Permutation of Indices of Supremum
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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 $\map R i$ be a propositional functions of $i \in I$.
Let $\ds \sup_{\map R i} a_i$ be the indexed supremum on $\family {a_i}$.
Then:
- $\ds \sum_{\map R i} a_i = \sum_{\map R {\map \pi i} } a_{\map \pi i}$
where $\pi$ is a permutation on the fiber of truth of $R$.
Proof
\(\ds \sup_{\map R {\map \pi j} } a_{\map \pi j}\) | \(=\) | \(\ds \sup_{j \mathop \in I} a_{\map \pi j} \sqbrk {\map R {\map \pi j} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sup_{j \mathop \in I} \paren {\sup_{i \mathop \in I} a_i \sqbrk {\map R i} \sqbrk {i = \map \pi j} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sup_{i \mathop \in I} a_i \sqbrk {\map R i} \sup_{j \mathop \in I} \sqbrk {i = \map \pi j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sup_{i \mathop \in I} a_i \sqbrk {\map R i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sup_{\map R i} a_i\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sup_{\map R j} a_j\) | Change of Index Variable of Supremum |
$\blacksquare$
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 $35$