What is $\sup \emptyset$ (if exists) w.r.t the usual ordering on ordinals?

Question

My textbook Introduction to Set Theory 3rd by Hrbacek and Jech defines the supremum of a set $X$ of ordinals as follows:

Clearly, If $X=\emptyset$, then $\bigcup \emptyset$ is undefined. So I would like to ask what is $\sup \emptyset$ (if exists).

IMHO, $\sup \emptyset = 0$ because

• $0$ is an upper bound of $\emptyset$. This is a vacuous truth (every element of $\emptyset$ is less than or equal to $0$).

• $0$ is the least ordinal. Thus if $\alpha$ is an upper bound of $\emptyset$, then $0 \le \alpha$.

Thank you for your help!

Answer 1

This elaborates on Greg Martin's comment:

$\bigcup\emptyset$ is in fact well defined. By definition it is $$\{x:\exists y\in\emptyset(x\in y)\},$$ which is simply $\emptyset$.

The intersection $\bigcap\emptyset$ is problematic - or, to put it precisely, the class $$\{x:\forall y\in\emptyset(x\in y)\}$$ is not a set (it's all of $V$).

• With a bit more precision: $\bigcup\mathcal{C}$ and $\bigcap\mathcal{C}$ always define a class for every class $\mathcal{C}$. Additionally we have that in case $\mathcal{C}$ is a set, the class $\bigcup\mathcal{C}$ is also a set. This is all precisely expressible and provable in, say, NBG; in ZFC (which can't directly talk about classes) we have to employ the usual classy circumlocutions.

So the standard definition works (and coincides with intuition): $sup(\emptyset)=\bigcup\emptyset=\emptyset=0$. The point where we have an issue is when we try to calculuate the infimum of the emptyset, since that corresponds to $\bigcap\emptyset$.