How to interpret the intersected information partition element of different actors?

In this paper, it said that given $$\omega$$ is the current state of the world and $$P_1(\omega)$$ is the unique element from the information partition $$\textit{P_1}$$ of actor 1 in which the player 1 is informed that $$P_1(\omega)$$ contains $$\omega$$. According to the aforementioned paper (p.1237), it says that "To say that 1 knows that 2 knows (Event) E means that E includes all $$P_2$$ in the information partition $$\textit{P_2}$$ that intersect $$P_1(\omega)$$." I wonder why is the aforementioned intersection in this passage so important, since intersection here only mean that two information partition elements share some states.

I think there are two thing I need to understand: The interpretation of the shared states of two information partition elements from different people; the interpretation of the case when two informational partition elements intersect.

Also why is it seem to be inapplicable for the case when the partition elements mentioned are partially in $$E$$

I'm not an expert in epistemic game theory but let me see if I can help with an example.

Suppose the state space is $$\Omega = \{1,2,3,4,5\}$$, and there are two players with information partitions $$\mathcal{P}_1 = \left\{\{1,2,3\},\{4,5\} \right\} \\ \mathcal{P}_2 = \left\{\{1,2\},\{3,4\},\{5\}\right\}$$ Suppose the true state is $$\omega = 5$$. Let us consider the event $$E = \{4,5\}$$.

Player 1 "knows" $$E$$, since $$P_1 = \{4,5\} \subset E$$.

Player 2 "knows" $$E$$, since $$P_2 = \{5\} \subset E$$.

However, does Player 1 know that Player 2 knows $$E?$$ No.

What are the elements of $$\mathcal{P}_2$$ that intersect $$P_1$$? $$\{3,4\}$$ and $$\{5\}$$.

However, $$\{3,4\} \not \subset E$$, so Player 1 does not know that Player 2 knows $$E$$.

On the other hand, does Player 2 know that Player 1 knows $$E$$? Yes.

What are the elements of $$\mathcal{P}_1$$ that intersect $$P_2$$? $$\{4,5\}$$, and $$\{4,5\} \subset E$$, so Player 2 knows that Player 1 knows $$E$$.

• I understand the concept algebraically. However, I still do not understand intuitively why (in your example) {3,4} also has to belong to E for Player 1 to know that Player 2 knows E. Also what about the case when a partition element is partially in $E$? Commented Nov 17, 2019 at 14:50

The intersection itself is not terribly interesting, what matters here is only that it is not empty- that it contains some state.

To know event $$E$$ means that one does not consider it possible that the event $$E$$ does not obtain and that $$E$$ does indeed obtain. Since the true state is always considered possible, that $$i$$ knows $$E$$ at $$\omega$$ means that $$\mathcal{P}_i(\omega)\subseteq E$$- all states considered possible by $$i$$ at $$\omega$$ are in $$E$$. Now, that states at which $$i$$ knows $$E$$ are an event itself. Write $$K_i(E)$$ for this event. Then $$K_i(E)=\{\omega\in\Omega\mid \mathcal{P}_i(\omega)\subseteq E\}$$ is the set of states at which $$i$$ knows $$E$$. We are now looking at the event $$K_1\big(K_2(E)\big),$$ the event that $$1$$ knows that $$2$$ knows that $$E$$ holds. Now, $$K_1\big(K_2(E)\big)=\{\omega\in\Omega\mid \mathcal{P}_1(\omega)\subseteq K_2(E)\}$$ $$=\{\omega\in\Omega\mid \mathcal{P}_1(\omega)\subseteq K_2(E)\}=\big\{\omega\in\Omega\mid \mathcal{P}_1(\omega)\subseteq \{\omega'\in\Omega\mid \mathcal{P}_2(\omega')\subseteq E\}\big\}$$ $$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}_1(\omega)\text{ then } \mathcal{P}_2(\omega')\subseteq E\big\}$$ $$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}_1(\omega), P\in\mathcal{P_2}, \omega'\in P\text{ then } P\subseteq E\big\}$$ $$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}_1(\omega)\cap P, P\in\mathcal{P_2}\text{ then } P\subseteq E\big\}$$ $$=\big\{\omega\in\Omega\mid \text{ if }\mathcal{P}_1(\omega)\cap P\neq\emptyset, P\in\mathcal{P_2}\text{ then } P\subseteq E\big\}.$$