# 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.

• My question was with the usual metric.Sorry for not mentioning that. – Vinod Jan 16 '11 at 6:48
• You will find this funny: youtube.com/watch?v=SyD4p8_y8Kw – Jack D'Aurizio Aug 15 '14 at 1:19
• Umm...every set is a subset of itself, isn't it? – TonyK Aug 2 '16 at 19:47
• "Singleton sets are open because {x} is a subset of itself. " um... so? All sets are subsets of themselves. What does that have to do with being open? "There are no points in the neighborhood of x". Um, yes there are $(x - \epsilon, x + \epsilon)$ have points. But $(x - \epsilon, x + \epsilon)$ doesn't have any points of ${x}$ other than $x$ itself so $(x- \epsilon, x + \epsilon)$ that should tell you that ${x}$ can !!!!NOT!!! be open. – fleablood Aug 2 '16 at 20:26
• @JackD'Aurizio This video gives me comfort that I am not the only one struggling to understand advanced mathematics... But if this is so difficult, I wonder what makes mathematicians so interested in this subject. I also like that feeling achievement of finally solving a problem that seemed to be impossible to solve, but there's got to be more than that for which I must be missing out. Honestly, I chose math major without appreciating what it is but just a degree that will make me more employable in the future. I am afraid I am not smart enough to have chosen this major. – user3000482 May 10 '17 at 18:27

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.

• :Excellent explanation! – Styles Aug 2 '16 at 20:23
• Arturo.Wonderful!! – Peter Szilas Feb 21 '19 at 15:42

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.

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.

• They are also never open in the standard topology. – Sean Tilson Jan 16 '11 at 7:29

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.

• This does not fully address the question, since in principle a set can be both open and closed. – Noah Schweber Sep 8 '15 at 6:12
• @NoahSchweber:What's wrong with chitra's answer?I think her response completely satisfied the Original post. – Styles Aug 2 '16 at 18:56

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.

• Welcome! See math notation guide. You can edit your post to improve its appearance. – user147263 Aug 15 '14 at 1:06

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.