Why is an empty function considered a function?
One thing to point out is that a function is not just any set of ordered pairs; a binary relation is a set of ordered pairs. However, a function $F$ in this case is a special type of binary relation such that for every $x\in\text{dom }F$, there is a unique $y\in\text{ran }F$, such that $xFy$, or $(x,y)\in F$. But $\text{dom }\varnothing=\varnothing=\text{ran }\varnothing$. Can you find some $x\in\text{dom }\varnothing$ such there is not a unique $y\in\text{ran }\varnothing$ for which $x\varnothing y$? Since you cannot, $\varnothing$ is indeed a function.
I believe that the answer lies hidden within the depth of "vacuously true argument".
An argument of the form $\forall x\varphi$ is true if and only if there is no $x$ such that $\lnot\varphi(x)$.
For example if our universe is the natural numbers with the usual $\ge$ order, then $\forall x(x\ge 0)$ is true because there are no negative numbers.
On the other hand, $\forall x(x\ge 0 \land x\neq 0)$ is false, simply because setting $x=0$ is a counterexample.
More generally, a sentence "If $p$ then $q$" ($p\implies q$, or $p\rightarrow q$) is true whenever the assumption is false, i.e. $p$ never occurs.
An example I often use is "If I am standing upside down from the ceiling right now, then you are all unicorns". It does not matter that I am talking to people, and not to unicorns, because I never stand upside down from the ceiling (it gives me a huge headache, you see).
The next point in our journey towards the empty function, is the bounded quantification. When we write $\forall x\in A(P(x))$ we actually write $\forall x(x\in A\rightarrow P(x))$, this means that we quantify over all the possible $x$, but if $x\notin A$ then we do not care about it anymore (the proposition is true since the assumption is false).
And lastly, the definition of a function $F$ is this: $$\begin{align} \forall z & (z\in F\rightarrow\exists x\exists y(z=\{\{x\},\{x,y\}\})\land\\ & \forall x(\exists z\exists y(z\in F\land z=\{\{x\},\{x,y\}\})\rightarrow \\ &\qquad(\forall u\forall v(\exists z\exists w((z\in F\land w\in F\land z= \{\{x\},\{x,v\}\}\land w=\{\{x\},\{x,u\}\})\rightarrow u=v) \end{align}$$
Let's read this long formula. It says that $F$ is such that every element of $F$ is an ordered pair, and for every $x$, if there is an ordered pair $z$ with $x$ for left coordinate, then there is only one such pair (i.e. given two pairs, if their right coordinate is equal then they are equal).
Informally, $F$ is a function if it is a set of ordered pairs, that for every $x\in Dom(f)$ there is a unique $y$ such that $\langle x,y\rangle\in F$.
An example is $F=\{\langle 1,2\rangle\}$ is a function, all its members are ordered pairs, and since there is only one member it automatically satisfies the requirement that the left-coordinate determines the pair.
On the other hand $R=\{\langle 1,2\rangle,\langle 1,3\rangle\}$ is a set that indeed all its members are ordered pairs, but there are two distinct ordered pairs with $1$ in the left coordinate, so it is not a function.
And even more $A=\{3,\langle 1,2\rangle\}$ is clearly not a function, since $3$ is not an ordered pair!
The definition of a function is vacuously true when it is applied to the empty set, let us see why:
- For all $z$ if $z\in\emptyset$ then $z$ is an ordered pair is vacuously true, as no $z$ is a member of the empty set.
- Next we have, that for all $x$, if there is some ordered pair in the empty set, with $x$ as left coordinate then the right coordinate is unique, this is also vacuously true since there is no ordered pair in the empty set with $x$ in the left coordinate. This is exactly the case $p\rightarrow q$ and $p$ is false.
The conjunction of two true statement is true as well, therefore the empty set satisfies the requirement that every element of it is an ordered pair, and if two ordered pairs have the same left coordinate then they are equal. Therefore, $\emptyset$ is a function.
The empty set is a set of ordered pairs. It contains no ordered pairs but that's fine, in the same way that $\varnothing$ is a set of real numbers though $\varnothing$ does not contain a single real number.