Let $(X, d)$ be a metric space. Let $E \subseteq X$ be a subset of $X$.
Let $p \in X$ be a accumulation point of $E$. So by definition for every neighbourhood $V \subseteq X$ of $p$ this intersection $(V \setminus \{p \} ) \cap E$ is not empty. I know how to show, that there is a sequence $(b_k)$, where for all $k \in \mathbb{N}: b_k \in E$ and $b_k \to p$. (by axiom of choice). My textbook states, that there also exists a sequence $(a_k)$, where for all $k \in \mathbb{N}: a_k \in E$ and $a_k \to p$ and $$\textbf{ for all } n,m \in \mathbb{N} \text{ with } n \neq m \textbf{ also } a_n \neq a_m .$$
So $(a_n)$ should be a sequence of distinct members of $E$.
My idea for a proof is to use $(b_k)$ and remove all members of the sequence that appear twice. Because $(b_k)$ is never constant, there will remain infinitely many points in the new sequence. Since it is a subsequence of a convergent sequence it will also have the limit $p$. Is this a good idea and how does one remove maybe countable infinite many different points from a set?
Best Answer
Let $n_1=1$. Now, let $n_2$ be the smallest natural $n$ such that $d(b_n,p)<d(b_{n_1},p)$. Then, let $n_3$ be the smallest natural $n$ such that $d(b_n,p)<d(b_{n_2},p)$, and so on. With this definition, $\lim_{k\to\infty}b_{n_k}=p$ and $k\ne l\implies b_{n_k}\ne b_{n_l}$.