I am pretty sure this is more than trivial, but I have a problem with the proof of a basic results in Aumann structures (this is related to a more general problem I have with proofs that involve equality chains).
Here, $K$ stands for the knowledge operator, and $E$ for an arbitrary event subset of a set $\Omega$ of states of the worlds.
Assume the following:
- $K E \subset E$
- $\sim K E \subset K \sim K E $
then $K E = KK E$ follows.
Aumann states that the result can be obtained in the following way:
$\sim KE = K \sim KE$, so $KE = \sim K \sim KE$, so $KKE = K \sim K \sim KE=\sim K \sim KE = KE$.
As I wrote I do have a general problem with proofs involving equality chains. Anyway, the way in which I try to establish a result like this is the following. Basically I have to prove that $KKE \subset KE$ and $KE \subset KKE$ both hold.
[For what I got, it's only after that you establish it, that you present it as a chain of equalities]-
Now, $KKE \subset KE$ is trivial, but I don't see how I can prove the converse.
What am I missing?
Any feedback will be greatly appreciated.
[The original article by Aumann, of which I use the notation, can be found here - the problem I have is between page 267 and page 270]
I think you can prove it this way.
$\forall w \in KE$, $I(w) \subset E$. $I$ is the information function induced by the knowledge operator $K$.
If $I(w) \not \subset KE$, then $\exists w^{'} \in I(w)$ s.t. $w^{'} \in \sim KE \subset K\sim KE$. Then $I(w^{'}) = I(w) \subset \sim KE$. Contradiction.
Thus, $I(w) \subset KE$. This means $w \in KKE$