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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)

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    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 $i$.2012-10-07

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$(L_k)_{k=1}^\infty$ 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 $L_k$.

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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 $L_k$ ...

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$L_k$ 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 $L$.

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    I thank you all for shredding some confusion over this question. But as it stands now, I am still confused2012-10-08
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"...$L_k$ 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 $(L_k)_{k=1}^\infty$.