1

Let $I, J$ be sets, such that $\emptyset \neq I, J$ and function $f : I \times J \longrightarrow \mathbb{R}$ bounded from above. I am supposed to show that:

$$ \sup\{\sup\{f(i,j) : i \in I\} : j \in J\} = \sup\{\sup\{f(i,j) : j \in J\} : i \in I\}. $$

Opposite to what's usually asked here, it seems like the equation is trivial as it is, from my thought process, it seems like it should be obvious that no matter what subset of the domain you take and get it's supremum, as long as the full domain is used up, the supremum always remains same. The equation looks a lot more like a proof on it's own than something to be proven...

What would be the thought process when proving statements like this? I can't grasp how this can be further simplified mathematically aside from writing it word for word like I did in this question.

Xander Henderson
  • 32,453
Mantas Kandratavičius
  • 175
  • 2
    Such exercises are usually given in order to help you familiarize yourself with basic definitions. Thus, I would probably start on the left by unpacking the definition of $\sup$. – Xander Henderson Dec 06 '22 at 19:54
  • If you think $\sup{\inf{f(i,j) : i \in I} : j \in J} = \inf{\sup{f(i,j) : j \in J} : i \in I}$ is just as trivial (hint it is generally wrong) then you may want to follow Xander Henderson's advice to prove it in detail with two $\sup$s. – Kurt G. Dec 06 '22 at 20:54

0 Answers0