Union of Empty Set

Theorem
Consider the set of sets $\mathbb S$ such that $\mathbb S$ is the empty set $\O$.

Then the union of $\mathbb S$ is $\O$:


 * $\mathbb S = \O \implies \ds \bigcup \mathbb S = \O$

Proof
Let $\mathbb S = \O$.

Then from the definition:
 * $\ds \bigcup \mathbb S = \set {x: \exists X \in \mathbb S: x \in X}$

from which it follows directly:
 * $\ds \bigcup \mathbb S = \O$

as there are no sets in $\mathbb S$.

Also presented as
Using the terminology of indexed families, this can also be written as:
 * $\ds \bigcup_{i \mathop \in \O} S_i = \set {x: \exists i \in \O: x \in S_i} = \O$