Is the empty set an open ball in a metric space? 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 $(X, d)$.

 A: 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.
A: Actually there are two related questions:


*

*Given the definition you cite, is the empty set an open ball?
This already has been answered by ajotatxe: The definition you cited explicitly excluded radius $0$, and open balls of positive radius are not empty.

*Should a reasonable definition include the empty set as open ball?
When looking at the definition in isolation, it seems arbitrary to require $r>0$. However the introduction of the open ball has a specific purpose: Namely to allow the more general definition of an open set as a set that contains an open ball around any of its points. If you'd include the empty set in the set of open balls, then this definition would result in every set to be open (as any set contains the empty set), which clearly is not wanted.
Now we could fix that be requiring that every point has a non-empty open ball around each point, but given that open balls are defined specifically for this purpose, the reasonable thing to do is to include that condition in the definition.
Also note that by excluding the empty set, we always have $x\in B_d(x,r)$, thus any open ball around $x$ is a neighbourhood of $x$. Again, we'd have to make an exception for the empty set if we'd include it in the definition of open balls.
