Why does $\bigcup_{i\in\emptyset}i=\emptyset$ (the empty union) It seems that the consensus/convention is that the empty union, $\bigcup_{i\in\emptyset}i=\emptyset$
At first glance this seems intuitive, however please consider my following arguments and let me know where I go wrong.
Let $a_0=\emptyset$, $a_{n+1}=\{a_n\}$. (e.g; $a_3=\{\{\{\emptyset\}\}\}$)
$$a_{n}\neq a_m \iff m \neq n$$
For ease of notation, let $f(S)=\bigcup_{i\in S}i$
$$\therefore f(a_{n+1})=\bigcup_{i\in a_{n+1}}i=\bigcup_{i\in \{a_n\}}i=a_n$$
$$\implies f(a_0)=a_{-1}=f(\emptyset)=\bigcup_{i\in \emptyset}i=\emptyset$$
$$\implies a_0=a_{-1}$$
Why should it be the case that the equivalence of $a_{n}\neq a_m \iff m \neq n$ fails when $m=0$ and $n=-1$? Would it not be more intuitive to define $a_{-1}$ as undefined, rather than accepting the convention?
If $f(\emptyset)$ describes the union of the elements of $\emptyset$, the set than contains nothing; which (at least) linguistically is equivalent to the union of nothing. I notice that the set containing nothing is distinct from nothing itself. Thus, why is the convention nondistinct? Would it not be more intuitive to instead define $\emptyset \neq f(\emptyset)$. 
Lastly, consider how there is a one-to-one correspondence between finitary pure sets and rooted identity trees (see A004111 on the OEIS). The previous series of $a_n$ would correspond to a unique tree of $n+1$ nodes. Hence, $a_{-1}$ corresponds to a tree of $0$ nodes, which is equivalent to no set at all, strictly distinct from the empty set.

Let $X$ be a set. The axiom of union (in ZFC) states that
$$\{u:(\exists v)[u\in v\in X]\}=: \bigcup X$$
is a set.
Now let $X$ be the empty set $\varnothing$. Then
$$\bigcup \varnothing=\{u:(\exists v)[u\in v\in\varnothing]\}=\varnothing.$$
Indeed, the first equality is simply a definition and the second equality follows because there does not exist an element in $\{u:(\exists v)[u\in v\in\varnothing]\}$.
