In the book of real mathematical analysis written by Pugh, there is a sentence about these two and condensation: Thus, S limits at p, clusters at p, or condenses at p according to whether each Mr(p)(r neighborhood) contains some, infinitely many, or uncountably many points of S.
Can anyone explain why? Or a counterexample that a point which is a limit point but not a cluster point will be great.
Terminology differs here, which might confuse you if see other answers online. But I'll follow Pugh here. We seem to be working in a metric space here, and using the notation $M_r(p)$ for an open ball of radius $r>0$ around $p$.
So whenever $S \subseteq X$, where $(X,d)$ is the metric space, then $p \in X$ is called a limit point of $S$ when for all $r>0$, $S \cap M_r(p) \neq \emptyset$; $p$ is called a cluster point of $S$ when for all $r > 0$ the set $S \cap M_r(p)$ is infinite, and $S$ condenses at $p$ (people also say that $p$ is a condensation point of $S$, which is more analogous to the previous names) whenever for all $r>0$ the set $S \cap M_r(p)$ is uncountable.
Note that if $p \in S$, then $M_r(p) \cap S$ will allways contain $p$ at least. So using this definition we see that all points of $S$ are themselves limit points of $S$, but there can be more, e.g. if $S = \{\frac{1}{n} \mid n \in \mathbb{N}^+\}$ (as a subset of the real numbers in the standard metric), then all points of $S$ are limit points of $S$, but $0 \notin S$ is too (and these are all the limit points of $S$). Now note that $1 \in S$ is not a cluster point of $S$, as $M_{\frac{1}{2}}(1)$ only intersects $S$ in $\{1\}$ and nowhere else. The same can be said for the other points $p$ of $S$: there exists some $r>0$ such that $S \cap M_r(p) = \{p\}$. Such a point is called an isolated point of $S$.
Now, in a metric space we have the following facts:
(1) if $p \in S$ is not an isolated point of $S$, then it is a cluster point of $S$.
Proof: suppose that $p \in S$ is not a cluster point of $S$. Then there is some $r>0$ such that $M_r(p) \cap S$ is finite, and so equals some set $\{p, p_1,\ldots,p_n\}$. Then $s = \min(r, d(p, p_1), \ldots, d(p, p_n)) > 0$ is well-defined (as a finite minimum of numbers > 0) and $M_{s}(p) \cap S = \{p\}$, as is easy to see. So $p$ is an isolated point of $S$. So all points of $S$ (which are all limit points of $S$) neatly divide into isolated points and cluster points of $S$.
(2) if $p \notin S$ is a limit point of $S$, then $p$ is a also a cluster point of $S$.
Proof: this is essentially the same argument. Suppose $p$ were not a cluster point, then for some $r>0$ we have $S \cap M_r(p) = \{p_1,\ldots,p_n\}$, where all $p_i \neq p$ (the set is not empty, as $p$ is a limit point by assumption). Defining $s = \min(d(p_1,p),\ldots,d(p_n,p))$, we see that again $s > 0$ and $M_s(p) \cap S = \emptyset$, contradiction.
So all limit points $p$ of $S$ are cluster points, except when $p \in S$ and $p$ is an isolated point of $S$. Another example where this occurs is for finite sets $S$ (no cluster points, and all points of the finite set are limit points and isolated points) and a set like $S = [0,1] \cup \{2\}$, where $2$ is a limit point, but isolated, so not a cluster point, and where all other points of $S$ are even condensation points of $S$.
