# 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

$$\sup_{i\in\mathbb{N}}\sup_{j\in\mathbb{N}}\beta_{i j} = \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j}$$

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 $$A = B$$