I have here a definition of accumulation point: A point $x$ in a metric space $M$ is called an accumulation point of $A \subset M$ if every neighbourhood of $x$ contains some point of $A$ distinct from $x$.
The definition seems vague to me; how to translate the condition in it?
I) For each neighbourhood $N$ of $x$, there exists some $y\in A$ such that $y\in N$ and $y\neq x.$
or
II) There exists some $y\in A$ such that for each neighbourhood $N$ of $x$, $y\in N$ and $y\neq x.$
Best Answer
I) is the appropriate choice. If you want to be completely formal you might want to add a definition of what a neighborhood of $x$ is (in terms of the metric on the space).
II) does not work because it requires that $y$ belongs to every neighborhood of $x$. By the definition of a metric this can only be if $y=x$ which conflicts with the rest of the definition.
EDIT: (Drawing a picture in these cases can help quite a bit) Let $d$ denote the metric on $M$ and $B_r:=\{y\in M:d(x,y)<r\}$ denote the open ball centred around $x$ of radius $r$. Suppose that $y\in B_r$ for all $r>0$. Then, $d(x,y)=0$. Otherwise there exists an $r>0$ such that $d(x,y)=r$. However, this contradicts our assumption that $y \in B_r$. Now, by the definition of the metric, $d(x,y)=0$ if and only if $x=y$.