I am studying zero-sum games from Robert Kleinberg's notes. In these notes, the author defines a normal form game as follows:
A normal form game is specified by
- A set of $\mathcal{I}$ players
- For each player $i \in \mathcal{I}$, a set of $A_{i}$ strategies.
- For each player $i \in \mathcal{I}$ a payoff function
$$ u_{i} \: : \prod_{i \in \mathcal{I}} A_{i} \to \mathbb{R} $$
and then goes on to specify a two-player zero-sum game:
A two-player zero-sum game is one in which $\mathcal{I} = \{1, 2\}$ and $u_{2}(a_{1}, a_{2}) = -u_{1}(a_{1}, a_{2})$ for all pure strategy profiles $(a_{1}, a_{2})$.
Mixed strategies are defined in the standard fashion:
For a finite set $A$ let $\Delta(A)$ denote the probability simplex generated by $A$ i.e.
$$ \Delta(A) = \left\{p : A \to [0, 1] \:\: \bigg| \:\: \sum_{a \in A} p(a) = 1 \right\} $$
The payoff functions for mixed strategies are extended naturally from the payoffs for pure strategies:
$$ u_{i}(p_{1}, p_{2}, \dots, p_{|\mathcal{I}|}) = \sum_{a} u_{i}(a)p_{1}(a_{1})p_{2}(a_{2})\dots p_{|\mathcal{I}|}(a_{|\mathcal{I}|})$$
where the sum runs over all pure strategy profiles $a = (a_{1}, a_{2}, \dots, a_{|\mathcal{I}|})$.
The main focus of the notes are von Neumann's minimax theorem. In the discussion, the author makes the following assertion:
Note that sets $B_{1} = \text{arg}\max_{p}\min_{q} u_{1}(p, q)$ and $B_{2} = \text{arg}\min_{q}\max_{p}u_{1}(p, q)$ are nonempty. This follows from compactness of $\Delta(A_{1})$ and $\Delta(A_{2})$, the continuity of $u_{1}$ and the finiteness of $A_{1}$ and $A_{2}$.
My question is rather simple. Why is this the case? This is my attempt so far:
I will only consider the case of $B_{1}$ in what follows.
$u_{1}$ is jointly continuous in $p$ and $q$, implying that the function $u_{1}(p) = \min_{q} u_{1}(p, q)$ is well-defined since $\Delta(A_{2})$ is compact. As a result, $u_{1}(p)$ is continuous (see this question). Hence, the maximum of $u_{1}(p)$ is attained and hence the $\text{arg}\max$ set is nonempty, which is exactly the claim we wanted to prove.
Is this logic correct? If so, then why do we need a finiteness assumption on $A_{1}$ and $A_{2}$? Maybe we don't, but the proof is made easier with this additional assumption?