Nonempty finite subset of $\mathbb{R}^n$ is closed: How to apply the definition of closed set? A nonempty finite subset of $\mathbb{R}^n$ is closed. Using the definition of a closed set, I cannot explain this rigorously: A closed set $E$ is one whose every limit point is also a member of the set. A limit point of the set $E$ is defined as a point $p$, whose every neighborhood contains a point $q$, which is $p\neq q$.
Now if $E$ contains finite elements of $\mathbb{R}^n$, then it also means that $E$ does not have any limit points. In that case how can we apply the definition of the closed set to $E$? Since $E$ does not have any limit points, does the definition of closed set becomes valid for it by the definition?
 A: 
A closed set $E$ is one whose every limit point is also a member of the set

Note that if $E$ doesn't have limit points, then this definition is trivialy satisfied and thus $E$ is closed. Yes, it is valid. We will show that finite sets don't have limit points.
What is a limit point?

A point $x\in \mathbb{R}^n$ (or any other topological space) is a limit point of $E$ if every neighbourhood of $x$ contains at least one point of $E$ different from $x$ itself.

First of all the special case: $E$ is empty. Somewhat weird but by the definition the empty set cannot have a limit point because the empty set is an open neighbourhood of itself. Thus it is closed (note that the only two open and closed sets are $\emptyset$ and $\mathbb{R}^n$).
So assume that $x\in\mathbb{R}^n$ is a limit point of a finite, nonempty subset $E\subseteq\mathbb{R}^n$ and $x\not\in E$. Write
$$E=\{a_1,\ldots, a_n\}$$
Thus
$$d = \min\big\{\lVert x-a_i\rVert\ \big|\ a_i\in E\big\}$$
is well defined and positive (as a minimum over a finite set of positive numbers) since $x\not\in E$.
Define
$$B=\big\{y\in\mathbb{R}^n\ \big|\ \lVert x-y\rVert < d\big\}$$
By definition $B$ is nonempty (because $d>0$), open neighbourhood of $x$.
Since $x\not\in E$ then $E\cap B=\emptyset$ and so $x$ cannot be a limit point of $E$.
Now note that if $x\in E$ then the same reasoning can be applied to $x$ and $E\backslash\{x\}$ to conclude that there is an open neighbourhood of $x$ that doesn't intersect with $E\backslash\{x\}$. In particular $E$ does not have limit points. Thus it is closed.
A: First show for any x that {x} is closed.
Now use the fact that the union of two closed sets is closed
to conclude that any finite set, {x,y}, {x,y,z}, ... is closed.
