Let $f$ be a function defined over $X\times \mathcal{C}$, where $\mathcal{C}$ is a finite collection of sets, and $X=\mathbb{R}$. Furthermore, Let $f$ be bounded on $X\times \mathcal{C}$. My question is:
Can we say that $$\sup_{t}\max_S f(t,S)=\max_S\sup_t f(t,S)?$$
I think the answer should be positive as indicated by the following quick analysis that I chalked out:
$$\sup_t\max_Sf(t,S)\ge f(t,S),\ \forall t\in X, Y\in \mathcal{C}\\\implies \sup_t\max_Sf(t,S)\ge \max_S\sup_t f(t,S).$$ Similarly establishing the reverse direction, we find the desired equality.
The above method seems legit to me, but I am not sure if I am missing something, probably some subtle aspect of $\sup$, or something related to compactness of the $X$. I really don't find anything wrong, but it will be really helpful if the experts in this site can kindly verify the validity of the statement, or provide counter-examples, and relevant references for this problem.
BTW, although this seems to me a very standard and classical problem, I could not find anything directly involving the $\sup$ and $\max$, whereas its relative, Sion's minimax theorem, which involves $\sup$ and $\inf$ and is true only under certain restrictions on $f$, is a rather famous and useful tool in many parts of analysis.
Any help is appreciated. Thanks in advance.
You can always switch the order in which supremum is taken. Of course maximum is same as supremum when it is sup over a finite set. The proof just involves verifying that each side is less than nor equal to the other.