The radius of an open ball

Question

This is a somewhat silly question. If X is any space and d is a metric on X, what exactly could be the values of the radius r of an open ball of a member of X? I know r>0 by definition. I'm just confused about the possible choices of r. Must r∈X? Or is r∈ℝ regardless of X?

Answer 1

A metric is, by definition, a function $d:X\times X\to \mathbb{R}_{\ge 0}$ satisfying the usual requirements (see for instance here). So you can choose any $r \in \mathbb{R}_{>0}$. This does not mean that the balls will be all different. For instance, if you put the trivial metric $$ d(x,y) = \begin{cases} 0 \quad x = y \\ 1 \quad x \neq y \end{cases}$$ on any set $X$, then

$$ B_r(x) = \{ y \in X \mid d(x,y) < r \} = \begin{cases} \{x\} \quad 0 < r \le 1 \\ X \quad \ \ \ r > 1 \end{cases}$$

Answer 2

As Ethan pointed out in the comment section, the radius must be a positive real number. The reason is that you should understand the distance as a function $d: X \times X \to \mathbb{R_{\geq 0}}.$ Recall that the definition of the open ball $B_r(x_0)$ of radius $r$ centered at $x_0$ is $$B_r(x_0):= \{x \in X | d(x,x_0) < r\}.$$ Therefore, $r$ could be any real value.