Definition. Let $(X,d)$ be a metric space. An open ball of radius $r>0$ around a point $x\in X$ is the set $B(x,r) = \{ y \in X \, | \, d(x,y)<r \}$.
Looking at the definition of an open ball, how is the empty set not an open ball? Surely we are not guaranteed the existence of $y$'s which fulfill the condition, are we? What prevents the ball from being empty?
The example I have in mind is the metric space of the natural numbers $\mathbb{N}$ with the Euclidean metric $|x-y|$. If I pick a ball of radius $r=1/2$ it will be empty around any point, wouldn't it?
Open sets are usually defined as sets which contain an open ball around all of their points$^1$. Nothing is specified about the open balls not being empty. I see how allowing the empty set be an open ball would result in any set being open. This is why I think there is something about the definition which I misunderstand.
Also, on a bit of a tangent, in this question (Is the empty set an open ball in a metric space?) one of the answers claims that an open ball contains the point around which it is centered. How can this be? If $d(x,y)=0 \Leftrightarrow x=y$ and $r>0$ then $x \notin B(x,r)$. Is this true and if yes, is it related to the previous issue I had?
$^1$ See for example E. Kreyszig, Functional Analysis, pp. 18, Definition 1.3-2.
It is indeed the case that every open ball contains its center: this is because, as long as $r>0$, we have $d(x,x)<r$.
Given that you specifically mention being confused by this claim, I suspect that what's going on is that you're misreading the definition of $B(x,r)$; it's $$B(x,r)=\{y: d(x,y)\color{red}{<}r\},$$ not $$B(x,r)=\{y: d(x,y)\color{red}{=}r\}\mbox{ or }B(x,r)=\{y: d(x,y)\color{red}{>}r\}.$$ We always have $d(x,x)<r$ (as long as $r>0$) so we always have $x\in B(x,r)$.