Let $(a_{nm})_{n,m=1}^\infty \in \mathbb{R}$. Show that
\begin{equation} \sup_{n}\sup_{m}a_{nm} = \sup_{m}\sup_{n}a_{nm} \end{equation}
This is a practice question for a test that I'm not really sure where to start. It seems fairly obvious, but I don't know what to do to make it rigorous.
You have $a_{mn} \le \sup_{n'} a_{mn'}$ for all $m,n$, hence $a_{mn} \le \sup_{m'}\sup_{n'} a_{m'n'}$ for all $m,n$. Then $\sup_m a_{mn} \le \sup_{m'}\sup_{n'} a_{m'n'}$ for all $n$ and so $\sup_n \sup_m a_{mn} \le \sup_{m'}\sup_{n'} a_{m'n'}$.
Now repeat the argument with the order switched.
We could just as easily make Milly's idea from the comment section rigorous: $$a_{mn}\leq \sup_{n'} a_{mn'}\leq \sup_{m'}\sup_{n'} a_{m'n'}\Rightarrow \sup_{m,n }a_{mn}\leq\sup_{m'}\sup_{n'} a_{m'n'} $$
Similarly:
$$a_{mn}\leq \sup_{m,n} a_{mn}\Rightarrow \sup_{n'}a_{m n'}\leq \sup_{m,n}a_{mn }\Rightarrow \sup_{m'}\sup_{n'} a_{m'n'}\leq \sup_{m,n} a_{mn}$$
Therefore, $\sup_{m,n}a_{m,n}=\sup_m \sup_n a_{m,n}$. Identical argument for the $\sup$ taken in inverted order.
