Double supremum of double sequence This is from "Measures, Integrals and Martingales" by R.L. Schilling, page 28, exercise 4.6ii.
Prove that for any double sequence $\beta_{ij}, i, j \in \mathbb{N}$ of real numbers, we have
$$ \sup_{i\in\mathbb{N}}\sup_{j\in\mathbb{N}}\beta_{i j} = \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j} $$
 A: In the solution manual to the textbook, we are given half of the proof. I couldn't figure out the other half, but then I did, so I decided to answer my own question:
Note that $\beta_{mn} \leq \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j} $ for all choices of $m,n \in \mathbb{N}$. The right hand side of the inequality is independent of $n$ so we can take the supremum over all $n$ i.e.
$$\sup_{n\in\mathbb{N}} \beta_{mn} \leq \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j}, \forall m \in \mathbb{N}$$
With the same argument, we can then take the supremum over all $m \in \mathbb{N}$ to get
$$\sup_{m\in\mathbb{N}} \sup_{n\in\mathbb{N}} \beta_{mn} \leq \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j}$$
Now, since the names of the indices on both sides of the equation are irrelevant, we can rewrite this as
$$\sup_{i\in\mathbb{N}} \sup_{j\in\mathbb{N}} \beta_{ij} \leq \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j}$$
Let 
$$A = \sup_{i\in\mathbb{N}} \sup_{j\in\mathbb{N}} \beta_{ij}, B = \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j}$$
Then
$$A \leq B$$
Now, for the other part of the proof: start off with $\beta_{mn} \leq \sup_{i\in\mathbb{N}}\sup_{j\in\mathbb{N}}\beta_{i j} $ for all choices of $m,n \in \mathbb{N}$ (i.e. we work with the other order of application of supremums). By the same argument, we can take the supremum over all $m \in \mathbb{N}$ i.e.
$$\sup_{m\in \mathbb{N}}\beta_{mn} \leq \sup_{i\in\mathbb{N}}\sup_{j\in\mathbb{N}}\beta_{i j}, \forall n \in \mathbb{N} $$
And then we take the supremum over all $n \in \mathbb{N}$
$$\sup_{n \in \mathbb{N}}\sup_{m\in \mathbb{N}}\beta_{mn} \leq \sup_{i\in\mathbb{N}}\sup_{j\in\mathbb{N}}\beta_{i j}$$
We again rename the indices on the left hand side to obtain
$$\sup_{j \in \mathbb{N}}\sup_{i\in \mathbb{N}}\beta_{ij} \leq \sup_{i\in\mathbb{N}}\sup_{j\in\mathbb{N}}\beta_{i j}$$
Which is
$$ B \leq A$$
Therefore $A = B$
