortable with the multiple indexing, maybe try an indirect approach: Assume there is no such subsequence, hence there is an $\epsilon>0$ such that only finitely many terms of the seuence are between $L-\epsilon$ and $L+\epsilon$. Now make use of $L_k\to L$ to find an $L_k$ closer than $\frac\epsilon2$ to $L$. Then make use of the subsequence of $(x_k)$ that converges to $L_k$ ...

-
add comment