Cartesian Product is Empty iff Factor is Empty
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Theorem
- $S \times T = \O \iff S = \O \lor T = \O$
Thus:
- $S \times \O = \O = \O \times T$
Family of Sets
Let $I$ be an indexing set.
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.
Then:
- $S = \O$ if and only if $S_i = \O$ for some $i \in I$
Proof
\(\ds S \times T\) | \(\ne\) | \(\ds \O\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists \tuple {s, t}\) | \(\in\) | \(\ds S \times T\) | Definition of Empty Set | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists s \in S\) | \(\land\) | \(\ds \exists t \in T\) | Definition of Cartesian Product | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds S \ne \O\) | \(\land\) | \(\ds T \ne \O\) | Definition of Empty Set | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \neg \leftparen {S = \O}\) | \(\lor\) | \(\ds \rightparen {T = \O}\) | De Morgan's Laws: Conjunction of Negations |
So by the Rule of Transposition:
- $S = \O \lor T = \O \iff S \times T = \O$
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 6$: Ordered Pairs
- 1964: William K. Smith: Limits and Continuity ... (previous) ... (next): $\S 2.1$: Sets: Exercise $\text{C} \ 5$
- 1965: Claude Berge and A. Ghouila-Houri: Programming, Games and Transportation Networks ... (previous) ... (next): $1$. Preliminary ideas; sets, vector spaces: $1.1$. Sets
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory: Exercise $1.2$
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 3$. Ordered pairs; cartesian product sets
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 8$: Cartesian product of sets