I'm reading through Functional Analysis by Bachman.
He defines a limit point as follows:
The point $x$ is said to be a limit point of $A \subset X$ iff for every $r$, $S_r(x) \cap A$ contains infinitely many points of $A$.
Then, later in the book, he gives the following example:
Suppose $A$ is a non-countably compact set with some infinite subset $B$ with no limit point in $A$. Let $M=\{x_1,x_2,…\}$ be a denumerable set of distinct points from $B$. Let $E_1, E_2,….$ be a sequence of $\epsilon_n – $ neighbourhoods of $x_1,x_2,…$ respectively, where the $\epsilon_n$ have been chosen to guarantee that $$E_n \cap M = \{x_n\}.$$ If it were not possible to construct such neighborhoods for every point, this would mean that for some $x_n$, every neighbourhood of it would contain some other point of $M$ distinct from $x_n$. Thus $x_n$ would be a limit point of the set $M$.
But didn't we require that each neighbourhood contain infinitely many points of the set? Here each neighbourhood may contain even only one of ther point of $M$, so wouldn't it be a point of closure, rather than a limit point?
EDIT: note that the definitions refer to metric spaces
Best Answer
By definition $p$ is a limit point of $E$ if every neighborhood of $p$ contains a point $q$ belonging to $E$ that is different from p , but using this definition we can also prove that if $p$ is a limit point of $E$ then every neighborhood of $p$ contains infintely many points of $E$.
Sketch of the proof is as follows : suppose there was a neighborhood with finitely many points take the minimum of distances between those points and $p$ then take a neighborhood with a diameter less than half of that minimum; this new neighborhood is empty contradicting the fact that $p$ is a limit point ! Excuse my long sentences I'm still working on my Latex .