Limit point intuition Quoting Rudin,
"A point $p$ is a limit point of the set $E$ if every neighborhood of $p$ contains a point $q\not=p : q \in E$."
This would imply that the points in an open ball would all be limit points, since for any $p$ in $E$ there are $q$ such that $d(p,q) < r$ for all $q \in E$. So E is also a neighborhood of the open ball.
Is my intuition correct? What can be improved about this statement?
 A: You can think of the set of limit points $L(S)$ of a set $S$ as all points which are "close to" $S$.  In the example of an open ball in $\mathbb{R}^n$, the limit points are all points of the open ball, plus all points lying on the boundary, since every punctured neighborhood of such points will intersect the set.
Note, however, that if $S$ is some set and $L(S)$ is the set of limit points, then it is not always true that $S \subseteq L(S)$.  For example, in $\mathbb{R}$ under the ordinary topology, the set of integers has no limit points.  (An element of a set which is not a limit point of the set is called an isolated point, which provides a good intuitive way of thinking about such points.)
A: The following is likely to be not relevant, since the space being discussed is undoubtedly $\mathbb{R}^n$ for some $n$. But it is too long for a comment.
Suppose that our space $S$ consists of the points $0$, and $\frac{1}{k}$, where $k$ ranges over the positive integers. Use the ordinary distance function $|x-y|$.
Then $0$ is a limit point of $S$, since every interval about $0$ contains a point of $S$ different from $0$. However, if we pick any $\frac{1}{k}$ in this interval, then there is an interval about $\frac{1}{k}$ that contains no point of $S$ other than $\frac{1}{k}$. 
