I am having trouble deriving the following basic result:
$\ast$) For every $\omega \in \Omega, \omega \in P (\omega),$
from the following axioms:
A1) $K (\Omega) = \Omega$,
A2) $K (A) \cap K (B) = K ( A \cap B)$.
Where, $\Omega$ is a state space, $K: 2^{\Omega} \to 2^{\Omega}$, and $P (\omega):= \bigcap \{ A \subseteq \Omega \ | \ \omega \in K (A) \}$.
My attempt was to prove it by contradiction, starting from the definition of $P$, that corresponds to
$$ \forall A \subseteq \Omega, \omega \in K (A) \Longrightarrow \omega \in P(\omega).$$
and having that, if $\{ \omega \} \in K (\{ \omega \})$, then $K (\{\omega\} ) = \varnothing$. However I am not going anywhere.
Any feedback is welcome.
Thank you for your time.
A counterexample to the claim is:
$$ K(A) = \Omega \text{ for all } A$$
Then $P(\omega_0)=\varnothing$ for all $\omega_0$ because it is the intersection of a class that includes $\varnothing$.
At first I read the claim as being $\omega\in K(P(\omega))$. But that is not true either; here's a counterexample to that:
$\Omega$ is infinte.
$\displaystyle K(A) = \begin{cases} \Omega & \text{if }\Omega\setminus A\text{ is finite} \\ \varnothing & \text{otherwise} \end{cases} $
Then $P(\omega_0)=\varnothing$ for all $\omega_0$ because it is the intersection of a class that includes $\Omega\setminus\{\omega\}$ for every $\omega$.
$\omega\in K(P(\omega))$ does hold under the additional assumption that $\Omega$ is finite.