Show that the singleton set is open in a finite metric spce.

Question

Let $(X,d)$ be a metric space such that $X$ has finitely many points. Prove that for every $x\in X$, the singleton set $\{x\}$ is open.

This is what I did: every finite metric space is a discrete space and hence every singleton set is open. But I don't know how to show this using the definition of open set(A set $A$ is open if for every $a\in A$ there is an open ball $B$ such that $x\in B\subset A$).

Hint: There is a smallest non-zero distance between all points. Take a ball with smaller radius. — Tobias Kildetoft, Nov 27 '15 at 10:27
@TobiasKildetoft: Let $\varepsilon$ be the smallest distance between $x$ and any point in $X$. Take the ball $B=B_{\frac{\varepsilon}{2}}$. Am right? — marya, Nov 27 '15 at 10:35

score 2 · Accepted Answer · answered Nov 27 '15 at 10:37

Define $r(x) = \min \{d(x,y): y \in X, y \neq x\}$. This is a minimum of finitely many strictly positive numbers (as all $d(x,y) > 0$ when $x \neq y$). So $r(x) > 0$.

Suppose $y \in B(x,r(x))$ and $y \neq x$. Then by definition of being in the ball $d(x,y) < r(x)$ but $r(x) \le d(x,y)$ by definition of $r(x)$. Contradiction. So $B(x, r(x)) = \{x\}$ and the latter set is open.

Show that the singleton set is open in a finite metric spce.

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