Why in open balls is radius $r>0$?

Question

the usual definition is the following:

Def.1: given $(a,f)$ a metric space, $c \in a$ and $r \in \Bbb{R}_{>0}$, the open ball of radius $r$ about $c$ is the set $$\mathcal{B}_f(c,r)=\{x \in a| f((c,x))<r\}$$

Why in open balls is radius $r>0$? And if in Def.1 I have $r \in \Bbb{R}$ then $$\mathcal{B}_f(c,r)=\emptyset \Leftrightarrow r \leq 0$$ ???

Answer 1

In a metric space, a distance (in your case $f$) is defined to be non-negative, i.e. $\forall x,y\ f(x,y) \geq 0$

So let $x \in \mathcal{B}(c,r)$ then $f(c,x) < r$ but if $r \leq 0$ then $0\leq f(c,x) < r \leq 0$ wich cannot be. Therefore $\nexists x \in \mathcal{B} \Rightarrow \mathcal{B} = \emptyset$

For other implication ($\mathcal{B}_f(c,r) = \emptyset \Rightarrow r \leq 0$) just note that $f(c,c) = 0$ so if $c \notin \mathcal{B}$ then $r \leq 0$

Answer 2

If $f(⋅,⋅)$ is the metric/distance which endows your metric space $A$ with a metric space structure, then $f(c,x)\geq 0$ for all $c,x\in A$ by definition of metric. This implies that the inequality $f(c,x)<r$, for $r<0$ has no solution.