# Limit of a subsequence

I studied a definition,and I didn´t find it in anyother book (but those I use). It´s like a point of closure for sequences.We call $a$ "value of closure" of $(x_n)$ when $a$ is the limit of a subsequence of $(x_n)$.

The question is:

For a real number $a$ be a "value of closure" is necessary and sufficient that $\forall \epsilon >0$ and $\forall k \in \mathbb{N}$ given ,there is $n > k$ such that $|x_n -a|< \epsilon$.

I could do the first part ($a \Rightarrow |x_n - a|<\epsilon$) but not the $\Leftarrow$.

Thanks for any help!

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Such an $a$ is sometimes called a subsequential limit of the sequence. – Brian M. Scott Jul 22 '12 at 21:00
oh!That´s it!Thanks!!!! – HipsterMathematician Jul 22 '12 at 21:39

To expand the comment, fix $n_1=1$ for example. We can find $n_2>1$ such that $|x_{n_2}-a|<\frac 12$. Assume that $n_1<n_2<\dots<n_k$ are construct. For $\varepsilon=2^{-(k+1)}$, we can find $n_{k+1}>n_k$ such that $$|x_{n_{k+1}}-a|\leq 2^{-(k+1)}.$$ Hence we have construct a subsequence $\{x_{n_k}\}$ such that $|x_{n_k}-a|\leq 2^{-k}$ for all integer $k$. This proves that $a$ is a value of closure of $\{x_n\}$.