# Open set on an arbitrary metric on any finite set

I know that for the standard metric on R, every singleton is closed. However, I have a claim that on a finite set for any arbitrary metric every singleton is open, how can I prove this?

• Hint: If $X$ is a finite metric space, then the set $\{d(x, y) \in (0, \infty) : x, y \in X, x \neq y\}$ is finite and hence has a (strictly positive) minimum. – Theo Bendit 2 days ago
• Great hint, I wil try to sketch a proof and update – Beaba 2 days ago
• How do I prove that {0} is not a subset of this set? – Beaba 2 days ago
• If $0$ were in this set, then $d(x, y) = 0$ for some $x, y \in X, x \neq y$, which violates one of the axioms. – Theo Bendit 2 days ago

If $$F$$ is finite, and $$x \in F$$, then the set $$D_x:=\{d(x,y) : y \in F, y \neq x\}$$ is also a finite set and does not contain $$0$$ (as $$d(x,y)=0$$ iff $$x=y$$ and $$y$$ is never equal to $$x$$) so $$r=\min D_x$$ is well-defined and $$>0$$ (set it to $$1$$ if $$D_x$$ is empty, which happens for $$F=\{x\}$$). Then for all $$y \neq x \in F$$, $$d(x,y) \ge r$$, so $$y \notin B(x,r) \cap F$$. It follows that $$B(x,r) \cap F = \{x\}$$ and so $$\{x\}$$ is relatively open in $$F$$.