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 settheoretic 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 settheoretic
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 oneagent model which employs a standard state space.
Xiong (2007) proposes using the “knowing whether” rather than
settheoretic 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 nonpartitional 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 singleagent 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 multiagent 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 multiagent 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 multiagent 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 multiagent 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 multiagent 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. AASelf Reﬂection ω ∈ A (E) ⇐⇒ ω ∈ trade. Using the multiagent 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. AKSelf Reﬂection ω ∈ A (E) ⇐⇒ ω ∈
V
on notrade 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. AIntrospection 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, AASelf Re
knowledge and therefore identical, there still can be
ﬂection, AKSelf Reﬂection and AIntrospection 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. AASelf 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. AKSelf 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. AIntrospection 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)
AASelf 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 nonempty 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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