Sum of Indexed Suprema

Theorem

Let $\sequence {a_i}_{i \mathop \in I}$ be a family of elements of the real numbers $\R$ indexed by $I$.

Let $\sequence {b_j}_{j \mathop \in J}$ be a family of elements of the real numbers $\R$ 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 $\sequence {a_i}$ and $\sequence {b_j}$ 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

Although this article appears correct, it's inelegant. There has to be a better way of doing it. In particular: The results upon which this depends need to be enhanced to encompass the full supremum, not just the max You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Improve}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

\(\ds \paren {\sup_{\map R i} a_i} + \paren {\sup_{\map S j} b_j}\)

\(=\)

\(\ds \sup_{\map R i} \paren {a_i + \sup_{\map S j} b_j}\)

Sum with Maximum is Maximum of Sum

\(\ds \)

\(=\)

\(\ds \sup_{\map R i} \paren {\sup_{\map S j} \paren {a_i + b_j} }\)

Sum with Maximum is Maximum of Sum

$\blacksquare$

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 $35$