How do I prove that if $L$ is sub-sequential limit of $a_{n_k}$ then $L$ is also sub-sequential limit of $a_n$? Could you please give me some hint how to solve this simple question: > Suppose $a_{n_k}$ is sub-sequence of $a_n$. Prove that if $L$ is sub-sequential limit of $a_{n_k}$ then $L$ is also sub-sequential limit of $a_n$. I know that if $L$ is sub-sequential limit of $a_{n_k}$, then there exist sub-sequence $a_{n_{k_m}}\to L$, therefore I need to construct some increasing sequence of indexes $n_s$ such as $a\in a_{n_s}$ if $a\in a_{n_{k_m}}$. But how to do this ? Thanks.

Let $(a_n)_{n\in I}$ be a sequence, and let $(a_{n_k})_{n_k\in J}$, for some subset $J\subseteq I$ of indices be a subsequence.

$L$ is a subsequential limit of $(a_{n_k})_{n_k\in J}$ if there exists a subsequence $(a_{n_{k_m}})_{n_{k_m}\in K}$ that converges to $L$. By the definition of a subsequence you have $K\subseteq J\subseteq I$ therefore $(a_{n_{k_m}})_{n_{k_m}\in K}$ is a subsequence of $(a_n)_{n\in I}$ that converges to $L$, and so $L$ is a subsequential limit of $(a_n)_{n\in I}$.

The general idea here is that "being a subsequence" is a transitive relation.