GCD of an empty set?
Yes, the convention $\gcd \emptyset =0 $ makes sense.
Every integer divides all elements of $\emptyset$ thus the "greatest" among them is $0$, when "greatest" is understood with respect to the partial order given by divisibility, which is the appropriate notion of 'size' in this context.
Define $\gcd(S)$ to be natural number $g$ (if it exists) such that:
$g \mid x$ for every $x \in S$.
For any $d$ such that $d \mid x$ for every $x \in S$, we have $d \mid g$.
Then indeed $\gcd(\{\}) = 0$, and moreover it can be proven that $\gcd(S)$ exists for every set $S \subset \mathbb{Z}$.
Also $\gcd(\{0,0\}) = 0$. (This differs from the other common definition of $\gcd$ as the greatest common divisor in terms of size, which would simply stipulate that $\gcd(0,0)$ is undefined.)