Why Is Every Singleton Set Open in a Finite Metric Space?

Recall the definitions. In a metric space $(X, d)$, the open ball of radius $r > 0$ around $x$ is

$$B(x, r) = \{\, y \in X : d(x, y) < r \,\}.$$

A set $U \subseteq X$ is open if every point of $U$ has some open ball around it that stays inside $U$.

For $U = \{x\}$ this means: we need a radius $r > 0$ with $B(x, r) = \{x\}$, so that the ball contains no other point.

The proof

Let $X$ be a finite metric space and $x \in X$.

Case 1: $X = \{x\}$. Then $\{x\} = X$, and the whole space is always open.

Case 2: $X$ has other points. Define

$$r = \min\{\, d(x, y) : y \in X,\ y \neq x \,\}.$$

This minimum exists because it is taken over a finite, non-empty set of numbers. Each of those numbers is positive, because $d(x, y) > 0$ whenever $y \neq x$. So $r > 0$.

Now if $y \neq x$, then $d(x, y) \ge r$, so $y \notin B(x, r)$. Hence

$$B(x, r) = \{x\} \subseteq \{x\},$$

and $\{x\}$ is open. $\blacksquare$

Consequence: every subset is open

Any subset $A \subseteq X$ is the union of its singletons, $A = \bigcup_{a \in A} \{a\}$. A union of open sets is open, so $A$ is open. Taking complements, every subset is also closed.

So a finite metric space always carries the discrete topology. In particular, a finite metric space with at least two points is never connected.

Where finiteness is used

The only step that needs finiteness is “the minimum exists and is positive”. In an infinite space the distances from $x$ to the other points can have infimum $0$, and then no ball isolates $x$:

• In $\mathbb{R}$, the distances from $0$ to the points $1/n$ are $1/n \to 0$. Every ball around $0$ contains some $1/n$, so $\{0\}$ is not open.
• Some infinite spaces still have open singletons. In $\mathbb{Z}$ with $d(m, n) = |m - n|$, the ball $B(n, 1/2) = \{n\}$, so $\mathbb{Z}$ is discrete too.

What matters is whether $x$ is an isolated point: whether $\inf_{y \neq x} d(x, y) > 0$. Finiteness guarantees this for every point.

Common mistakes

• Writing “let $r = \min d(x,y)$” without saying why it is positive. The proof depends on $y \neq x \Rightarrow d(x,y) > 0$, which is one of the metric axioms.
• Using a closed ball. The closed ball $\{y : d(x,y) \le r\}$ with this $r$ contains the nearest neighbour. Either use the open ball or a smaller radius such as $r/2$.
• Forgetting the one-point space, where the minimum is over an empty set.