Are Singleton sets in $\mathbb{R}$ both closed and open? I think singleton sets $\{x\}$ where $x$ is a member of $\mathbb{R}$ are both open and closed.
Singleton sets are open because $\{x\}$ is a subset of itself. There are no points in the neighborhood of $x$.
I want to know singleton sets are closed or not.
 A: Every singleton set is closed. It is enough to prove that the complement is open. Consider $\{x\}$ in $\mathbb{R}$. Then $X\setminus \{x\} = (-\infty, x)\cup(x,\infty)$ which is the union of two open sets, hence open. Since the  complement of $\{x\}$ is open, $\{x\}$ is closed.
A: It depends on what topology you are looking at. For $T_1$ spaces, singleton sets are always closed. So for the standard topology on $\mathbb{R}$, singleton sets are always closed.
A: As has been noted, the notion of "open" and "closed" is not absolute, but depends on a topology. So in order to answer your question one must first ask what topology you are considering.
A topological space is a pair, $(X,\tau)$, where $X$ is a nonempty set, and $\tau$ is a collection of subsets of $X$ such that:


*

*$\emptyset$ and $X$ are both elements of $\tau$;

*If $A$ and $B$ are elements of $\tau$, then $A\cap B$ is an element of $\tau$;

*If $\{A_i\}_{i\in I}$ is an arbitrary family of elements of $\tau$, then $\bigcup_{i\in I}A_i$ is an element of $\tau$.


The elements of $\tau$ are said to be "open" (in $X$, in the topology $\tau$), and a set $C\subseteq X$ is said to be "closed" if and only if $X-C\in\tau$ (that is, if the complement is open).
In $\mathbb{R}$, we can let $\tau$ be the collection of all subsets that are unions of open intervals; equivalently, a set $\mathcal{O}\subseteq\mathbb{R}$ is open if and only if for every $x\in\mathcal{O}$ there exists $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq\mathcal{O}$. You may want to convince yourself that the collection of all such sets satisfies the three conditions above, and hence makes $\mathbb{R}$ a topological space. This topology is what is called the "usual" (or "metric") topology on $\mathbb{R}$.
If you are working inside of $\mathbb{R}$ with this topology, then singletons $\{x\}$ are certainly closed, because their complements are open: given any $a\in \mathbb{R}-\{x\}$, let $\epsilon=|a-x|$. Then $x\notin (a-\epsilon,a+\epsilon)$, so $(a-\epsilon,a+\epsilon)\subseteq \mathbb{R}-\{x\}$; hence $\mathbb{R}-\{x\}$ is open, so $\{x\}$ is closed.
The reason you give for $\{x\}$ to be open does not really make sense. Every set is a subset of itself, so if that argument were valid, every set would always be "open"; but we know this is not the case in every topological space (certainly not in $\mathbb{R}$ with the "usual topology"). So that argument certainly does not work.
So: is $\{x\}$ open in $\mathbb{R}$ in the usual topology? Well, $x\in\{x\}$. Does there exist an $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq \{x\}$? If so, then congratulations, you have shown the set is open. If there is no such $\epsilon$, and you prove that, then congratulations, you have shown that $\{x\}$ is not open.
A: Assume for a Topological space $(X,\mathcal{T})$ that the singleton sets $\{x\} \subset X$ are closed. Then every punctured set $X/\{x\}$ is open in this topology. If these sets form a base for the topology $\mathcal{T}$ then $\mathcal{T}$ must be the cofinite topology with $U \in \mathcal{T}$ if and only if $|X/U|$ is finite. This is because finite intersections of the open sets will generate every set with a finite complement. Equivalently, finite unions of the closed sets will generate every finite set.
A: In the space $\mathbb R$,each one-point {$x_0$} set is closed,because every one-point set  different from $x_0$ has a neighbourhood not intersecting {$x_0$},so that {$x_0$} is its own closure.
NOTE:This fact is not true for arbitrary topological spaces.
Consider the topology $\mathfrak F$ on the three-point set X={$a,b,c$},where $\mathfrak F=${$\phi$,{$a,b$},{$b,c$},{$b$},{$a,b,c$}}.
Since  X\ {$b$}={a,c}$\notin \mathfrak F$ $\implies $ In the topological space (X,$\mathfrak F$),the one-point set {$b$} is not closed,for its complement is not open.
A: If you are giving $\{x\}$ the subspace topology and asking whether $\{x\}$ is open in $\{x\}$ in this topology, the answer is yes.  There is only one possible topology on a one-point set, and it is discrete (and indiscrete).  However, if you are considering singletons as subsets of a larger topological space, this will depend on the properties of that space.  As Trevor indicates, the condition that points are closed is (equivalent to) the $T_1$ condition, and in particular is true in every metric space, including $\mathbb{R}$.  When $\{x\}$ is open in a space $X$, then $x$ is called an isolated point of $X$.  If all points are isolated points, then the topology is discrete.  In the real numbers, for example, there are no isolated points; every open set is a union of open intervals.
In summary, if you are talking about the usual topology on the real line, then singleton sets are closed but not open.
A: You may just try definition to confirm.
Definition of closed set :
Let E be a subset of metric space (x,d).
E is said to be closed if E contains all its limit points.
Now cheking for limit points of singalton set E={p},
for r>0 ,
N(p,r) intersection with (E-{p})  is empty equal to phi
i.e. set of limit points of {p}= phi
so, set {p} has no limit points
so clearly {p} contains all its limit points (because phi is subset of {p})
therefor {p} is closed.
