IIRC, that’s how Aumann’s paper was describing it.
No, not as I read it.
My C & E_i correspond to blocks in the partition that Aumann assigns to agent 1. Aumann has agent 1 evaluating the posterior probability of A in world ω by computing p(A | P₁(ω)), where P₁(ω) is the element of 1′s partition containing ω. So the agent will never condition on a union P₁(ω₁) ∪ P₁(ω₂) of distinct blocks. (Here I’m following Aumann’s notation as closely as possible.) Correspondingly, my agent never conditions on a disjunction [(C & E₁) ∨ (C & E₂)].
But I think that you’re right about the generalization being trivial. It should just involve the standard procedure for turning a union A ∪ B into a disjoint union A ∪ (B∖A).
No, not as I read it.
My C & E_i correspond to blocks in the partition that Aumann assigns to agent 1. Aumann has agent 1 evaluating the posterior probability of A in world ω by computing p(A | P₁(ω)), where P₁(ω) is the element of 1′s partition containing ω. So the agent will never condition on a union P₁(ω₁) ∪ P₁(ω₂) of distinct blocks. (Here I’m following Aumann’s notation as closely as possible.) Correspondingly, my agent never conditions on a disjunction [(C & E₁) ∨ (C & E₂)].
But I think that you’re right about the generalization being trivial. It should just involve the standard procedure for turning a union A ∪ B into a disjoint union A ∪ (B∖A).