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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.

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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

5 Answers 5

up vote 16 down vote accepted

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 $\mathop{\cup}\limits_{i\in I}A_i$ 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.

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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.

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

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.

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Assume for a Topological space (X,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 T then T must be the cofinite topology with U \in 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.

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

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.

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