Can we say that all singletons are closed by this way? We know that a closed ball in space $E$:
$\bar{B}_{\epsilon}(x)=$ {$ y\in E: d(x,y) \leq \epsilon$} is a closed set.
If we consider, for example, $\epsilon = 0$,  $\bar{B}_{0}(x)= $ {$x$} , so every singleton {$x$} can be represented by a closed ball of radius zero and center $x$, which's a closed set. However, I know that singletons are open in some sets like  $\mathbb{N}$, so are singletons closed in all sets but open is some of them? Or the reason I put is incorrect?
Thanks!
 A: By definition, the radius of a ball (open or closed) is greater than $0$. However, it is easy to prove that, in a metric space, all singletons are closed: if $x\in E$,$$E\setminus\{x\}=\bigcup_{y\in E\setminus\{x\}}B_{d(y,x)}(y),$$and this proves that $E\setminus\{x\}$ is an open set.
Yes, in some cases, some singletons are open . So what?
A: To prove that any singleton $\{x\}$ is closed in space $X$, we can prove that $X\setminus\{x\}$ is open. For that to be true, it is enough to prove that for all $y\in X\setminus\{x\}$ there exists open $U_y\ni y$ such that $U_y\subseteq X\setminus\{x\}$; in that case we could write $$X\setminus\{x\} = \bigcup_{y\in X\setminus\{x\}} U_y.$$
On the other hand, condition $y\in U_y$ and $U_y\subseteq X\setminus\{x\}$ is equivalent to $y\in U_y,\ x\not\in U_y$. Thus, singleton $\{x\}$ is closed if and only if $$(\forall y\neq x)(\exists\ \text{open}\ U)\ y\in U,\ x\not\in U.$$
This is closely related to the notion of $T_1$-space, where every two distinct points $x,y$ have open neighbourhoods not containing the other point. Equivalent way to say this is to say that all singletons are closed.
A stronger condition than $T_1$ is $T_2$-space (more commonly referred to  as Hausdorff space) which says that every two distinct points $x,y$ have disjoint open neighbourhoods. It is obvious that $T_2$ implies $T_1$.
Finally, every metric space is a Hausdorff space. Let $x,y\in (X,d)$. Define $U = B(x,d(x,y)/2)$ and $V = B(y,d(x,y)/2)$. Then $U\cap V=\emptyset$ by triangle inequality which completes the proof. Thus, every metric space is Hausdorff and every singleton in a metric space is closed.
Considering $\mathbb N\subseteq \mathbb R$ with subspace topology, it is again a metric space with metric inherited from $\mathbb R$. Thus, all singletons are closed. However, all singletons are open as well, since an open ball with radius $1$ in $\mathbb N$ is given by $B(x,1) = \{y\in\mathbb N\mid d(x,y)<1\} = \{x\}$. This is an example of discrete space.
