Sub-models for interactive unawareness

Abstract

We propose a notion of a sub-model for each agent at each state in the Heifetz et al. (2006) model of interactive unawareness. Presuming that each agent is fully cognizant of his sub-model causes no difficulty and fully describes his knowledge and his beliefs about the knowledge and awareness of others. We use sub-models to motivate the HMS conditions on possibility correspondences.

Appendix: Proofs

Proof of Proposition 1

First, we show that ${\mathcal {M}}^{i,\omega }$ satisfies M1. It follows by Definition 2.2. and M1.2 on ${\mathcal {M}}$ that M1.2 holds for ${\mathcal {L}}^{i,\omega } \equiv ({\mathcal {S}}^{i,\omega },\preceq ^{i,\omega })$. Let us see that M1.1 holds as well. Let $R\subseteq {\mathcal {S}}^{i,\omega }$. By C and Definition 2, there is a unique base space $S\in {\mathcal {S}}^{i,\omega }$ containing $\varPi _{i}(\omega )$ and $T\preceq S$ for all $T\in R$. By completeness of ${\mathcal {L}}$, the set $R$ has a supremum, $\sup (R)\in {\mathcal {S}}$, and infimum, $\inf (R)\in {\mathcal {S}}$. By the definition of supremum, $\sup (R)\preceq S$. By Definition 2 , $\sup (R)$ is also the supremum of $R$ in ${\mathcal {S}}^{i,\omega }$. Similarly, we find that $\inf (R)$ is also the infimum of $R$ in ${\mathcal {S}}^{i,\omega }$. Thus, we have shown that M1.2 holds. That ${\mathcal {M}}^{i,\omega }$ inherits M2, M3 and C follows from inspection of Definition 2. $\square$

Proof of Proposition 2

(If-part): Let $(B^{\uparrow },S)\in {\mathcal {E}}$. Then, $B\subseteq S$, and as such, $(\{\omega ^{\prime }\in S:\varPi _{i}(\omega ^{\prime })\subseteq B\}^{\uparrow },S)\in {\mathcal {E}}$. We will show that $K_{i}(B^{\uparrow })=\{\omega ^{\prime }\in S:\varPi _{i}(\omega ^{\prime })\subseteq B\}^{\uparrow }$. First, let us see that $K_{i}(B^{\uparrow })\subseteq \{\omega ^{\prime }\in S:\varPi _{i} (\omega ^{\prime })\subseteq B\}^{\uparrow }$. Let $\omega \in K_{i}(B^{\uparrow })$. Then $\omega \in S^{\prime \prime }$ for some $S^{\prime \prime } \in {\mathcal {S}}$, and by (2), $\varPi _{i}(\omega )\subseteq B^{\uparrow }$. By C, $\varPi _{i}(\omega )\subseteq S^{\prime }$ for some $S^{\prime }\in {\mathcal {S}}$ and $S^{\prime }\preceq S^{\prime \prime }$. Since $\varPi _{i} (\omega )\subseteq B^{\uparrow }\cap S^{\prime }$, and $B\subseteq S$, it follows that $S\preceq S^{\prime }$ and $(\varPi _{i}(\omega ))_{S}\subseteq B$. So, we have $S\preceq S^{\prime }\preceq S^{\prime \prime }, \omega \in S^{\prime \prime }$ and $\varPi _{i}(\omega )\subseteq S^{\prime }$. By PPK, $\varPi _{i}(\omega _{S} )=(\varPi _{i}(\omega ))_{S}\subseteq B$. This implies that $\omega _{S}\in \{\omega ^{\prime }\in S:\varPi _{i}(\omega ^{\prime })\subseteq B\}$, which in turn implies $\omega \in \{\omega ^{\prime }\in S:\varPi _{i}(\omega ^{\prime })\subseteq B\}^{\uparrow }$.

Next, let’s see that $\{\omega ^{\prime }\in S:\varPi _{i}(\omega ^{\prime })\subseteq B\}^{\uparrow }\subseteq K_{i}(B^{\uparrow })$. Let $\omega \in \{\omega ^{\prime }\in S:\varPi _{i}(\omega ^{\prime })\subseteq B\}^{\uparrow }$. Then, $\omega \in S^{\prime }$ for some $S^{\prime }\in {\mathcal {S}}$ with $S\preceq S^{\prime }$ and $\varPi _{i}(\omega _{S})\subseteq B\subseteq S$. By C, $\varPi _{i}(\omega )\subseteq S^{\prime \prime }$ for some $S^{\prime \prime }\preceq S^{\prime }$. By NIA, $S\preceq S^{\prime \prime }$. So, we have $S\preceq S^{\prime \prime }\preceq S^{\prime }, \omega \in S^{\prime }$ and $\varPi _{i}(\omega )\subseteq S^{\prime \prime }$. It follows by PPK that $(\varPi _{i}(\omega ))_{S}=\varPi _{i}(\omega _{S})\subseteq B$. Since $S\preceq S^{\prime \prime }$ and $\varPi _{i}(\omega )\subseteq S^{\prime \prime }$, we find that $\varPi _{i}(\omega )\subseteq B^{\uparrow }$. Hence, $\omega \in K_{i}(B^{\uparrow })$.

(Only if-part): We prove the contrapositive, that is, if PPK or NIA fails, then KE fails. Observe that PPK can be broken into two conditions:

$\hbox {PPK}^{\supseteq }$ :

If $S\preceq S^{\prime }\preceq S^{\prime \prime }, \omega \in S^{\prime \prime }$ and $\varPi _{i}(\omega )\subseteq S^{\prime }$, the $(\varPi _{i}(\omega ))_{S}\supseteq \varPi _{i}(\omega _{S})$.

$\hbox {PPK}^{\subseteq }$ :

If $S\preceq S^{\prime }\preceq S^{\prime \prime }, \omega \in S^{\prime \prime }$ and $\varPi _{i}(\omega )\subseteq S^{\prime }$, then $(\varPi _{i}(\omega ))_{S}\subseteq \varPi _{i}(\omega _{S})$.

We break up the proof into three cases: Case 1: ${\mathcal {M}}$ violates $\hbox {PPK}^{\supseteq }$; Case 2: ${\mathcal {M}}$ satisfies $\hbox {PPK}^{\supseteq }$, but violates $\hbox {PPK}^{\subseteq }$; Case 3: ${\mathcal {M}}$ satisfies PPK, but violates NIA. For each case, we will construct an event $(E,S)\in {\mathcal {E}}$ and show that $(K_{i}(E),S)\notin {\mathcal {E}}$.

• Case 1: ${\mathcal {M}}$ violates $\hbox {PPK}^{\supseteq }$.

In this case, we have $S\preceq S^{\prime }\preceq S^{\prime \prime }, \omega \in S^{\prime \prime }$ and $\varPi _{i}(\omega )\subseteq S^{\prime }$, but $\varPi _{i}(\omega _{S})\nsubseteq (\varPi _{i}(\omega ))_{S}$. Consider the event $(E,S)=(\varPi _{i}(\omega ))_{S}^{\uparrow },S)\in {\mathcal {E}}$. We will show that $(K_{i}((\varPi _{i}(\omega ))_{S}^{\uparrow }),S)\notin {\mathcal {E}}$. Suppose, on the contrary, that $(K_{i}((\varPi _{i}(\omega ))_{S}^{\uparrow }),S)\in {\mathcal {E}}$. Then, there must be some set $B$, such that $B\subseteq S$, and $B^{\uparrow }=K_{i}((\varPi _{i}(\omega ))_{S}^{\uparrow })$. Since $\varPi _{i} (\omega )\subseteq (\varPi _{i}(\omega ))_{S}^{\uparrow }$, it follows that $\omega \in K_{i}((\varPi _{i}(\omega ))_{S}^{\uparrow })=B^{\uparrow }$. Thus, $\omega _{S}\in B\subseteq B^{\uparrow }=K_{i}((\varPi _{i}(\omega ))_{S}^{\uparrow })$. Since $\varPi _{i}(\omega _{S})\nsubseteq (\varPi _{i}(\omega ))_{S}$, it follows by C that $\omega _{S}\notin K_{i}((\varPi _{i}(\omega ))_{S}^{\uparrow })$, which is a contradiction. Hence, we conclude that $(K_{i}((\varPi _{i}(\omega ))_{S} ^{\uparrow }),S)\notin {\mathcal {E}}$.

• Case 2: ${\mathcal {M}}$ satisfies $\hbox {PPK}^{\supseteq }$, but violates $\hbox {PPK}^{\subseteq }$.

In this case, we have $S\preceq S^{\prime }\preceq S^{\prime \prime }, \omega \in S^{\prime \prime }$ and $\varPi _{i}(\omega )\subseteq S^{\prime }$, but $(\varPi _{i}(\omega ))_{S}\nsubseteq \varPi _{i}(\omega _{S})$. Consider the event $(E,S)=(\varPi _{i}(\omega _{S})^{\uparrow },S)\in {\mathcal {E}}$, which is an event since ${\mathcal {M}}$ satisfies $\hbox {PPK}^{\supseteq }$. We will show that $(K_{i} (\varPi _{i}(\omega _{S})^{\uparrow }),S)\notin {\mathcal {E}}$. Suppose, on the contrary, that $(K_{i}(\varPi _{i}(\omega _{S})^{\uparrow }),S)\in {\mathcal {E}}$. Then, there must be some set $B\subseteq S$, and $B^{\uparrow }=K_{i}(\varPi _{i}(\omega _{S})^{\uparrow })$. Since $\varPi _{i}(\omega _{S})\subseteq \varPi _{i}(\omega _{S})^{\uparrow }, \omega _{S}\in K_{i}(\varPi _{i}(\omega _{S})^{\uparrow })=\{\hat{\omega }\in \varSigma :\varPi _{i}(\hat{\omega })\subseteq \varPi _{i}(\omega _{S})^{\uparrow }\}$. Since, by assumption, $B^{\uparrow } =K_{i}(\varPi _{i}(\omega _{S})^{\uparrow })$, it follows that $\omega \in B^{\uparrow }$. However, since $(\varPi _{i}(\omega ))_{S}\nsubseteq \varPi _{i} (\omega _{S}), \omega \notin K_{i}(\varPi _{i}(\omega _{S})^{\uparrow } )=B^{\uparrow }$, a contradiction. Hence, we conclude that$(K_{i}(\varPi _{i}(\omega _{S})^{\uparrow }),S)\notin {\mathcal {E}}$.

• Case 3: ${\mathcal {M}}$ satisfies PPK, but violates NIA.

In this case we have $\omega \in S,\varPi _{i}(\omega )\subseteq S^{\prime }, S^{\ell }\preceq S$, and $\varPi _{i}(\omega _{S^{\ell }})\subseteq S^{\prime \prime }$ but $S^{\prime \prime }\npreceq S^{\prime }$. Consider the event $(E,S^{\prime \prime })=(\varPi _{i}(\omega _{S^{\ell }})^{\uparrow },S^{\prime \prime })\in {\mathcal {E}}$, and suppose that there is some set $B\subseteq S^{\prime \prime }$, and $B^{\uparrow }=K_{i}(\varPi _{i}(\omega _{S^{\ell } })^{\uparrow })$. Since $\varPi _{i}(\omega _{S^{\ell }})\subseteq \varPi _{i} (\omega _{S^{\ell }})^{\uparrow }, \omega _{S^{\ell }}\in K_{i}(\varPi _{i} (\omega _{S^{\ell }})^{\uparrow })=B^{\uparrow }$. It follows that $\omega \in B^{\uparrow }$. However, since $\varPi _{i}(\omega )\subseteq S^{\prime }$ and $S^{\prime \prime }\npreceq S^{\prime }, \varPi _{i}(\omega )\nsubseteq \varPi _{i}(\omega _{S^{\ell }})^{\uparrow }$. Hence, $\omega \notin K_{i}((\varPi _{i}(\omega _{S})^{\uparrow })$, a contradiction. Hence, we conclude that $(K_{i}(\varPi _{i}(\omega _{S})^{\uparrow }),S)\notin {\mathcal {E}}$. $\square$

Proof of Proposition 4

(a) First, we show that the function $f$ is an injection. Suppose $(B^{\uparrow },S)$ and $(C^{\uparrow },S^{\prime })$ are two distinct events in ${\mathcal {E}}_{A_{i},\omega }$. Then, either $B^{\uparrow }\ne C^{\uparrow }$, or $S\ne S^{\prime }$. In either case, $(B^{\uparrow ^{i,\omega }},S)$ is distinct from $(C^{\uparrow ^{i,\omega }},S^{\prime })$. Now, let’s see that $f$ is a surjection. Let $(B^{\uparrow ^{i,\omega }},S)\in {\mathcal {E}}^{i,\omega }$. Then $S\in {\mathcal {S}}^{i,\omega }, (B^{\uparrow },S)\in {\mathcal {E}}$ and $f(B^{\uparrow },S)=(B^{\uparrow ^{i,\omega }},S)$. We need to show that $(B^{\uparrow },S)\in {\mathcal {E}}_{A_{i},\omega }$, which is equivalent to showing that $\omega \in A_{i}(B^{\uparrow })$. Let $S^{\prime }$ denote the base space containing $\varPi _{i}(\omega )$. Since $S\in {\mathcal {S}}^{i,\omega }$ it follows from R1, that $S\preceq S^{\prime }$. Hence, $\varPi _{i}(\omega )\subseteq S^{\prime }\in S^{\uparrow }$. By (2), $\omega \in K_{i}(S^{\uparrow })$.

(b) Let $(B^{\uparrow ^{i,\omega }},S)$ be an event in ${\mathcal {E}} ^{i,\omega }$ and suppose that $\omega ^{\prime }\in K_{j}^{i,\omega } (B^{\uparrow ^{i,\omega }})$. Then $\omega ^{\prime }\in \varSigma ^{i,\omega } \subseteq \varSigma$ and $\varPi _{j}(\omega ^{\prime })\subseteq B^{\uparrow ^{i,\omega }}\subseteq B^{\uparrow }$. Hence, $\omega ^{\prime }\in K_{j}(B^{\uparrow } )\cap \varSigma ^{i,\omega }$. Conversely, suppose that $\omega ^{\prime }\in K_{j}(B^{\uparrow })\cap \varSigma ^{i,\omega }$. Then, $\varPi _{j}(\omega ^{\prime })\subseteq B^{\uparrow }$ and $\omega ^{\prime }\in \varSigma ^{i,\omega }$. By C, $\varPi _{j}(\omega ^{\prime })\subseteq \varSigma ^{i,\omega }$. Since $B^{\uparrow ^{i,\omega }}=B^{\uparrow }\cap \varSigma ^{i,\omega }$, we have $\varPi _{j} (\omega ^{\prime })\subseteq B^{\uparrow ^{i,\omega }}$. By (2), $\omega ^{\prime }\in K_{j}^{i,\omega }(B^{\uparrow ^{i,\omega }})$. $\square$

Proof of Propostion 5

(If-part) Fix $\omega \in \varSigma$ and $i\in N$. First we show $\varPi _{i}(\omega )\subseteq \{\omega ^{\prime }\in \varSigma ^{i,\omega }:\varPi _{i}(\omega ^{\prime })=\varPi _{i} (\omega )\}$. Let $\hat{\omega }\in \varPi _{i}(\omega )$ Then, $\hat{\omega }\in \varSigma ^{i,\omega }$. By ST, $\varPi _{i}(\hat{\omega })=\varPi _{i}(\omega )$. Hence, $\hat{\omega }\in \{\omega ^{\prime }\in \varSigma ^{i,\omega }:\varPi _{i}(\omega ^{\prime })=\varPi _{i}(\omega )\}$. Next we show $\{\omega ^{\prime }\in \varSigma ^{i,\omega } :\varPi _{i}(\omega ^{\prime })=\varPi _{i}(\omega )\}\subseteq \varPi _{i}(\omega )$. Let $\hat{\omega }\in \varSigma ^{i,\omega }$ such that $\varPi _{i}(\hat{\omega })=\varPi _{i}(\omega )$. By C, $\varPi _{i}(\hat{\omega })\subseteq S(\hat{\omega })$. Hence, by RGA, $\hat{\omega }\in \varPi _{i}(\hat{\omega })=\varPi _{i}(\omega )$.

(Only if-part) Suppose first that RGA fails. Then, for some $\omega \in \varSigma$ and $i\in N$, we have $\hat{\omega }\in \varSigma ^{i,\omega }$ such that $\varPi _{i}(\hat{\omega })\subseteq S(\hat{\omega })$, but $\hat{\omega }\notin \varPi _{i}(\hat{\omega })$. Then, $\hat{\omega }\in \varSigma ^{i,\hat{\omega }}$, so $\hat{\omega }\in \{\omega ^{\prime }\in \varSigma ^{i,\hat{\omega }}:\varPi _{i} (\omega ^{\prime })=\varPi _{i}(\hat{\omega })\}$. Hence, $\varPi _{i}(\hat{\omega })\ne \{\omega ^{\prime }\in \varSigma ^{i,\hat{\omega }}:\varPi _{i}(\omega ^{\prime })=\varPi _{i}(\hat{\omega })\}$, that is, ${\mathcal {M}}$ violates SMIC.

Finally, suppose that ST fails, i.e., there are $\omega ,\hat{\omega }\in \varSigma$ such that $\hat{\omega }\in \varPi _{i}(\omega )$ but $\varPi _{i}(\hat{\omega })\ne \varPi _{i}(\omega )$. Then $\hat{\omega }\notin \{\omega ^{\prime }\in \varSigma ^{i,\omega }:\varPi _{i}(\omega ^{\prime })=\varPi _{i}(\omega )\}$. Hence, ${\mathcal {M}}$ violates SMIC. $\square$

The proof of Theorem 1 is based on the following Lemma.

Lemma 1

Fix a model ${\mathcal {M}}$ satisfying PPK. ${\mathcal {M}}$ satisfies RGA if and only if ${\mathcal {M}}$ satisfies GR.

Proof

The if-part is immediate from the definitions of RGA and GR. We prove the only if-part. Suppose $\omega \in \varSigma$. If $\varPi _{i}(\omega )\subseteq S(\omega )$, then by RGA, we have $\omega \in \varPi _{i}(\omega )$. Since $\varPi _{i}(\omega )\subseteq \varPi _{i}(\omega )^{\uparrow }$, GR holds. If, alternatively, $\varPi _{i}(\omega )\nsubseteq S(\omega )$, then by C, $\varPi _{i}(\omega )\subseteq S$ for some $S\prec S(\omega )$. Observe that we have $S\preceq S\preceq S(\omega ), \omega \in S(\omega )$ and $\varPi _{i} (\omega )\subseteq S$. Hence by PPK, $(\varPi _{i}(\omega ))_{S}=\varPi _{i}(\omega _{S})$. Since $(\varPi _{i}(\omega ))_{S}\subseteq S$, we have $\varPi _{i}(\omega _{S})\subseteq S$. Applying RGA to $\omega _{S}$, we find $\omega _{S}\in \varPi _{i}(\omega _{S})$. As observed already, $(\varPi _{i}(\omega ))_{S}=\varPi _{i} (\omega _{S})$, so $\omega _{S}\in (\varPi _{i}(\omega ))_{S}$. But, since $\varPi _{i}(\omega )\subseteq S, (\varPi _{i}(\omega ))_{S}=\varPi _{i}(\omega )$, we have $\omega _{S}\in \varPi _{i}(\omega )$ and $\omega \in \varPi _{i}(\omega )^{\uparrow }$, that is GR holds. $\square$

Proof of Theorem 1

(Only if-part): PPK and NIA follow from KE and Proposition 2. GR and ST follow from SMIC and Proposition 5 and Lemma 1. It remains only to show PPI.

Let $\omega \in S^{\prime \prime }$ and $S\preceq S^{\prime \prime }$. We need to show that $\varPi _{i}(\omega )^{\uparrow }\subseteq \varPi _{i}(\omega _{S})^{\uparrow }$. By C, $\varPi _{i}(\omega )\subseteq S^{*}$ for some $S^{*}\preceq S^{\prime \prime }$, and $\varPi _{i}(\omega _{S})\subseteq S^{\prime }$ for some $S^{\prime }\preceq S$. By NIA, $S^{\prime }\preceq S^{*}$. We will show the following two things:

• (a) $\varPi _{i}(\omega )^{\uparrow }\subseteq \varPi _{i}(\omega _{S^{\prime }} )^{\uparrow }$;

• (b) $\varPi _{i}(\omega _{S^{\prime }})=\varPi _{i}(\omega _{S})$.

First, let’s see (a). From our results above, we have $S^{\prime }\preceq S^{*}\preceq S^{\prime \prime }, \omega \in S^{\prime \prime }$, and $\varPi _{i}(\omega )\subseteq S^{*}$. Hence, $\varPi _{i}(\omega )^{\uparrow } \subseteq (\varPi _{i}(\omega ))_{S^{\prime }}^{\uparrow }$. It follows by $\hbox {PPK}^{\prime }$ that $(\varPi _{i}(\omega ))_{S^{\prime }}=\varPi _{i}(\omega _{S^{\prime }})$, whence $\varPi _{i}(\omega )^{\uparrow }\subseteq \varPi _{i}(\omega _{S^{\prime } })^{\uparrow }$ as required.

Next, let’s see (b). Since $\varPi _{i}(\omega _{S})\subseteq S^{\prime }$, it follows by GR that $\omega _{S^{\prime }}\in \varPi _{i}(\omega _{S})$. Hence, by ST, $\varPi _{i}(\omega _{S^{\prime }})=\varPi _{i}(\omega _{S})$ and (b) has been proved. It follows immediately from (a) and (b) that $\varPi _{i}(\omega )^{\uparrow } \subseteq \varPi _{i}(\omega _{S})^{\uparrow }$, that is, we have proved PPI.

(If-part): As noted by HMS in their remark 2 on p.83, PPI and C imply NIA. Hence, KE follows from PPK and NIA using Proposition 2. SMIC follows from Proposition 5 and Lemma 1 using GR, ST, and PPK. $\square$