Interchange of Suprema

Question

Let $A$ and $B$ be sets and $f:A \times B \to \mathbb{R}$. Then is it true that suprema interchange ? i.e. $$\sup_{s \in A}\sup_{t \in B} f(s,t)\overset{?}{=}\sup_{t \in B}\sup_{s \in A} f(s,t)$$

Answer 1

Consider the chain of inequalities: $$f(s,t)\leq\sup_{(s,t)}f(s,t)\Rightarrow\sup_s f(s,t)\leq\sup_{(s,t)}f(s,t)\Rightarrow\sup_t\sup_s f(s,t)\leq\sup_{(s,t)}f(s,t)$$ and similar chain of inequalities: $$f(s,t)\leq\sup_s f(s,t)\leq\sup_t\sup_s f(s,t)\Rightarrow\sup_{(s,t)}f(s,t)\leq\sup_t\sup_s f(s,t)$$ So both together: $$\sup_t\sup_s f(s,t)=\sup_{(s,t)} f(s,t)$$

The analogous result then holds for interchanged indices: $$\sup_s\sup_t f(s,t)=\sup_{(s,t)} f(s,t)$$

Concluding: $$\sup_t\sup_s f(s,t)=\sup_{(s,t)} f(s,t)=\sup_s\sup_t f(s,t)$$