Union of Empty Set

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Theorem

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


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

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


Proof

Let $\mathbb S = \O$.

Then from the definition:

$\bigcup \mathbb S = \set {x: \exists X \in \mathbb S: x \in X}$

from which it follows directly:

$\bigcup \mathbb S = \O$

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

$\blacksquare$


Also presented as

Using the terminology of indexed families, this can also be written as:

$\ds \bigcup_{i \mathop \in \O} S_i = \set {x: \exists i \in \O: x \in S_i} = \O$


Also see


Sources

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