Product of Indexed Suprema of Non-Negative Numbers
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Theorem
Let $\family {a_i}_{i \mathop \in I}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $I$.
Let $\family {b_j}_{j \mathop \in J}$ be a family of elements of the non-negative real numbers $\R_{\ge 0}$ indexed by $J$.
Let $\map R i$ and $\map S j$ be propositional functions of $i \in I$, $j \in J$.
Let $\ds \sup_{\map R i} a_i$ and $\ds \sup_{\map S j} b_j$ be the indexed suprema on $\family {a_i}$ and $\family {b_j}$ respectively.
Then:
- $\ds \paren {\sup_{\map R i} a_i} \paren {\sup_{\map S j} b_j} = \sup_{\map R i} \paren {\sup_{\map S j} \paren {a_i b_j} }$
Proof
\(\ds \paren {\sup_{\map R i} a_i} \paren {\sup_{\map S j} b_j}\) | \(=\) | \(\ds \sup_{\map R i} \paren {a_i \times \sup_{\map S j} b_j}\) | Product with Supremum is Supremum of Product of Non-Negative Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \sup_{\map R i} \paren {\sup_{\map S j} \paren {a_i b_j} }\) | Product with Supremum is Supremum of Product of Non-Negative Numbers |
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $35$