Are Singleton Sets in ℝ Both Open and Closed?

Throughout, $\mathbb{R}$ has its usual topology: a set $U$ is open if around every point of $U$ there is an interval $(x - \varepsilon, x + \varepsilon)$ contained in $U$.

$\{a\}$ is closed

A set is closed when its complement is open. The complement of $\{a\}$ is

$$\mathbb{R} \setminus \{a\} = (-\infty, a) \cup (a, \infty).$$

Each piece is an open interval: if $x < a$, the interval $(x - \varepsilon, x + \varepsilon)$ with $\varepsilon = a - x$ stays below $a$, and the case $x > a$ is symmetric. A union of open sets is open, so the complement is open and $\{a\}$ is closed.

Another way to see it: $\{a\}$ contains all of its limit points, because it has none. No other point can be a limit of points of $\{a\}$, since the only sequence available is the constant one.

$\{a\}$ is not open

For $\{a\}$ to be open we would need some $\varepsilon > 0$ with $(a - \varepsilon, a + \varepsilon) \subseteq \{a\}$. But that interval contains $a + \varepsilon/2$, which is not $a$. No $\varepsilon$ works, so $\{a\}$ is not open.

Why no proper non-empty subset of ℝ is both

A set that is both open and closed is called clopen. In $\mathbb{R}$ the only clopen sets are $\varnothing$ and $\mathbb{R}$. That is exactly the statement that $\mathbb{R}$ is connected.

Sketch of the proof: suppose $A$ is clopen with $\varnothing \ne A \ne \mathbb{R}$. Pick $p \in A$ and $q \notin A$, say with $p < q$. Let $s = \sup\{x \in [p, q] : x \in A\}$.

• Since $A$ is closed, $s \in A$.
• Since $A$ is open and $s \in A$, a small interval around $s$ lies in $A$. If $s < q$, that gives points of $A$ greater than $s$ inside $[p, q]$, contradicting the definition of the supremum. If $s = q$, then $q \in A$, which contradicts the choice of $q$.

Either way we reach a contradiction. So in particular a singleton cannot be clopen.

Compare with other spaces

The answer depends on the topology, not just on the set:

Space	$\{a\}$ open?	$\{a\}$ closed?
$\mathbb{R}$, usual topology	No	Yes
$\mathbb{Z}$ as a subspace of $\mathbb{R}$	Yes	Yes
Any finite metric space	Yes	Yes
$\mathbb{R}$ with the discrete topology	Yes	Yes

In $\mathbb{Z}$, $\{n\} = \mathbb{Z} \cap (n - \tfrac12, n + \tfrac12)$, so it is open in the subspace topology even though it isn’t open in $\mathbb{R}$.

Common mistakes

• Treating open and closed as opposites. Most subsets of $\mathbb{R}$, such as $[0, 1)$ or $\mathbb{Q}$, are neither.
• Forgetting which space you are in. “Open” always means open relative to a given space.