Exchange of Order of Supremum Operators

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:

$\ds \sup_{\map R i} \paren {\sup_{\map S j} a_{i j} } = \sup_{\map S j} \paren {\sup_{\map R i} a_{i j} }$

For any chosen $i$ and $j$, we have:

$\ds a_{ij} \le \sup_{\map R i} a_{ij}$

Taking the supremum over $\map S j$ yields:

$\ds \sup_{\map S j} a_{ij} \le \sup_{\map S j} \sup_{\map R i} a_{ij}$

for all $i$.

Hence the right hand side remains an upper bound after taking the supremum over $\map R i$ on the left hand side.

That is:

$\ds \sup_{\map R i} \paren {\sup_{\map S j} a_{ij} } \le \sup_{\map S j} \paren {\sup_{\map R i} a_{ij} }$

Applying the same argument with $\ds a_{ij} \le \sup_{\map S j} a_{ij}$, we obtain the reverse inequality.

$\blacksquare$

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 $35$