Problem
Let $(X,d)$ be a metric space where $X$ is a non-empty set. Is the empty set an open ball in $X$?
I think that it is true because if $X=\mathbb{R}$ with the usual metric then for all $a\in\mathbb{R}$ we can say that the set $(a,a)$ is an open interval in $\mathbb{R}$.
But I can't devise a proof of this. Can anyone help?
By the way I am using the following definition of an open ball,
Open Ball (in a Metric Space)
Let $(X, d)$ be a metric space and let $r\in\mathbb{R}^+$. Then the set,
$B_d(x, r) := \{y \in X : d(x, y) < r\}$
will be said to be the open ball of radius $r$ centered at $x$ in the metric space $(X, d)$.
Best Answer
It is not a matter of "thinking", "considering" or "debating". Your professor perhaps has given a definition of open ball. Or, at least, he must have assumed some definition.
That definition should specify if the radius of the ball must be a positive number or null radii are allowed.
From my experience, most books that include a definition of open ball say that the radius must be positive; in this case, the empty set is not a ball in any metric space, since the center must belong to the ball.
In any case, topologic and metric properties are not affected in any way.