So I am having a trouble with this game theory proof. I feel pretty good with my answer for part 1, but I am not really sure how to get started on the rest of it. Any help would be appreciated.
Let $A$ be an $m \times n$ matrix game, and consider the two player, zero sum, matrix game. For each pair of mixed strategies $(X,Y)$ define
$B_I(X,Y)=max_{1\leq i \leq m}(e_iAY^t)-XAY^t$ (that is the most P1 can improve if P2 plays strategy Y)
$B_{II}(X,Y)=max_{1\leq j\leq n}(XAe_j^t)-XAY^t$
$B=B_I+B_{II}$
Since $S_m \times S_n$ is compact, one can find $(X^*,Y^*)$ such that $B(X^*,Y^*)$ is minimal.
1) Explain why if $B(X^*,Y^*)=0$ then $(X^*,Y^*)$ is a saddle point in mixed strategies.
I am thinking that since $B_I$ is the most P1 can improve if P2 plays strategy Y, and $B_{II}$ is the most P2 can improve if P1 plays strategy X, that if $B(X^*,Y^*)=0$ this implies that:
$$B(X^*,Y^*)=B_I+B_{II}=max_{1\leq i \leq m}(e_iAY^t)-XAY^t+max_{1\leq j\leq n}(XAe_j^t)-XAY^t=max_{1\leq i \leq m}(e_iAY^t)+max_{1\leq j\leq n}(XAe_j^t)-2(XAY^t)=0$$
$$max_{1\leq i \leq m}(e_iAY^t)+max_{1\leq j\leq n}(XAe_j^t)=2(XAY^t)$$
Which implies that P1's strategy and P2's strategy cannot be improved, and therefore $(X^*,Y^*)$ is an optimal strategy, so it is a saddle point in mixed strategies.
2) Assume $B(X^*,Y^*)\neq 0$. Assume there are unique $i,j$ such that $e_iAY^t-XAY^t=B_X, XAe_j^t-XAY^t=B_Y$. Show that for small enough $\epsilon$ either $B((1-\epsilon)X^*+\epsilon e_i,Y^*)<B(X^*,Y^*)$ or $B(X^*,(1-\epsilon)Y^*+\epsilon e_j)<B(X^*,Y^*)$. This is a contradiction.
3) Use induction on the size of the matrix to show that, even if $i,j$ are not unique in part 2, you still reach a contradiction. Conclude that saddle points in mixed strategies always exist.