Union of Union of Cartesian Product

Theorem

Let $A$ and $B$ be sets such that $A \ne \O$ and $B \ne \O$.

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

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

Then:

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

where:

$\cup$ denotes union

$\times$ denotes Cartesian product.

Proof

\(\ds \bigcup \bigcup \paren {A \times B}\)

\(=\)

\(\ds \bigcup \bigcup \set {\tuple {a, b}: a \in A, b \in B}\)

Definition of Cartesian Product

\(\ds \)

\(=\)

\(\ds \bigcup \paren {\bigcup \set {\set {\set a, \set {a, b} }: a \in A, b \in B} }\)

Definition of Kuratowski Formalization of Ordered Pair

\(\ds \)

\(=\)

\(\ds \bigcup \set {\set {a, b}: a \in A, b \in B}\)

Definition of Set Union: $\ds \bigcup \set {\set a, \set {a, b} } = \set {a, b}$

\(\ds \)

\(=\)

\(\ds \set {x: x \in A \text { or } x \in B}\)

Definition of Set Union

\(\ds \)

\(=\)

\(\ds A \cup B\)

Definition of Set Union

$\blacksquare$

Also see

• Union of Union of Cartesian Product with Empty Factor

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 $4 \ \text {(a)}$