There is something wrong with a homework exercise that was given to me about limit points.
Def. We say that a point $p$ in a metric space $(E,p)$ is a limit point of a sequence $\{p_n\}$ if $p \notin \{p_n\}$ and there is a subsequence $\{p_{n_k}\}$ such that $p_{n_k} \rightarrow p$.
It was given to me to prove the following result.
Exercise. Show that $p$ is a limit point of a sequence $\{p_n\}$ iff $\forall \epsilon>0,$ there is infinitely many $k \in \mathbb{N}$ satisfaying $d(p_k,p)<\epsilon$.
My attempt
$(\rightarrow)$ By the definition of limit point, there is a subsequence $\{p_{n_k}\}$ in $\{p_n\}\setminus \{p\}$ so that, given $\epsilon>0, \exists K\in \mathbb{N}: \forall k>K, 0<d(p_{n_k},p)<\epsilon$. It clearly implies that $d(p_k,p)<\epsilon$ for infinitely many points of $\{p_n\}$.
My problem is with the other direction. I can see the existence of convergent subsequence(s) to $p$, but I am not able to show that $p$ is not in this (these) subsequences. If I choose a constant sequence, the statement does not hold. It means that it cannot be shown.
It would be correct if he had stated:
" $p$ is a limit point of a sequence $\{p_n\}$ iff $\forall \epsilon>0,$ there is infinitely many $k \in \mathbb{N}$ satisfaying $0<d(p_k,p)<\epsilon$".
What do you think?
UPDATE Having accpted the exercise is wrong, a new question arose.
Suppose, given $\epsilon>0$, infinitely many points satisfy $0<d(p_k,p)<\epsilon$ and denote the set of these indices $k$ by $I$. Clearly, $\{p_k\}_{k\in I}$ is a subsequence of $\{p_n\}_{n\in \mathbb{N}}$ and $p\notin \{p_k\}_{k\in I}$. But what guarantees that $p\notin \{p_n\}_{n\in \mathbb{N}}$, which is our aim?
As long as I know, for arbitrary metric spaces, the limit point need not to be unique. Furthermore, the sequence $\{p_n\}_{n\in \mathbb{N}}$ need not to be convergent. It may be the case, in addition to impose $0<d(p_k)<\epsilon$, to suppose $x\notin \{p_n\}$, to prove $(\leftarrow)$.
Any ideas?
If you want that $p$ is not an element of the sequence you need to add, $0<d(p,p_n)<\epsilon$ otherwise you can take the constant sequence $p_n=p$.
You can take $\epsilon={1\over n}$ this insures the existence of infinite and (at least) an element $p_n$ such that $d(p,p_n)<{1\over n}$. This implies that $p$ is a limit point.