Let $\displaystyle A$ and $\displaystyle B$ be nonempty finite sets. For $\displaystyle f:A \rightarrow B$ to be a function, it must satisfy this condition:

$\displaystyle x\in A \Rightarrow (x, f(x)) \in f$.

Now, if I make $\displaystyle A=\emptyset$ and $\displaystyle B=\emptyset$, the statement of implication becomes

$\displaystyle x\notin A \Rightarrow (x,f(x)) \notin f$, from which we know that the statement of implication is true, since

False $\displaystyle \Rightarrow$ False is a true statement. In this regard, I have a question:

Is $\displaystyle f:\emptyset \rightarrow \emptyset$ a function?