In a two-person zero-sum game, infinite repetition with contingent strategies does not generate new subgame perfect equilibria nor nash equilibria. Is this true?
My idea. For example:
A set of players $N=\left\{1,2 \right\}$
A set of strategies $S_i=\left\{s_{i,1}, s_{i,2} \right\}$ for each $i\in N$.
A pay-off functión:
$$\begin{array}{cccc} U: & S & \longrightarrow & \mathbb{R}^{2}\\ & (s_{1,1},s_{2,1}) & \rightarrow & (2,-2)\\ & (s_{1,1},s_{2,2}) & \rightarrow & (3,-3)\\ & (s_{1,2},s_{2,1}) & \rightarrow & (0,0)\\ & (s_{1,2},s_{2.2}) & \rightarrow & (-1,1) \end{array}$$
In matrix form:
$$\mathbf{U}=\left(\begin{array}{cc} (2,-2) & (3,-3)\\ (0,0) & (-1,1) \end{array}\right)$$
$BR_{1}(s_{2,1})=\left\{ s_{1,1}\right\} $, $BR_{1}(s_{2,2})=\left\{ s_{1,1}\right\} $, $BR_{2}(s_{1,1})=\left\{ s_{2,1}\right\} $ and $BR_{2}(s_{1,2})=\left\{ s_{2,2}\right\} $.
Therefore:
$NE=\left\{ (s_{1,1},s_{2,1})\right\} $
This is true.
Infinite repetition allows only for payoffs that are feasible (they lie in the convex hull of the payoff matrix) and individually rational (must at least be better than minmax payoffs). In zero-sum games such as yours, this set is a singleton.
For details, see the various folk theorems for repeated games. These are also a nice series of lecture notes on the subject.