Let $A$ be a commutative ring and $\mathfrak{p} \subset$ Spec($A$).
Then for an ideal $\mathfrak{a} \subset A$, we have a closed set in the Zariski topology defined by
$V(\mathfrak{a})$:={$\mathfrak{p} \in$ Spec($A$) | $\mathfrak{a} \subset \mathfrak{p}$} $\subset$ Spec($A$).
My question is, why do we allow that $V(\mathfrak{a})= \emptyset$? As there is no empty prime ideal the inclusion $\emptyset = V(\mathfrak{a}) \subset$ Spec($A$) does not make sense to me.