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!
$\begingroup$
$\endgroup$
2
-
3$\begingroup$ My initial impression is that they should both be equal to the supremum over all $(a,b) \in A \times B$ $\endgroup$– Jordan HardyCommented Apr 9, 2016 at 7:05
-
$\begingroup$ Instead of proving your equation $x=y$ in a single stroke, can you prove $x\leq y$? $\endgroup$– Chris CulterCommented Apr 9, 2016 at 7:24
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
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 $$\forall n\in N\; (M^*\geq \sup_{a\in A^*(n)}G(a)>n) \text {...... implying } M^*=\infty =M.$$
answered Apr 9, 2016 at 8:08
-
1$\begingroup$ Now, I understand sup{f(a,b):a∈A∧b∈B} = supa∈Asupb∈Bf(a,b). Then, I should set another H(a) = supa∈A f(a,b) & new M`` =supb∈Bsupa∈Af(a,b), and then, residual process is trivial and we got our proof is that right? $\endgroup$– DkdgCommented Apr 9, 2016 at 8:24
-
$\begingroup$ Yes.Just interchange A and B throughout. $\endgroup$ Commented Apr 9, 2016 at 12:22
-
$\begingroup$ Perfect! Thank you again~ have a nice weekend $\endgroup$– DkdgCommented Apr 9, 2016 at 13:37