Reading the proof of the first theorem here (page 1), it says Let $$V(x,y)=\sum_i\sum_jp_iA_{ij}q_j$$ and let $(x^*,y^*)$ bean equalibrium pair. Than $$v_B=min_ymax_xV(x,y)\le max_xV(x,y^*)=V(x^*,y^*)=min_y V(x^*,y)\le max_xmin_yV(x,y)=v_A$$
So far so good. But I do not understand the following.
But, since we always have $v_A\le v_B$....
Why do we always have $v_A\le v_B$?
Excuse the conflicting bound & unbound variables.
Work through the following sequence.
$\min_y V(x,y) \le V(x,y)$, so $\min_y V(x,y) \le \max_x V(x,y)$, so $\max_x \min_y V(x,y) \le \max_x V(x,y)$ and so $\max_x \min_y V(x,y) \le \min_y \max_x V(x,y)$