TITLE: Switching order of supremum for doubly indexed sequence?
QUESTION [6 upvotes]: Suppose you have a doubly indexed sequence of reals, $(\alpha_{ij})$. Why is
$$
\sup_i \;\sup_j\ \alpha_{ij}=\sup_j\;\sup_i\ \alpha_{ij}?
$$
I know one approach is to note $\alpha_{mn}\leq\sup_j\;\sup_i\alpha_{ij}$ for any $m$ and $n$. Why is this exactly? I don't really know what $\sup_j\sup_i\alpha_{ij}$ means. Does it mean first fix some $j$ and find the supremum of $\alpha_{ij}$ as $j$ remains fixed as $i$ runs over $\mathbb{N}$? And then after that, find the supremum of $\sup_i\alpha_{ij}$ as $j$ runs over $\mathbb{N}$? It's not clear how I would find $\sup_i\alpha_{ij}$ for fixed arbitrary $j$ first. Thanks.

REPLY [4 votes]: To directly argue:
$$
a_{ij} \leq \sup_j a_{ij}\implies \sup_{i}a_{ij} \leq \sup_i\sup_j a_{ij} \implies
\sup_j\sup_{i}a_{ij} \leq \sup_i\sup_j a_{ij}
$$