Theorems and Unawareness

Spyros Galanis

Department of Economics

University of Rochester

Rochester, NY 14627

Abstract tion structures. However, Dekel, Lipman, and Rusti-

chini (1998) propose three intuitive properties for un-

awareness and show that they are incompatible with

This paper provides a set-theoretic model

1

the use of a standard state space. On the other hand,

of knowledge and unawareness, in which

Fagin and Halpern (1988), Halpern (2001), Modica

reasoning through theorems is employed.

and Rustichini (1994, 1999) and Halpern and Re ˆgo

A new property called Awareness Leads to

(2005) construct syntactic models. Two papers that try

Knowledge shows that unawareness of the-

to circumvent the problem and provide a set-theoretic

orems not only constrains an agent’s knowl-

generalization of the standard model are Li (2006)

edge, but also, can impair his reasoning

and Heifetz, Meier, and Schipper (2006). They de-

about what other agents know. For exam-

part from the standard model in that they use multiple

ple, in contrast to Li (2006), Heifetz, Meier,

state spaces, one for each state of awareness. Fein-

and Schipper (2006) and the standard model

berg (2004, 2005), Sadzik (2005), Copic and Gale-

of knowledge, it is possible that two agents

otti (2006), Li (2006b), Heifetz, Meier, and Schip-

disagree on whether another agent knows a

per (2007), Filiz (2006) and Ozbay (2006) model un-

particular event.

awareness in the context of games.

This paper provides a model of knowledge and aware-

1 INTRODUCTION

ness, using multiple state spaces. In order to illustrate

its main difference with the models of Heifetz, Meier,

A common assumption in economics is that agents

and Schipper (2006) and Li (2006), consider the fol-

who participate in a model perceive the “world” the

lowing example, depicted in the ﬁgure below. There

same way the analyst does. This means that they un-

are two agents, Holmes and Watson, and two relevant

derstand how the model works, they know all the rele-

dimensions or questions: “Did the dog bark?” and

vant theorems and they do not miss any dimension of

“Was there an intruder?”. Holmes is always aware of

the problem they are facing. In essence, agents are as

both questions, so his subjective state space is the full

educated and as intelligent as the analyst and they can

state space, containing the four states(ω ,ω ,ω ,ω )

1 2 3 4

make the best decision, given their information and

on the plane. At state ω which speciﬁes that there

4

preferences. Modeling unawareness aims at relaxing

was no intruder and no barking, Holmes knows that

this assumption, so that agents may perceive a more

there is no intruder because he knows that the dog

simpliﬁed version of the world.

did not bark and he is also aware of and knows the

The standard model of knowledge without unaware-

ness was introduced into economics by Aumann 1

Ely (1998) argues against one of these properties and sug-

(1976). Its simplicity and the fact that it was purely

gests a one-agent model which employs a standard state space.

Xiong (2007) proposes using the “knowing whether” rather than

set-theoretic led to many economic applications. One

the “knowing that” operator and suggests two different unaware-

of the ﬁrst attempts to model unawareness is by

ness operators that circumvent the impossibility result of Dekel,

Geanakoplos (1989), using non-partitional informa- Lipman, and Rustichini (1998).

1452

theorem “no barking implies no intruder”. Hence, The example shows that unawareness can restrict Wat-

H

P (ω ) =ω . son’s reasoning about Holmes’ knowledge about an

4 4

event that both are aware of. This is not captured

in other papers that model unawareness. Moreover,

Watson formally makes no mistake. It is true that

with Watson’s awareness, Holmes would not know

that there is no intruder and Watson can reason only

up to his awareness. Essentially, there are two dif-

ferent views of Holmes’ knowledge. This is formally

captured in this model by creating one knowledge op-

erator for each state of awareness. If Watson is aware

of questions V then his view of Holmes’ knowledge

1

is K . But Holmes is aware of more questions, V ,

V 2

1

4

so his view of Holmes’ knowledge is K . The re-

V

2

lationship between the two is given by the property

Awareness Leads to Knowledge which states that ifV

2

containsV thenK will contain (even strictly)K

1 V V

2 1

when both are projected to the same state space. That

Figure 1

is, higher states of awareness give a more complete

Watson is aware only of the question “Was there an description of one’s knowledge. Heifetz, Meier, and

intruder?” and he is unaware of the theorem “no Schipper (2006) specify one knowledge operator K

barking implies no intruder”. His subjective state so that there is always one objective view of Holmes’

space consists of states ω and ω on the horizon- knowledge.

5 6

tal axis. The property “Projections Preserve Knowl-

One can argue that another way of accommodating

edge” in Heifetz, Meier, and Schipper (2006) and the

the example is with a model that allows false be-

construction in Li (2006) prescribe that when Watson

liefs. Such a (syntactic) model is provided by Halpern

reasons at ω about Holmes’ knowledge, he projects

6

and Re ˆgo (2005), who extend that of Heifetz, Meier,

H

P (ω ) = ω to his state space. Therefore, he rea-

4 4

and Schipper (2006). However, allowing false be-

H

sons thatP (ω ) =ω and that Holmes knows atω

6 6 6

liefs would have stronger implications - that agents

that there is no intruder. We argue that this is restric-

may make mistakes about any event, not just events

tive. Since Watson is unaware of any theorem that

which describe other agents’ knowledge. In order to

could lead someone to know whether there is an in-

allow for unawareness of theorems without allowing

truder, he should not be able to correctly deduce that

for agents to have false beliefs in general, we retain

Holmes knows atω .

6

the truth property but index knowledge, K , with a

V

In order to accommodate the example so that Watson set of questionsV .

reasons that Holmes does not know whether there is

The paper proceeds as follows. Section 2 introduces

an intruder, we have to abandon projections. Mod-

the basic single-agent model, while its main results are

eling reasoning through theorems does exactly that.

presented in Section 3. We conclude in Section 4. The

When Watson reasons about Holmes atω , he is un-

6

Appendix contains the proofs and the construction of

aware of the theorem “no barking implies no intruder”

gests that Watson can never be certain that Holmes does not know

and therefore he cannot reason that Holmes is aware

an event. The reason is that Watson can always think that Holmes

H

of it. As a result, P (ω ) = { ω ,ω } and Watson

6 5 6

is smarter, more aware and therefore could know. But this sug-

3

reasons that Holmes does not know. gests that in an environment with unawareness an agent can never

be certain that another agent does not know an event, which is

2

That is, he considers ω to be impossible. Specifying clearly not true.

3

H

4

P (ω ) = { ω } is irrelevant for the example since this state

3 3 In other words, Watson is only aware of the formula “Holmes,

never occurs.

up to awarenessV , knows that there is no intruder”. He is un-

1

3

One argument against this reasoning is that Watson could be aware of the respective formula when V is substituted for V .

2 1

aware that Holmes is smarter than him, so that he could always More importantly, the formula “Holmes knows that there is no in-

think that Holmes could know, even though Watson cannot de- truder” is not expressed in this model because knowledge,K , is

V

scribe exactly how this can happen. But this argument also sug- always indexed with a set of questionsV .

146the state space for the multi-agent model. 2.2 THE FULL STATE SPACE

For a subset of questions V ⊆ Q , where Q is the

0 0

set of basic questions, the resulting Cartesian prod-

5

uct of their answers is × A . DeﬁnemV to be the

q

2 THE MODEL

q∈V

question “What subjective states in × A does the

q

q∈V

agent consider impossible?”. The collection of possi-

2.1 PRELIMINARIES

ble answers for question mV is the collection of all

proper subsets of × A . The questions mV capture

q

q∈V

Consider a set of questions Q and denote by A the

q

the agent’s knowledge of theorems, as shown in Sec-

set of possible answers for question q ∈ Q. The set

tion 2.4.

A can contain one, two, or more answers. The notion

q

For each question q, where q ∈ Q , or q = mV ,

0

of awareness that will be deﬁned in the following sec-

for V ⊆ Q , deﬁne aq to be the question “Is the

0

tions requires that if an agent is aware of a question,

agent aware of question q?”. This question captures

then he is aware of all possible answers. The full state

∗

the agent’s awareness of questions, as shown in Sec-

space Ω is a subset of the Cartesian product × A .

q

q∈Q

tion 2.3. In a multi-agent model, it will also capture

In the example, the full state space consists of the four

the agent’s knowledge about each agent’s awareness.

states on the plane, ω ,ω ,ω andω . Given any set

1 2 3 4

The possible answers for this type of questions are just

of questionsV ⊆ Q, a subjective state space Ω is the

two: “yes” and “no”. Questions of the typeaaq,aaaq,

∗

projection of Ω to the Cartesian product × A . For

q

aa...aq are not deﬁned. Justiﬁcation for this restric-

q∈V

instance, Watson’s subjective state space consists of tion will be given in Section 2.3, where awareness of

ω and ω . It is the projection of the full state space questions will be deﬁned.

5 6

to the question he is aware of. An event E is a sub-

The set of all questionsQ contains the basic questions

∗

set of a subjective state space Ω (and given Ω , there

Q , together with the epistemic questions of the type

0

is a unique subjective state space Ω satisfying this in-

mV , where V ⊆ Q , and of the type aq, where q ∈

0

clusion). DeﬁneV to be the unique set of questions

E

Q , or q = mV , for V ⊆ Q . The full state space

0 0

such thatE⊆ Ω⊆ × A . If the event is{ ω } , then

q 5 ∗

Ω is a subset of the Cartesian product of the answers

q∈V

E

∗

V is the question “Was there an intruder?”. Deﬁne of all questions in Q: Ω ⊆ × A . DeﬁneS to be

q

{ ω }

5

q∈Q

the negation ofE to be the complement ofE with re- S

Q

∗

the union of all state spaces: S = { Π (Ω ) : ∅ 6=

V

spect to the subjective state space Ω of which it is a

V ⊆ Q} . The construction of the full state space in

subset. Denote the complement ofΩ by the empty set

the multi-agent case is more complicated, as an agent

associated with it,∅ .

V

Ω

has to reason about other agents’ reasoning as well.

′

Take two sets of questions,V ⊆V ⊆Q, and letΩ to

The details are given in the appendix.

′

be the subjective state space generated byV , andΩ to

′

be the subjective state space generated by V . There

2.3 AWARENESS

V ′

exists a surjective projection Π : Ω→ Ω . For any

′

V

V

subjective stateω ∈ Ω,Π (ω) is the restriction ofω The awareness of an agent is given by W , which is

′

V

′

to the smaller set of questions V . For instance, the a mapping fromS to sets of questions. For any state

ω∈S,

restriction ofω to Watson’s state space isω . For an

2 6

′ V [

eventE ⊆ Ω, its restriction toV is Π (E). Its en-

′

V

W(ω) = { q,aq}⊆V :ω = “yes”

{ ω} aq

′′

largement to the bigger set of questionsV is denoted

′′

V −1

denotes the questions, of which the agent is aware if

by (Π ) (E). The restriction of { ω ,ω } is the

2 3

V

ω ∈ S occurs. If ω speciﬁes “yes” to question aq,

event { ω ,ω } while the enlargement of { ω ,ω } is

5 6 5 6

{ ω ,ω ,ω ,ω } . To save on notation and only when then the agent is aware of questionq atω. We then as-

1 2 3 4

it is unambiguous of which state spaceE is a subset, sume that he is also aware of question aq. Questions

we abbreviate restrictions and enlargements by E ′ 5

V

The basic questions describe the physical world but not the

′′

andE , respectively. agents’ knowledge or awareness.

V

147of the typeaaq, aaaq,aa...aq are not permitted by of theorems is given by the function M, which maps

the model. The ﬁrst reason for this restriction stems S to subsets ofS. For anyω∈S,

for the deﬁnition ofW , which speciﬁes that the agent

′

M(ω) ={ ω ∈ Ω(ω) :

is aware of q and aq if ω speciﬁes “yes” to question

aq. Therefore, question aaq, which would also spec-

′

{ ω} ∈ω ,{ mV}∪ V ⊆W(ω)}

V mV

ify whether the agent is aware of question aq, is not

needed. Another reason why these higher orders of

denotes the set of subjective states that the agent con-

questions would seem necessary is to express that if an

siders impossible at ω, and expresses what theorems

′

agent is aware of something, then he is aware that he is

he knows at that state. An element ω ∈ Ω(ω) of

aware of it. One of the results of Theorem 2 is exactly

the agent’s state space at ω is considered impossible

this property and it does not require these higher order

if two conditions are met. Firstly, at ω the agent is

questions. In the multi-agent case however, questions

aware of question mV and all questions in V . That

i j k

of the typea a a q wherei6= j andj 6= k arise nat-

is, he can formulate the Cartesian product × A and

q

q∈V

urally when common knowledge is deﬁned and thus

ask the questionmV : “What states in × A does the

q

will be included in the formal model. The agent’s sub-

q∈V

∗

jective state space atω∈S is Ω(ω) = Ω , which agent consider impossible?”. Secondly, the projection

W(ω)

∗ ′

is the projection of the full state spaceΩ to the set of

of ω to the set of questions V is contained in ω ,

mV

6

questions he is aware of atω. which is the answer thatω speciﬁes for questionmV .

This answer, ω , is an event, a subset of the Carte-

mV

Take an eventE and deﬁneU(E) to be the set of states

sian product × A .

q

ω∈S that describe that the agent is unaware of it:

q∈V

U(E) ={ ω∈S :V *W(ω)} .

E

2.5 IMMEDIATE PERCEPTION

The agent is unaware of eventE if he is not aware of

It is assumed that for some questions q ∈ Q that the

all questionsV that generate this event.

E

agent is aware of, he always knows the answer. For

example, questions that describe what the agent sees

Given a set of questionsV that generate the state space

∗ ∗

or hears. Denote by X the set of all such questions.

Ω , we deﬁneU (E) = Ω ∩U(E) to be the states

V

V V

The following axiom is assumed throughout the paper.

of that particular state space, which describe that the

∗

DeﬁneE to be the set that contains all epistemic ques-

agent is unaware of E. Hence, U (E) ⊆ Ω is an

V

V

tionsaq∈Q forq∈Q and anymV ∈Q, forV ⊆Q:

event. Denote the complement ofU (E) byA (E).

V V

E ={ aq∈Q :q∈Q}∪{ mV ∈Q :V ⊆Q} .

It is natural to require that V be big enough so that

∗

the generated state space Ω can adequately express

Axiom 1. E⊆X.

V

E and the agent’s awareness of it. Hence, we ﬁrst re-

The axiom states thatX contains at least all the epis-

quire thatV should contain all questions inV . Sec-

E

temic questions that belong toQ.

ondly, we require that for each question q ∈ V , V

E

contains its respective counterpartaq. Denote this set

7

of questions by α(V ). Then, the condition is that

2.6 POSSIBILITY AND KNOWLEDGE

E

V ∪α(V )⊆V .

E E

For anyω∈S,

2.4 THEOREMS AND IMPOSSIBLE STATES

′

P(ω) ={ ω ∈ Ω(ω) :

A theorem of the form “A implies B” can equivalently

′

ω =ω ,q∈W(ω)∩X}\ M(ω)

q

q

be expressed as the impossibility of the state that spec-

iﬁes “A is true but B is false”. The agent’s knowledge

denotes the subjective states the agent considers pos-

6

sible ifω occurs. More speciﬁcally, atω the agent is

IfW(ω) = ∅, then deﬁne Ω(ω) = ∅. In that case, Ω(ω) is

not an event and carries no awareness. aware of questions that belong toW(ω) and his sub-

7

The respective counterpart ofaq isaq itself, since question

jective state space isΩ(ω). For the questions inW(ω)

aaq is not allowed in the model. Formally, for any V ⊆ Q,

S

′ ′ that also belong toX, he knows the answer. This is the

α(V) = { aq : q ∈ V,q 6= aq for allq ∈ Q} { q ∈ V :

′ ′

q =aq,q ∈Q} . answer thatω speciﬁes for that question. For all other

148questions inW(ω) he does not know the answer, but The conditionV ∪α(V )⊆ V ,V ensures that the

E E 2 1

he can utilize his knowledge of theorems by exclud- state spaces generated byV andV are rich enough to

1 2

ing the impossible statesM(ω). The following axiom describe the agent’s knowledge ofE, so thatK (E),

V

2

states that the agent never excludes the true state. K (E) are well deﬁned, as explained in Section 2.6.

V

1

The condition V ⊆ V says that the state space gen-

∗ ∗ ∗ ∗

2 1

Axiom 2. For allω ∈ Ω ,{ ω } ∗ ∈P(ω ).

W(ω )

erated by questions V is richer than that generated

1

by questions V . The property then states that state

Axiom 2 implies that for allω∈S such thatW(ω)6=

2

spaces which are generated by more questions give a

∅,{ ω} ∈P(ω).

W(ω)

more complete description of the agent’s knowledge

Take an event E and deﬁne K(E) to be the set of

of an event E. In other words, if a more complete

statesω∈S that describe that the agent knowsE:

description of the world ω speciﬁes that the agent

knows eventE, (ω ∈ K (E)), the less complete de-

V

1

K(E) ={ ω∈S :V ⊆W(ω) and(P(ω)) ⊆E} .

E V

E

scription{ ω} may specify that he does not know it

V

2

({ ω} ∈/ K (E)).

V V

2 2

The agent knows E if he is aware of it and in all the

The next theorem veriﬁes properties that have been

states he considers possible, it obtains. Given a set of

∗ proposed in the literature, or are generalizations of

questions V that generate state space Ω , we deﬁne

V

∗ properties of the standard model.

K (E) = Ω ∩K(E) to be the event of that partic-

V

V

ular state space, which describes that the agent knows Theorem 2. SupposeV ∪α(V )∪V ∪α(V )∪

E E F F

8

E.

α(V)⊆V . Then:

1. Subjective Necessitation Suppose axiom 2

3 RESULTS

∗

holds. Then, for allω∈ Ω ,ω∈K (Ω(ω)).

V

V

The following Theorem generalizes properties

2. Generalized Monotonicity E ⊆

V ∪V

E F

P1,P2 and P3 of the standard model without

F ,V ⊆V =⇒ K (E)⊆K (F).

V ∪V F E V V

E F

unawareness. All the results of this section are valid

for the multi-agent case as well.

3. Conjunction K (E) ∩ K (F) =

V V

Theorem 1.

K (E ∩F ).

V V ∪V V ∪V

E F E F

4. The Axiom of Knowledge Suppose axiom 2

9

1. { ω} ∈/ M(ω) ⇐⇒ { ω} ∈P(ω).

W(ω) W(ω)

holds. Then,K (E)⊆E .

V V

′ ′

2. ω ∈P(ω) impliesP(ω ) =P(ω).

5. The Axiom of Transparencyω∈K (E) ⇐⇒

V

ω∈K (K (E)).

V

W(ω)

The next property is the most important departure

from other models dealing with unawareness, and

6. The Axiom of Wisdom Suppose axiom 2 holds.

stems from the explicit use of reasoning through the-

Then, ω ∈ A (E) ∩ ¬ K (E) ⇐⇒ ω ∈

V V

orems in the construction.

K (A (E)∩¬ K (E)).

V

W(ω) W(ω)

Property 1. Awareness Leads to Knowledge

7. Plausibility U (E) ⊆ ¬ K (E) ∩

V V

Suppose axiom 2 holds. For any event E, if V ∪

E

¬ K (¬ K (E)).

V V

α(V )⊆V ⊆V , then

E 2 1

8. Strong Plausibility U (E) ⊆

V

• K (E)⊆ (K (E)) and

V V V ¬ K (E) ∩ ¬ K (¬ K (E)) ∩ ... ∩

2 1 2

V V V

¬ K (¬ K (...¬ K (E))).

V V V

• K (E)⊇ (K (E)) is not necessarily true.

V V V

2 1 2

9. AU IntrospectionU (E)⊆U (U (E)).

V V V

8

As with the unawareness operator U (E), we impose the

V

restrictionV ∪α(V )⊆V .

E E

9 ∗ 10. KU IntrospectionK (U (E)) =∅ .

V V V

The following property is also true. Suppose M ⊆ Ω is

V

∗

a set of impossible states,ω ∈ Ω ,ω ∈/ M andM(ω) ⊆ M.

V

V

Then{ ω} ∈P(ω). 11. SymmetryU (E) =U (¬ E).

W(ω) V V

14912. AA-Self Reﬂection ω ∈ A (E) ⇐⇒ ω ∈ trade. Using the multi-agent version of this model

V

A (A (E)). we show that asymmetric information due to asym-

V W(ω)

metric awareness can allow for trade. The literature

13. AK-Self Reﬂection ω ∈ A (E) ⇐⇒ ω ∈

V

on no-trade theorems stems from the result of Au-

A (K (E)).

V W(ω)

mann (1976) that if agents have common priors and

their posteriors about an event are common knowl-

14. A-Introspection Suppose axiom 2 holds. Then,

edge, then these posteriors must be identical. It is

ω∈A (E) ⇐⇒ ω∈K (A (E)).

V V W(ω)

shown that in an environment with unawareness the

same result is true only for common priors and poste-

The conditionV ∪α(V )∪V ∪α(V )∪α(V)⊆V

E E F F

riors which are deﬁned on a “common” state space,

only ensures that the eventsU (E),K (E),U (F),

V V V

which is the state space that not only everyone is

K (F), and U (U (E)) are well deﬁned. The ﬁrst

V V V

aware of, but it is also common knowledge that ev-

six properties are generalizations of the six proper-

eryone is aware of. However, as the property Aware-

ties of the standard model. Some of these generaliza-

ness Leads to Knowledge suggests, state spaces which

tions are proposed by Li (2006). Plausibility, Strong

carry more awareness give a more complete descrip-

Plausibility, AU Introspection and KU Introspection

tion of one’s knowledge and posteriors. In an example

are the properties used by Dekel, Lipman, and Rusti-

with two agents we show that although the posteriors

chini (1998) to show that unawareness precludes the

deﬁned on this “common” state space are common

use of a standard state space. Symmetry, AA-Self Re-

knowledge and therefore identical, there still can be

ﬂection, AK-Self Reﬂection and A-Introspection have

trade because one agent’s higher awareness implies

been proposed by Modica and Rustichini (1999) and

that his actual posterior is different and beyond the

Halpern (2001).

other agent’s reasoning. Heifetz, Meier, and Schipper

(2006, 2007) also provide examples where trade takes

4 CONCLUDING REMARKS

place. Comparison between the different approaches

is provided in the companion work.

In this paper we argue that with unawareness of theo-

rems it is possible that two agents disagree on whether

A Appendix

a third agent knows a particular event. This disagree-

ment does not arise because agents make logical mis-

Proof of Theorem 1.

takes or have false beliefs but because they have dif-

ferent awareness, which implies that they reason dif-

1(a). The proof is immediate from the deﬁnition of

ferently about the knowledge of others. The idea

P(ω).

that differences in awareness may specify different

views of one’s knowledge is captured by formulating, 1(b). For footnote 9 we have that ω ∈/ M =⇒

V −1 V

for each state of awareness V , a knowledge operator ω ∈/ (Π ) (M(ω)) =⇒ Π (ω) ∈/

W(ω) W(ω)

V

K . The relation between knowledge expressed with

V

M(ω) =⇒ Π (ω)∈P(ω).

W(ω)

awareness V and knowledge expressed with aware-

′

2. First, we prove the following proposition.

nessV is captured by the property Awareness Leads

to Knowledge. These connections between aware-

Proposition 1. ω∈P(ω ) implies

1

ness and knowledge are not accommodated in Heifetz,

Meier, and Schipper (2006) and Li (2006). In particu- i) W(ω ) =W(ω).

1

lar, Heifetz, Meier, and Schipper (2006) specify an ob-

ii) M(ω ) =M(ω).

1

jective knowledge operatorK, so that there can never

be two different views of one’s knowledge, because of Proof.

differences in awareness.

i) Supposeq ∈ W(ω ). There are two cases.

1

′ ′ ′

In a companion work we show that unawareness of Eitherq 6= aq for any q ∈ Q, orq = aq

′

theorems has interesting implications. In particular, for some q ∈ Q. In the ﬁrst case, we

one of the results of the standard model of knowledge have that ω = “yes” and aq ∈ W(ω ).

1aq 1

is that asymmetric information alone cannot explain In the second case, ω ′ = “yes” and

1

aq

150′ ′

aq ∈ W(ω ). The proof is identical in Next, we need to show that ω ∈/ M({ ω} ). Sup-

1 V

2

′

both cases, so we just illustrate the ﬁrst case. pose that ω ∈ M({ ω} ). Then, there exist V

V

2

From axiom 1, aq ∈ X ∩ W(ω ). Since and mV such that V ∪ { mV} ⊆ W({ ω} ) and

1 V

2

′ ′

ω ∈ P(ω ), we have ω = “yes”, which, { ω} ∈ ω . Since { ω } = ω and

1 aq V mV 1

W({ ω} )

V

2

together with{ q,aq} ⊆ W(ω ) = V im- V ∪{ mV} ⊆ W({ ω} ) ⊆ W(ω), we have that

1 ω V

2

plies{ q,aq}⊆ W(ω). The other direction { ω } ∈ ω , which implies that ω ∈ M(ω) and

1 V mV 1

′

is immediate sinceV =W(ω ). ω ∈/ P(ω), a contradiction. Hence,ω ∈/ M({ ω} )

ω 1 1 V

2

′

andω ∈P({ ω} ).

V

ii) Suppose ω ∈ M(ω ). Then, there exist 2

2 1

{ mV} , V such that{ mV}∪ V ⊆ W(ω )

1

We have shown that (P(ω)) ⊆ P({ ω} )⊆

W({ ω} ) V

V 2

2

W(ω )

1

andΠ (ω )∈ω . Fromi) we have

2 1mV E , and V ⊆ W({ ω} ) ⊆ W(ω).

V E V

W({ ω} ) 2

V

2

W(ω ) = W(ω), which implies { mV}∪

1

Therefore, P(ω) ⊆ E , which implies that

W(ω)

V ⊆ W(ω). Moreover, from axiom 1 we

(K (E)) ⊆ K (E). Finally, since V ⊆ V , we

V V V 2 1

2 1 1

have that mV ∈ X ∩W(ω ). Thus, ω ∈

1 also have thatK (E)⊆ (K (E)) . For the second

V V V

2 1 2

P(ω ) impliesω = ω and therefore

1 mV 1mV bullet, a counter example is provided in the Holmes

ω ∈ M(ω). The other direction is identi-

2

and Watson example.

cal.

Proof of Theorem 2.

SetsP(ω ) andP(ω) are repeated below:

1

1. Subjective Necessitation First, note

P(ω ) ={ ω ∈ Ω(ω ) :

1 2 1

that K (Ω(ω)) is well deﬁned becasue

V

W(ω)∪ α(W(ω)) ⊆ V . Subjective necessi-

ω =ω ,q∈W(ω )∩X}\ M(ω ),

2q 1q 1 1

tation then follows from V = W(ω) and

Ω(ω)

P(ω) ={ ω ∈ Ω(ω) :

2 ∅ = 6 P(ω)⊆ Ω(ω).

ω =ω ,q∈W(ω)∩X}\ M(ω).

2q q

2. Generalized Monotonicity Suppose ω ∈

K (E). Then, V ⊆ W(ω) and∅ = 6 P(ω) ⊆

From Proposition 1 we have W(ω ) = W(ω) V E

1

E . Also, V ⊆ W(ω) which implies that

andM(ω ) = M(ω). Sinceω ∈ P(ω ) implies F

1 1 W(ω)

E ⊆F . Therefore,ω∈K (F).

thatω =ω for allq∈W(ω )∩X =W(ω)∩ W(ω) W(ω) V

q 1q 1

X, we have thatP(ω ) =P(ω).

1

3. Conjunction We have that V ⊆ W(ω) and

E

V ⊆ W(ω) if and only ifV ∪V ⊆ W(ω).

F E F

Also, ∅ = 6 P(ω) ⊆ E and ∅ 6= P(ω) ⊆

W(ω)

F if and only if ∅ 6= P(ω) ⊆ E ∩

W(ω) W(ω)

Proof of Property 1.

F = (E ∩ F ) . The lat-

V ∪V V ∪V

W(ω) W(ω)

E F E F

First we prove that if V ⊆ V , then (K (E)) ⊆

2 1 V V ter equality follows because ω ∈ (E ∩

2 1 1 V ∪V

E F

K (E). Supposeω ∈ (K (E)) . Then,{ ω} ∈ F ) ⇐⇒ { ω } ∈ E ∩

V V V V

1 2 1 2 V ∪V W(ω) 1 V ∪V V ∪V

E F E F E F

K (E), which implies that ∅ 6= P({ ω} ) ⊆

F ⇐⇒ ω ∈E ∩F .

V V

2 2 V ∪V 1 W(ω) W(ω)

E F

E andV ⊆ W({ ω} ). We have to show

W({ ω} ) E V

2

V

2

4. The Axiom of Knowledgeω ∈ K (E) implies

V

thatV ⊆ W(ω) and∅ = 6 P(ω) ⊆ E . Firstly,

E

W(ω)

V ⊆ W(ω) and∅ = 6 P(ω) ⊆ E . Axiom

E

since V ⊆ V we also have W({ ω} ) ⊆ W(ω). W(ω)

2 1 V

2

2 implies{ ω} ∈ P(ω). Hence,{ ω} ∈

Therefore, V ⊆ W(ω). Non emptiness of P(ω) is W(ω) W(ω)

E

E , which impliesω∈E .

guaranteed by axiom 2. W(ω) V

We next show that (P(ω)) ⊆ P({ ω} ).

V

W({ ω} ) 2 5. The Axiom of Transparency Suppose ω ∈

V

2

′

Suppose that ω ∈ (P(ω)) . Then, there ex-

K (E). Then, V ∪ α(V ) ⊆ W(ω) and

W({ ω} )

V E E

V

2

′

istsω ∈ P(ω) such that{ ω } = ω . More- ∅ 6= P(ω) ⊆ E . We have to show that

1 1 W({ ω} ) W(ω)

V

2

over, ω ∈ P(ω) implies that ω = ω for all q ∈ ∅ = 6 P(ω)⊆K (E), or thatω ∈P(ω) im-

1 1q q W(ω) 1

′

W(ω)∩X, henceω =ω for allq∈W({ ω} )∩X. pliesV ⊆ W(ω ) and∅ 6= P(ω ) ⊆ E .

q V E 1 1

q 2 W(ω )

1

151From Proposition 1, we have thatω ∈P(ω) im- fore, V ⊆ W(ω ). But ω ∈ U (E) implies

1 E 1 1 V

pliesW(ω ) = W(ω). Hence,V ⊆ W(ω ) = thatV *W(ω ), a contradiction.

1 E 1 E 1

W(ω). From Theorem 1 we have that ω ∈

1

11. Symmetry Follows fromV =V .

E ¬ E

P(ω) implies P(ω) = P(ω ). Thus, ∅ 6=

1

P(ω )⊆E =E .

1

W(ω) W(ω )

1 12. AA-Self Reﬂectionω∈A (E) impliesW(ω)∪

V

α(W(ω)) ⊆ V and V ∪ α(V ) ⊆ W(ω).

E E

Supposeω∈K K (E), which implies that

V

W(ω)

Therefore, A (A (E)) is well deﬁned and

V W(ω)

∅ = 6 P(ω) ⊆ K (E). Hence, for all ω ∈

1

W(ω)

ω ∈ A (A (E)). For the other direc-

V

P(ω), we have that ω ∈ K (E), W(ω) = W(ω)

1 W(ω)

tion, suppose that ω ∈ A (A (E)). Since

V

W(ω ), P(ω) = P(ω ) and ∅ 6= P(ω ) ⊆ W(ω)

1 1 1

A (E) is deﬁned only if V ∪ α(V ) ⊆

W(ω) E E

E . Therefore, ∅ = 6 P(ω) ⊆ E and

W(ω) W(ω)

W(ω), we have thatω∈A (E).

ω∈K (E). V

V

13. AK-Self Reﬂection The proof is similar.

6. The Axiom of Wisdom Supposeω ∈ A (E)∩

V

¬ K (E). Then,V ∪α(V ) ⊆ W(ω) and ei-

V E E 14. A-Introspection Suppose ω ∈ A (E). Then,

V

ther P(ω) = ∅ or ∅ = 6 P(ω) * E . Ax-

W(ω) V ∪ α(V ) ⊆ W(ω) ⊆ V and W(ω) =

E E

iom 2 implies thatP(ω)6=∅, so we just have to

V , so we just have to show that ∅ 6=

A (E)

W(ω)

show that P(ω) ⊆ A (E)∩¬ K (E).

W(ω) W(ω)

P(ω) ⊆ A (E). That P(ω) is non empty

W(ω)

Suppose ω ∈ P(ω). Proposition 1 implies

1

follows from axiom 2. Suppose that ω ∈

1

that W(ω ) = W(ω). Hence, V ⊆ W(ω )

1 E 1

P(ω). From Proposition 1, we have W(ω) =

and ω ∈ A (E). Theorem 1 implies that

1

W(ω)

W(ω ) which implies V ⊆ W(ω ) and ω ∈

1 E 1 1

P(ω) = P(ω ). Thus P(ω ) * E =

1 1 W(ω)

A (E). For the other direction, suppose that

W(ω)

E andω ∈¬ K (E).

W(ω ) 1 W(ω)

1 ω ∈ K (A (E)). This implies that ω ∈

V W(ω)

A (A (E)) and ω ∈ A (E) follows from

Supposeω∈K (A ∩¬ K (E)). Then, V W(ω) V

V W(ω) W(ω)

AA-Self Reﬂection.

∅ 6= P(ω) ⊆ A ∩ ¬ K (E). Since

W(ω) W(ω)

A (E) is deﬁned only if V ∪ α(V ) ⊆

E E

W(ω)

W(ω), we have that ω ∈ A (E). It remains

V

to show that ω ∈ ¬ K (E), or that P(ω) *

V

A.1 THE FULL STATE SPACE

E . We know that for all ω ∈ P(ω),

W(ω) 1

ω ∈¬ K (E), which implies thatP(ω )*

1 1

W(ω)

This section gives a detailed construction of the full

E . Since P(ω) = P(ω ), we have that

1

W(ω)

state space, which is the state space of the analyst or

P(ω)*E .

W(ω)

of a fully aware agent. The construction is similar to

that of a beliefs space: starting from an initial state

8. Strong Plausibility By assumption, V ⊆

E

spaceS, deﬁne each player’s ﬁrst order beliefs onS,

V = V = V =

¬ K (E) ¬ K (¬ K (E))

V V V

then each player’s second order beliefs on S and all

V . Suppose ω ∈ U (E).

V

¬ K (¬ K (...¬ K (E)))

V V V

other players ﬁrst order beliefs, and so on. The dif-

Then, V * W(ω) and therefore V * W(ω).

E

ference with this formulation is that instead of beliefs

Hence,ω∈¬ K (E)∩¬ K (¬ K (E))∩...∩

V V V

i i

we have the epistemic questions a q and m V , that

¬ K (¬ K (...¬ K (E))).

V V V

describe the awareness of questions and knowledge of

theorems for each agenti.

9. AU Introspection Suppose ω ∈ U (E), Then,

V

V * W(ω) and since V ⊆ V = V ,

i

E E

U (E)

V

For any state space Ω = × A , letE (Ω) be the set

q

q∈V

we have V * W(ω), which implies ω ∈

U (E)

V

of epistemic questions of agent i about Ω. This set

U (U (E)).

V V

i i

will consist of questions of the typea q andm V . In

particular, suppose Ω = × A is generated from a

10. KU Introspection Suppose ω ∈ K (U (E)). q

V V

q∈V

Then, W(ω) = V and there exists ω ∈

1

set of questionsV . The set of all questions of the type

P(ω) ⊆ U (E). From Proposition 1 we have i

V

m V , for all nonempty subsetsV ofV is

1 1

that W(ω ) = W(ω) = V . Moreover, the def-

1

i

inition of U (E) implies thatV ⊆ V . There- m V :∅ = 6 V ⊆V . (1)

V E 1 1

152These questions represent all the theorems that agent Continuing inductively, we deﬁne for allk≥ 1,

i can potentially have about state spaceΩ.

j

i i

Ω = Ω × T ,

k+1 k k

The set

j6=i

i i

a q :q∈V ∪ m V :∅6=V ⊆V (2)

1 1

i

T = × A .

q

k+1

i i

i i

q∈E (Ω )\E (Ω )

i

k+1 k

contains all the a q questions, for all questions in V

i

and in { m V : ∅ = 6 V ⊆ V} . Denote the union

1 1

i

i

Note that T is non-empty for all k, as new epis-

of the two sets of questions in (1) and (2) byE (Ω).

k+1

i

temic questions are created in each step. Deﬁne T

An element that gives an answer to all questions in

∞

i

i i

E (Ω) describes agenti’s awareness of questions and

to be the Cartesian product × T . An element inT

n

n=1

knowledge of theorems, about state spaceΩ.

contains an answer for all epistemic questions about

∗

agenti. In particular, it gives an answer to only ques-

To construct the full state space Ω , we begin with an

i i

tions of the type a q, or of the type m V , where q

initial state space S = × A , which is generated

q

q∈Q

0

can be either a basic question or an epistemic ques-

from a ﬁnite or countably inﬁnite set of basic ques-

j k i ′

tion about another agent (e.g. q = a a aq ), while

tionsQ . A state of natures∈S gives a detailed de-

0

V can contain both basic and epistemic questions

scription of the world, but not what agents are aware

for all other agents. Note that questions of the type

i

of or know. Let Ω = S be agent i’s ﬁrst order state

i i i

1

a a ...aq are not created. Summarizing, an element

i i

space. Questions inE (Ω ) describe agenti’s aware-

1 i

in T describes agent i’s awareness of questions and

ness of questions and knowledge of theorems about

knowledge of theorems for each successively bigger

i

state space Ω . Deﬁne the set of all combinations of

1

state space Ω , wherek≥ 1.

k

i

answers for these questions to beT :

1

i

Interpreting T as the set of all types for agent i, we

i

T = × A ,

q

1 can deﬁne a full state to specify a state of nature s ∈

i

q∈E (Ω )

1

S, together with a type for each playeri∈I. The full

∗

state spaceΩ is then a subset of the Cartesian product

which we interpret as the ﬁrst order type of agent i.

i

The second order state space for agenti is S× T :

i∈I

j ∗ i

i

Ω =S × T . Ω ⊆S× T .

2 1

i∈I

j6=i

i The set of all questions that generate the full state

An element inΩ describes the state of natures∈S,

2

∗

space Ω is denoted by Q. Formally, V ∗ = Q and

Ω

together with the awareness of questions and knowl-

∗

Ω ⊆ × A .

q

edge of theorems about S, for all agents besides i.

q∈Q

i i

The setE (Ω ) contains all the epistemic questions of

2

i

agenti about state spaceΩ . Note that there are some

2

Acknowledgments

i i i i

questions in E (Ω ) that also belong toE (Ω ). For

2 1

example, if q is a basic question and belongs to Q ,

0

I am grateful to Larry Epstein and Paulo Barelli

i i i i i

then a q belongs to E (Ω )∩E (Ω ). To avoid any

1 2

for their continuous guidance and encouragement

duplication of questions, we deﬁne the second order

throughout this project. I would also like to thank

type of agenti to be

Jan Eeckhout, Jeffrey Ely, Yossi Feinberg, Dino Ger-

i

ardi, Aviad Heifetz, Jing Li, Bart Lipman, Alessan-

T = × A .

q

2

i i i i

q∈E (Ω )\E (Ω )

dro Pavan, Burkhard Schipper, William Thomson,

2 1

Ga ´bor Vira ´g, the seminar participants at British

i i

An element inT × T speciﬁes the questions the agent

1 2

Columbia, Colorado at Boulder, Maastricht, Queen

is aware of and the theorems he knows in state space

Mary, Rochester, Southampton, Western Ontario,

i

Ω . Accordingly, the third order state space of agenti

2

the 2006 NBER/NSF/CEME Conference on General

is

Equilibrium Theory and the 2006 Midwest Economic

i i j

Ω = Ω × T .

3 2

2

Theory Conference. All remaining errors are mine.

j6=i

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