In Real analysis:

Let $$S$$ $$\in$$ $$\mathbb{R}$$ be any set, and let $$x$$ $$\in$$ $$\mathbb{R}$$ then:
the point $$x$$ is said to be limit point or a cluster point or an accumulation point or condense point of $$S$$ if $$\forall \epsilon > 0$$ , $$(x-\epsilon,x+\epsilon)\cap$$ $$S \setminus {x} \neq \phi$$

In Topological Space:

Let $$A$$ be a subset of a topological space $$(X,\tau)$$. A point $$x \in X$$ is said to be limit point of $$A$$ if every open set, $$U$$, containing $$x$$ contains a point of $$A$$ different from $$x$$.

Closed Sets:
A Set $$S$$ is said to be closed if one of these conditions achieved:
•  It's complement is open.
• Or, it contains all its limit points.
Notions:
1. The set of limit points of a set $$S$$ is denoted by $$L(S)$$
2. The subset $$X$$ is a closed subset of itself.
3. The Empty set is closed.
4. Any Finite set is closed.
5. A Subset of $$\mathbb{R}$$ with no limit points is a closed set. ( See Example #6 ).
6. In the metric space, the open disc $$\left \{ \left ( a,b \right ) \in R^{2} \mid a^{2}+b^{2} < 1 \right \}$$ has the closed disc $$\left \{ \left ( a,b \right ) \in R^{2} \mid a^{2}+b^{2} \leqslant 1 \right \}$$ as limit points.
Examples:

Example #1: Consider the topological space $$(X,\tau)$$ where the set $$X = \left \{ a,b,c,d,e \right \}$$, the topology $$\tau =$$ $$\left \{X,\phi, \left \{ a\right \}, \left \{ c,d \right \}, \left \{ a,c,d \right \}, \left \{ b,c,d,e \right \}\right \}$$, and $$A=\left \{ a,b,c \right \}$$. Then $$b$$, $$d$$, and $$e$$ are limit points of $$A$$ but $$a$$ and $$c$$ are not limit points of $$A$$. Explain ?

Solution:
-The set $$\left \{ a \right \}$$ is open and contains no other points of A. Hence, $$a$$ is not a limit point of $$A$$ even though the set $$\left \{ a,c,d \right \}$$ contains other points of $$A$$, because not every open set containing $$a$$ contains a point of $$A$$ different from $$a$$.
-Now, $$b$$ is a limit point of $$A$$ because the only open sets containing $$b$$ are $$X$$ and $$\left\{b,c,d,e \right\}$$ contains at other element of $$A$$ which is $$c$$. Thus, $$b$$ is a limit point of $$A$$.
-Choosing $$\left \{c,d \right\}$$ as an open set containing $$c$$, will exclude $$c$$ from being limit point of $$A$$, because it contains no other point of $$A$$ other than $$c$$. Hence, $$c$$ is not limit point of $$A$$.
-Moreover, an interesting fact that the two elements $$e$$ and $$d$$ are limit points of $$A$$ even they are not in $$A$$. And this is because every open set in $$\tau$$ containing either $$d$$ or $$e$$ will contain point of $$A$$ other than these points. Thus, $$d$$ and $$e$$ are limit points of $$A$$.

Example #2: Let $$S=\left (a,b \right )$$ and $$x \in \left (a,b \right )$$. Show that $$x$$ is a limit point of $$S$$.

Solution:
Let $$\epsilon > 0$$ and consider the interval $$\left (x-\epsilon,x+\epsilon \right )$$. Then, this interval will contain points of $$\left (a,b \right )$$ other than $$x$$ .i.e $$\forall \epsilon > 0$$ , $$(x-\epsilon,x+\epsilon)\cap$$ $$S \setminus {x} \neq \phi$$. In fact, it contains many infinite points other than $$x$$. Similarly, we can show that $$a$$ and $$b$$ are also a limit point of the open interval $$\left (a,b \right )$$ and the closed interval $$\left [ a,b\right ]$$.
In General, $$L\left (\left (a,b \right )\right )$$ = $$\left [ a,b\right ]$$.

Example #3: Let $$S=\left (0,1 \right ) \cup \left \{ 2 \right \}$$ , then show that:
1. $$2$$ is close to $$S$$
2. $$2$$ is not limit point of $$S$$.
3. What is the closure of $$S$$ ?
4. Find the limit points of $$S$$, .i.e $$L(S)$$ ?
Solution:
1. It is obvious that $$\forall \epsilon > 0$$, $$\left \{ 2 \right \}$$ $$\subseteq$$ $$\left (2-\epsilon,2+\epsilon \right ) \cap S$$ such that $$\left (2-\epsilon,2+\epsilon \right ) \cap S$$ $$\neq\phi$$.
2. Applying the definition above, $$\forall \epsilon > 0$$,  $$\left (2-\epsilon,2+\epsilon \right ) \cap S$$ $$\setminus \left \{ 2 \right \}=\phi$$. Thus $$2$$ is not a limit point of $$S$$.
3. To find the closure of $$S$$ , .i.e $$\bar{S}$$, we need to find the smallest closed interval that contain $$S$$. In tacit, the set $$\left [ 0,1\right ] \cup \left \{ 2 \right \}$$ is the closure of $$S$$.
4. We can show that the points $$a$$ and $$b$$ are both limit points of $$S$$. Also, any point belong to the open interval $$\left(a,b\right )$$ is limit point of $$S$$. Thus, all limit points of $$S$$ belong to the closed interval $$\left[a,b \right]$$. Hence, $$L(S)=\left[a,b \right]$$.
Example #4: Show that the set of all integer numbers has no limit points, .i.e $$L\left( \mathbb{Z} \right) = \phi$$.

Solution:
Let $$x\in\mathbb{R}$$. We say that $$x$$ is limit point of $$\mathbb{Z}$$ iff $$\forall \epsilon > 0$$ the interval $$\left(x-\epsilon,x+\epsilon \right )$$ contains points of $$\mathbb{Z}$$ other than $$x$$. But $$\forall x \in \mathbb{R}$$, $$\exists n \in \mathbb{Z}$$ such that $$n-1 \leq x < n$$. so, if $$\ epsilon < min \left ( \left | x-n+1 \right |,\left | x-n \right | \right )$$ then the interval $$\left (x-\epsilon , x+\epsilon \right )$$ will contain no integer. Therefore, $$\left (x-\epsilon , x+\epsilon \right ) \cap \mathbb{Z}\setminus \left \{ x \right \}= \phi$$. Thus, no real number can be limit point of $$\mathbb{Z}$$.
In analogy with $$\mathbb{N}$$, we can show that $$L\left( \mathbb{N} \right) = \phi$$ either.

Example #5: Show that the set of all rational number has infinitely many limit points, .i.e $$L\left(\mathbb{Q} \right) = \mathbb{R}$$.

Solution:
By the definition, $$\forall \epsilon > 0$$, the interval $$\left (x-\epsilon , x+\epsilon \right )$$ contains infinitely many points of $$\mathbb{Q}$$ other than $$x$$. Hence,  $$\left (x-\epsilon , x+\epsilon \right ) \cap \mathbb{Q}\setminus \left \{ x \right \} \neq \phi$$. Therefore, every real number is limit point of $$\mathbb{Q}$$.

Example #6: Show (in term of limit points) that the set of all natural number $$\mathbb{N}$$ is closed set.

Solution:
To prove that $$\mathbb{N}$$ is closed, we have to show that it contains all its limit points. And we knew that $$L\left( \mathbb{N} \right) = \phi$$. Now, since $$\phi \subset \mathbb{N}$$ then $$\mathbb{N}$$ contains all of limit point. i.e, $$\phi \cap \mathbb{N} \setminus \phi \neq \phi$$.

Example #7: Let $$S=\left \{ \frac{1}{n}:n \in \mathbb{Z}^{+} \right \}$$. Show that $$L \left ( S \right ) = \left \{ 0 \right \}$$.

Solution:
Implement the following set on the real line to have the following sketch:

The only limit point appear here is the $$0$$ point, since $$\forall \epsilon > 0$$, the interval $$\left (0-\epsilon , 0+\epsilon \right )$$ contains points of $$S$$ other than the zero point itself. Therefore, using the difinition of the limit point, we can say that  $$\left (0-\epsilon , 0+\epsilon \right ) \cap S \setminus 0 \neq \phi$$. Hence the only limit point of $$S$$ is the Zero point.