Union of Union of Cartesian Product
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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
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 $4 \ \text {(a)}$