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.

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    $\begingroup$ If you want $$\bigcup_{i \in S}\,i=\bigcup_{i\in S\cup \emptyset} \,i=\left(\bigcup_{i\in S}\,i\right)\cup\left(\bigcup_{i\in \emptyset}\,i\right)$$ for all sets $S$, then you need to define $$\bigcup_{i\in \emptyset}\,i:=\emptyset\,.$$ Likewise, If you want $$\bigcap_{i \in S}\,i=\bigcap_{i\in S\cup \emptyset} \,i=\left(\bigcap_{i\in S}\,i\right)\cap\left(\bigcap_{i\in \emptyset}\,i\right)$$ for all sets $S$, then you need to define $$\bigcap_{i\in \emptyset}\,i:=\mathcal{U}\,,$$ where $\mathcal{U}$ is the universe. $\endgroup$ – Batominovski Apr 15 '20 at 11:51
  • $\begingroup$ @Batominovski Ahah, this is a very strong argument indeed. $\endgroup$ – Graviton Apr 15 '20 at 11:58

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 $\{u:(\exists v)[u\in v\in\varnothing]\}$.


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