Union of Union of Cartesian Product with Empty Factor
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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
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 $4 \ \text {(b)}$