In Rubinstein's Electronic Mail Game, Player I and Player II's strategies take the form as $s_i : \mathbb{N} \to \{A,B\}$, $(i =1,2)$.
Rubinstein shows that the pair of constant functions, $s_1(t_1) = A$ and $s_2(t_2) = A$, for all $t_1,t_2 \in \mathbb{N}$ constitute a equilbirium by induction.
When $t_1 = 0$, $B$ is weakly dominated by $A$. Thus there's no harm to let $s_1(0) = A$. If $s_1(0) = A$, then for player II, $B$ is strictly dominated by $A$, when $t_2 = 0$. So $s_2(0) = A$. In general, $s_1(t_1) = A$ and $s_2(t_2) = A$ for all $t_1, t_2 < m$ imply that $s_1(m) = A$. $s_2(n) = A$ is implied by $s_1(t_1) = A$ and $s_2(t_2) = A$ for all $t_1, t_2 < n$ plus $s_1(n) = A$. By strong induction, it's a equilibrium as desired.
I try to show whether $s_1(t_1) = B$ and $s_2(t_2) = B$, for all $t_1,t_2 \in \mathbb{N}$ constitute a equilbirium, which end up with a backward induction without an end.
when $t_1 = 0$, player I is indifferent between $A$ and $B$, only when$s_2(0)= B$.(In other words, if there's a pair of equilibrium strategy $s_1'$ and $s_2'$ that statisfy that $s_1'(0) = B$, then $s_2'(0)$ has to equal $B$.)But if $s_2(0)= B$ is an optimal action for player II, it has to require that $s_1(1) = B$. $\ldots$ Such reasoning continues ad infinitum. In general if we want $s_2(m) = B$, we have to require $s_1(m+1) = B$. If we want $s_1(n) =B$, we have to require that $s_2(n) = B$.
I'm confused whether $s_1(t_1) = B$ and $s_2(t_2) = B$, for all $t_1,t_2 \in \mathbb{N}$ are equilibrium strategies. To me they're kind of like a Ponzi scheme. What went wrong with this kind of reasoning?
Added: Thanks to Trurl. I notice a difference in the payoff matrices between Rubinstein's original paper and the textbook, Osborne and Rubinstein(1994).
In the paper I referred to:
In the textbook that Trurl mentioned:
Noticeably, my question doesn't arise in the textbook version as shown in Trurl's answer.
My reading of the electronic mail game is that $(A,A)$ is a unique Nash equilibrium: this is Proposition 83.1 in Osborne and Rubinstein. I think he assumes that when $t_1=0$, $A$ is strictly dominant, and that gets you off on the road to uniqueness. The point is that even when both players know it is better to play B, that is not common knowledge, so they do not both choose B, and $(A,A)$ is better than $(A,B)$. Your reasoning shows that an attempt to use induction on B fails. Hope this helps.