# Points where a sequence accumulates

Let $(x_n)$ be a sequence of reals.

$p \in \mathbb{R}$ is an accumulation point of the sequence if for every $\epsilon > 0$, $x_n$ lies in $(p-\epsilon, p+\epsilon)$ for infinitely many $n$ (stricten "infinitely many" to "all but finitely many", and we get definition of limit of a sequence).

A subsequence of $(x_n)$ is just a sequence $x_{n_1}, x_{n_2}, \ldots$ with $n_1 < n_2 < \ldots$

Notice $\{$ limits of convergent subsequences of $(x_n) \}$ = $\{$ accumulation points of $(x_n) \, \}$ :
A subsequential limit is clearly an accumulation point of seq. Say $p \in \mathbb{R}$ is an accumulation point of seq. There is an $n_1$ with $|x_{n_1} - p | < 1$. Now as $x_n$ lies in $(p-\frac{1}{2}, p+\frac{1}{2})$ for infinitely many $n$, we can pick an $n_2 > n_1$ with $|x_{n_2} - p | < \frac{1}{2}$. And now we can pick an $n_3 > n_2$ with $|x_{n_3} - p | < \frac{1}{3}$, so on. This gives a subsequence $(x_{n_k})$ converging to $p$.