Union of Union of Cartesian Product with Empty Factor

Theorem

Let $A$ and $B$ be sets such that either $A = \O$ or $B = \O$.

Let the ordered pair $\tuple {a, b}$ be defined using the Kuratowski formalization:

$\tuple {a, b} := \set {\set a, \set {a, b} }$

Then:

$\ds \bigcup \bigcup \paren {A \times B} = A \cup B \iff A = B = \O$

where:

$\cup$ denotes union

$\times$ denotes Cartesian product.

That is, if either $A$ or $B$ is empty:

$\ds \bigcup \bigcup \paren {A \times B} = A \cup B$

holds if and only if they are both empty

Proof

Let $A = \O$ or $B = \O$.

From Cartesian Product is Empty iff Factor is Empty:

$A \times B = \O$

Hence from Union of Empty Set:

$\ds \bigcup \bigcup \paren {A \times B} = \O$

However, from Union is Empty iff Sets are Empty:

$A \cup B = \O \iff A = \O \text { and } B = \O$

The result follows.

$\blacksquare$

Also see

• Union of Union of Cartesian Product

Sources

• 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $4 \ \text {(b)}$