Switching the order of supremum

Question

Let $f:[0, T]\times\{1, \cdot, N\}$ a function.

Can I say that $$\sup_{t\in [0, T]}\sup_{ x\in \{1, \cdots, N\}}f(t, x)=\sup_{ x\in \{1, \cdots, N\}}\sup_{t\in [0, T]}f(t, x)$$

This should be possible since of of the two spaces is finite.

Thank you

Answer 1

Yes. Even if both spaces are infinite, you can always swap the suprema (but not a supremum and an infimum).

Let $a(t) := \sup_{x\in\{1,\dots,N\}}f(t,x)$ and $b(x) := \sup_{t\in[0,T]}f(t,x)$. Suppose by contradiction $\sup a > \sup b$. Then there exists $t\in[0,T]$ and $\varepsilon>0$ such that $$\forall x\in\{1,\dots,N\}, a(t)\geq b(x)+\varepsilon$$ In fact, $\varepsilon = a(t)-\sup b$. However, $$\forall x\in\{1,\dots,N\},f(t,x)\leq \sup_{s\in[0,T]} f(s,x) = b(x)$$ and therefore by taking the supremum $$a(t)\leq \sup b$$ which contradicts the first statement (using a sequence $(x_n)\in\{1,\dots,N\}^\mathbb N$ such that $\lim_{n\to\infty}b(x_n)=\sup b$). With a similar reasoning, you get a contradiction supposing $\sup a < \sup b$ too.