TITLE: Is it possible to interchange order of supremum and supremum?
QUESTION [7 upvotes]: Prove that if A and B are arbitrary sets and f is a bounded real-valued
function on $A\times B$, then
$$ \sup_{a \in A} \sup_{b \in B} f(a,b) = \sup_{b \in B} \sup_{a \in A} f(a,b) . $$
If it is possible, then does it change into one supremum like $\sup f(a,b)$? Thanks for help!

REPLY [10 votes]: Assuming  $A\ne \phi\ne B.$ $$\text {Let }\;  M=\sup \{f(a,b): a\in A\land b\in B\}.$$ $$\text {Let }\; M^*=\sup_{a\in A} \sup_{b\in B}f(a,b).$$ $$\text {For } a\in A \;\text {let }\; G(a)=\sup_{b\in B}f(a,b).$$ (1) For  $M<\infty$: For $n\in N$ let $$A(n)=\{a\in A:\exists b\in B\;(f(a,b)>M-1/n)\}.$$      We have $\forall a\in A\;(G(a)\leq M).$  So $M^*=\sup_{a\in A} G(a)\leq M.$
And  $\forall n\in N\;( A(n)\ne \phi)$, so $\forall n\in N\;(M^*\geq \sup_{a\in A(n)}G(a)>M-1/n).$  Therefore $$\forall n\in N\;(M-1/n< M^*\leq M)$$ which implies $M^*=M.$
(2) For  $M=\infty$: For $n\in N$ let $$A^*(n)=\{a\in A:\exists b\in B\;(f(a,b)>n)\}.$$ Then $\forall n\in N\;(A^*(n)\ne \phi)$ , so $$\forall n\in N\; (M^*\geq \sup_{a\in A^*(n)}G(a)>n) \text {...... implying } M^*=\infty =M.$$