Terminal Type Structures∗
Amanda Friedenberg†
Extremely Preliminary
October 9, 2006
Abstract
Say a type structure is terminal if, for every hierarchy of beliefs satisfying a certain coherency
condition, there is a type in that structure associated with that hierarchy of beliefs. When is a
type structure terminal? This paper attempts to characterize terminal structures. In doing so,
it relates terminal structures to other concepts of “large type structures” found in the literature.
It goes on to apply these results to game theory. In particular, the characterization sheds light
on the epistemic conditions for the iteratively undominated strategies.
Many game theoretic analyses require specifying players’ hierarchies of beliefs. This is true of
both the epistemic program and so-called games of incomplete information.
Harsanyi [12, 1967-1968] introduced the concept of a type structure and argued that it is a model
for hierarchies of beliefs. For a type structure to be such a model, it should be able to represent
all hierarchies of beliefs. This raises the question: Does there exist a terminal type structure—a type
structure that “contains all hierarchies of beliefs”? The question goes back to Böge-Eisele [4, 1979],
who also used the phrase “terminal type structure” and constructed one such structure. (Section
6.a discusses the relationship between this question and the question of the existence of a “universal
type structure.”)
This paper asks: Which type structures are terminal? That is, can we characterize the class of
terminal type structures? The paper also introduces the concept of a finitely terminal structure and
asks about the characterization of these structures. In addressing these questions, the paper treats
the relationship between terminal (resp. finitely terminal) structures and other notions of “large
type structures” found in the literature.
∗ This paper owes much to joint research and many conversations with Adam Brandenburger. Indeed, I believe
the paper would not have existed without Adam’s important input. I also thank the Olin School of Business for
financial support. t e r m i n a l - 1 0 - 0 9 - 0 6 . r a p
† Olin School of Business, Washington University, 1 Brookings Drive, Campus Box 1133, St. Louis, MO 63130,
friedenberg@wustl.edu.
1
Terminal type structures have important implications for game theoretic analyses. In particular,
we will see that the results here shed light on the epistemic conditions for the iteratively undominated
strategies.
Section 1 begins with a heuristic treatment of the main results. It can be read either before or
simultaneously with the formal treatment. The formal treatment begins in Section 2, which defines
terminal and finitely terminal structures. Sections 3 and 4 provide a characterization. Section 5
gives the application to game theory. Finally, Section 6 concludes with a discussion of conceptual
and formal aspects of the paper.
1
Heuristic Treatment
Consider a game situation where Nature can choose one of two actions, namely Heads or Tails.
Suppose there are two players, Ann or Bob, each of whom faces uncertainty about this choice of
Nature. Ann (resp. Bob) may also face uncertainty about what Bob (resp. Ann) believes Nature
chooses. And so on.
1.1
Type Structure
We want a model to represent the hierarchies of beliefs associated with such a situation. Following
Harsanyi [12, 1967-1968], we use a type structure as such a model. Figure 1.1 depicts one possible
type structure associated with this situation.
λa(ta)
Heads
Tails
λb(tb)
ta
Heads
1
Tails
0
tb
Figure 1.1
The type structure has two components. First, for each player, there is a set of types. In our
example, this is given by T a = {ta } and T b = tb . Second, there are maps, viz. λa and λb , from
the types of a given player to measures about the choice of Nature and the types of the other player.
In our example, λa (ta ) assigns probability 12 to (Heads, tb ) and probability 12 to (Tails, tb ). Also,
λb tb assigns probability 1 to (Heads, ta ). At times, we will be loose and say that “type tb assigns
probability 1 to (Heads, ta )” instead of referring to the measure λb tb .
2
The type structure implicitly models hierarchies of beliefs. To see this, notice that type ta of
Ann assigns probability 12 : 12 to Heads : Tails. This will be called type ta ’s first-order belief.
Similarly, type tb has a first-order belief that assigns probability 1 to Heads. Using this, type ta
assigns probability 12 to the event “Nature chooses Heads, and Bob assigns probability 1 to Nature
choosing Heads” and probability 12 to the event “Nature chooses Tails, and Bob assigns probability
1 to Nature choosing Heads.” This is type ta ’s second-order belief. And so on.
We point to a feature of the hierarchies of beliefs induced by types ta and tb . These hierarchies
of beliefs are coherent: Type ta ’s second-order belief agrees with her first-order belief, in the sense
that both assign probability 12 : 12 to Heads : Tails. More formally, suppose type ta is associated
with a hierarchy of beliefs δ a (ta ) = (δ a1 (ta ) , δ a2 (ta ) , . . .), where δ am is the “mth order belief” on the
“mth space of uncertainty” (Section 2 gives a precise definition.) Then the marginal of δ am+1 (ta )
on the “mth space of uncertainty” is δ am (ta ).
More generally, type structures induce hierarchies of beliefs that are coherent, assign probability
1 to coherent beliefs, etc. (See Lemma 3.3.) Following Brandenburger-Dekel [7, 1993], say that
types induce hierarchies of beliefs that satisfy coherency and common belief of coherency.1
1.2
Large Type Structures
At this point, a natural question arises: Does there exist a “large” type structure—i.e., a type structure
that has a type associated with every possible hierarchy of belief? Quite a number of papers in the
literature have given positive answers. (See Section 6.)
This paper asks a related but distinct question: Which type structures represent all hierarchies
of beliefs (satisfying coherency and common belief of coherency)?
Return to the example in Figure 1.1. In that type structure, Ann has one type, and so there
is only one associated hierarchy of belief. So, for instance, since there is no type of Ann, viz. ua ,
with λa (ua ) {Heads} × T b = 1, there is also no type of Ann with a first-order belief that assigns
probability 1 to Heads.
In part, the example fails to represent all hierarchies of beliefs because Ann’s map λa is not
surjective. Of course, it can’t be in our example since T a contains only a single point. So, imagine
that we expand T a so that it is now M {Heads, T ails} × T b —i.e., the set of probability measures
on {Heads, T ails} × T b —and take λa to be the identity map. Then λa is surjective and so the new
set T a is associated with all possible first-order beliefs of Ann.
Still, the new expanded structure may fail to “contain” all hierarchies of beliefs. While λa is
surjective, λb is not. This causes two problems. First, as above, there are first-order beliefs of
Bob that are not associated with Bob’s unique type. But, because such first-order beliefs of Bob
are missing, there are also second-order beliefs of Ann that are missing from this structure. So, we
will want to expand Bob’s type set too, so that the map λb can be surjective. But, when we do so,
1 Brandenburger-Dekel [7, 1993] used the term “common knowledge of coherency.” Given the modern usage of the
terms “knowledge” and “belief,” we stick to the latter.
3
there are more possible measures of Ann, and so we will need to expand her type space again, as
well. And so on.
Such reasoning naturally leads to the notion of a complete type structure: A type structure where
both λa , λb are surjective. The idea is due to Brandenburger [5, 2003]. Complete type structures
are important for a number of epistemic analyses. (See, for instance, Battigalli-Siniscalchi [3, 2002]
and Brandenburger-Friedenberg-Keisler [8, 2004].) We return to the game theory application in
Section 5.
One natural conjecture is that a complete type structure represents all possible hierarchies of
beliefs. We will show that this is the case, provided the complete structure satisfies certain conditions. That is, under certain conditions, completeness is a sufficient condition for terminality.
Moreover, these conditions would appear to be tight. We will also see that completeness is not
a necessary condition for terminality. However, we will establish certain instances under which a
terminal structure must be complete.
To state these results formally, we will need to introduce further terminology.
1.3
Terminal Type Structures
We begin by stating the condition that a type structure represent all hierarchies of beliefs (that
satisfy coherency and common belief of coherency). There are two conditions:
(a) A type structure Λ is finitely terminal if, for each type ua (resp. ub ) that occurs in some type
structure and for each m, there is some type tam (resp. tbm ) in Λ so that the hierarchies induced by
ua and tam (resp. ub and tbm ) agree up to level m.
(b) A type structure Λ is terminal if, for each type ua (resp. ub ) that occurs in some type structure,
there is some type ta (resp. tb ) in Λ so that ua and ta (resp. ub and tb ) induce the same hierarchies
of beliefs.
Condition (a) is new. Condition (b) has a long history in game theory, going back to Böge-Eisele [4,
1979], who also refer to this property as terminal. (Our use of the term also fits with BrandenburgerKeisler’s [10, 2006] recent taxonomy. See also Meier [21, 2006]. Section 6 discusses the relationship
between the terminal property and the so-called “universal type structure.”)
We will show:
Theorem: Fix a complete type structure Λ with metrizable type sets T a , T b and
measurable maps λa , λb .
(i) If the type sets T a , T b are analytic subsets of a Polish space, then the structure is
finitely terminal.
4
(ii) If the type sets T a , T b are compact metrizable and λa , λb are continuous, then the
structure is terminal.
To understand the need for these assumptions, suppose Ann and Bob face the same underlying
set of uncertainty, viz. X, and this set is finite. In our example, X is the set {Heads, Tails}, and
so finite. The main text treats a more general case, where X is Polish.
Begin with part (i) of the result. Fix a complete structure. Also fix a hierarchy of beliefs for
Ann (that satisfies coherency and common belief thereof), viz. h = (µ1 , µ2 , . . .). We want to argue
that, for each m, there is some type tam in the complete structure so that the hierarchies induced by
tam agree with h = (µ1 , µ2 , . . .) up to level m.
Tb
µ1
µ1(x1)
x1 x2
µ1(xk)
X
xk
ρ1
X
xk
Figure 1.2
First, suppose m = 1 and refer to Figure 1.2. It suffices to show that there is some measure ν 1
on X × T b associated with the same measure on X as µ1 . If so, by completeness, there is some type
ta1 with λa (ta1 ) = ν 1 , and this type must be associated with the first-order belief µ1 .
To construct the measure ν 1 : Begin with the projection map ρ1 : X × T b → X. Construct ν 1 by
assigning (ρ1 )−1 (xk ) = {xk } × T b the same probability as µ1 assigns to xk , for each point xk ∈ X.
Clearly, ν 1 is well-defined and so we have established the result for m = 1.
Tb
B’s 1st-order
beliefs
ρ2
X
X
xk
xk
Figure 1.3
5
Next, suppose m = 2 and refer to Figure 1.3. Here, we are concerned with Ann’s second-order
space of uncertainty. This is X cross the first-order beliefs of Bob. We can write a map ρ2 from
X × T b to Ann’s second-order space of uncertainty. Specifically, ρ2 maps xk , tb to (xk , δ b1 (tb )),
i.e., to xk and the first-order belief induced by tb .
Tb
B’s 1st-order
beliefs
ν2((ρ2)-1(E))
µ2(E)
ρ2
X
X
xk
xk
Figure 1.4
Now refer to Figure 1.4. We want to find a measure ν 2 on X × T b associated with the same
measure on “Ann’s second-order space of uncertainty” as µ2 . (If so, by completeness, an appropriate
type exists.) To do so, follow a construction similar to the case of m = 1. Now, for any event E
in “Ann’s second-order space of uncertainty,” let ν 2 assign (ρ2 )−1 (E) the same probability as µ2
assigns to E. Doing this gives a probability to every event in the sub-σ-algebra generated by ρ2 .
That is, as constructed, ν 2 is a probability measure on a sub-σ-algebra, not a probability measure
on all the Borel sets. So, the question is: Can we extend ν 2 to a probability measure on all the
Borel sets?
The question we have here is an instance of a general mathematical question: Fix a topological
space Ω and a sub-σ-algebra of the Borel σ-algebra. Given a measure on the sub-σ-algebra, can
the measure be extended to a measure on all the Borel sets? In general, the answer is no. (See
Aldaz-Render [1, 2000] for examples.) But, when Ω is an analytic subset of a Polish space (and,
in particular, Polish spaces are analytic subsets of Polish spaces), any probability measure on the
sub-σ-algebra can be extended to a probability measure on the Borel sets. This result is due to
Lubin [18, 1974]. (See, also, Landers-Rogge [17, 1974] and Yershov [26, 1974]. Aldaz-Render [1,
2000] provide other topological conditions.)
With this, we complete the analysis for m = 2. An induction argument establishes the result
more generally.
Now turn to part (ii) of the result and continue to fix Ann’s hierarchy of beliefs, viz. h =
(µ1 , µ2 , . . .). Refer to Figure 1.5. By part (i) of the result, for each m, there is some type ta whose
induced hierarchy agrees with h up to level m. Let T [h|m] be the set of all such types and note
that each T [h|m] is non-empty. Also note that T [h|1] , T [h|2] , . . . is a decreasing sequence of sets.
If a type’s induced hierarchy agrees with h, then the type must be contained in m T [h|m]. Of
6
course, in general, a decreasing sequence of non-empty sets may have an empty intersection. But,
in the particular case where this decreasing sequence of sets is also closed and in a compact space,
the intersection is non-empty. This is why the stated result assumes that the set T a is compact (so
that the sets lie in a compact space) and that the map λa is continuous (to insure that the sets are
closed).
Ta T[h|2]
T[h|1]
Figure 1.5
Currently, this paper does not offer an example that violates the conclusion of parts (i) (resp.
(ii)) in the absence of analyticity (resp. compactness and continuity). That said, the results seem
tight, in the sense that the mathematical steps used appear to be fundamentally related to the
question at hand.
1.4
Are Terminal Structures Complete?
Now let’s ask about a converse: Given a terminal structure Λ, is Λ also complete? The answer is
no.
To see this, again consider a game situation with two players Ann and Bob. They each face
uncertainty about a choice of Nature, but now this uncertainty must be degenerate. Specifically,
Nature has only one choice, namely Heads. It follows that there is only one possible hierarchy of
beliefs for each player, i.e., Ann (resp. Bob) assigns probability 1 to Heads, assigns probability 1 to
“Heads and ‘Bob (resp. Ann) assigns probability 1 to Heads,’” and so on.
One example of a type structure is as follows: Let T a = {ta , ua } and T b = tb . Also let
λa (ta ) (Heads, tb ) = λa (ua ) (Heads, tb ) = 1
λb tb (Heads, ta ) = 1.
Indeed, it does induce the one hierarchy of beliefs—as it must. So it is a terminal structure. But
it is not complete since λb is not surjective. For instance, there is no type of Bob that assigns
probability 1 to (Heads, ua ).
7
A key feature of this example is that the structure has two types of Ann that induce the same
hierarchy of beliefs, namely ta and ua . As such, the structure may contain all hierarchies of beliefs
for Bob, even though no type of Bob considers ua possible. One conjecture is that, when no two
types induce the same hierarchy of beliefs, a terminal structure must be complete. We will see that
a bit more is required:
Theorem: Fix a terminal type structure Λ where T a , T b are Borel subsets of a Polish
space. If no two types induce the same hierarchies of beliefs, then the structure Λ is
also complete.
Section 4 discusses the requirement that T a , T b be Borel subsets of a Polish space. In particular, this requirement is quite related to the reason we require that no two types induce the same
hierarchies of beliefs.
2
Type Structures
Throughout the paper, we will restrict attention to topological spaces, viz. (Ω, T (Ω)). If (Ω1 , T (Ω1 ))
and (Ω2 , T (Ω2 )) are topological spaces with Ω1 ⊆ Ω2 , we will always take T (Ω1 ) to be the topology
induced by (Ω2 , T (Ω2 )). Analogously, we will always restrict attention to the product topology
when relevant.
A topological space (Ω, T (Ω)) induces a measurable space (Ω, B (Ω)) where B (Ω) is the Borel
σ-algebra on Ω. Given a measurable space (Ω, B (Ω)), write M (Ω) for the set of (Borel) probability
measures on (Ω, B (Ω)). We will endow M (Ω) with the topology of weak convergence so that it
is again a topological space. Of course, if (Ω, T (Ω)) is separable (resp. Polish, resp. compact
metrizable), so is M (Ω) endowed with the topology of weak convergence. Given a map f : Ω → Φ
write f : M (Ω) → M (Φ), where f (µ) is the image measure of f under µ.
For notational simplicity, we restrict attention to two player game situations with players a and
b.2 Let (X a , T (X a )) (resp. (X b , T (X b ))) be a Polish space. Think of X a as player a’s (primitive)
set of uncertainty. We will write c for an arbitrary player amongst a, b. Given an arbitrary player
c, we will write d for the other player.
Definition 2.1 An X a , X b -based type structure is given by
Λ = X a , X b ; (T a , T (T a )), (T b , T (T b )); λa , λb
where (T a , T (T a )) and (T b , T (T b )) are each metrizable and
λa
λb
2 The
: T a → M Xa × T b
: T b → M Xb × T a
results readily follow for game situations with three or more players.
8
are Borel measurable. The sets T a and T b are type sets. A state is some quadruple xa , ta , xb , tb ∈
X a × T a × X b × T b.
Definition 2.1 defines a type structure where T a , T b are metrizable spaces. We will also be
interested in cases where T a , T b are more specific topological spaces. Now we mention some spaces
that will be of interest. Say that a set is an analytic subset of a Polish space if it is the continuous
image of a Polish space. (So a Polish space is an analytic subset of itself.)
Definition 2.2 Call the type structure Λ analytic if (T a , T (T a )) and (T b , T (T b )) are each analytic
subsets of Polish spaces.
Definition 2.3 Call the type structure Λ compact if (T a , T (T a )) and (T b , T (T b )) are each compact.
Definition 2.4 Call the type structure Λ Borel if (T a , T (T a )) and (T b , T (T b )) are each Borel
subsets of Polish spaces.
Definition 2.5 Call the type structure Λ continuous if λa and λb are continuous.
We are also interested in a particular “large” type structure:
Definition 2.6 An X a , X b -based type structure, viz.
Λ = X a , X b ; (T a , T (T a )) , T b , T T b ; λa , λb ,
is complete if λa , λb are onto.
We will want to relate a complete type structure to structures that contain all possible hierarchies
of beliefs. For that, we will need to specify how types induce hierarchies of beliefs.
d
c
c
c
Begin by defining Z1c = X c and Zm+1
= Zm
× M Zm
= Z1c ×
. For each m ≥ 2, Zm+1
d
m
c
c
n=1 M Zn (Lemma A1). Note that each of the sets Zm is Polish since X is Polish.
c
Now, define maps ρcm+1 : X c × T d → Zm+1
so that
Then, for each m,
ρc1 xc , td = xc
ρcm+1 xc , td = (ρcm xc , td , ρdm (λd (td ))).
ρcm+1 xc , td = (xc , δ d1 (td ), . . . , δ dm (td ))
c
(Lemma A1). Define δ cm : T c → M (Zm
) so that δ cm = ρcm · λc . The maps ρcm and δ cm are
measurable. They are continuous when the structure is continuous. (See Lemma A4.)
9
c c
c c
c c
c
Define δ c : T c → ∞
Then δ c (tc ) is the
m=1 M (Zm ) so that δ (t ) = (δ 1 (t ) , δ 2 (t ) , ...).
hierarchy of beliefs induced by tc . The map δ c is measurable and continuous when the structure is
continuous.
We can now state the condition that a type structure contains all hierarchies of beliefs. In the
definitions below, take
X a , X b ; (T a , T (T a )) , T b , T T b ; λa , λb
a
a
b
b
a
b
Λ =
X a , X b ; (T , T (T ), (T , T (T )); λ , λ
Λ =
to be X a , X b -based type structures.
Definition 2.7 An X a , X b -based type structure Λ is finitely terminal if, for each X a , X b c
c
based type structure Λ, each type t ∈ T , and each m, there is a type tc ∈ T c with
c
c
c
c
(δ c1 (tc ) , . . . , δ cm (tc )) = (δ 1 (t ), . . . , δ m (t )).
c
(Note that, in Definition 2.7, tc can depend on both t and m.)
Definition 2.8 An X a , X b -based type structure Λ is terminal if, for each X a , X b -based type
c c
c
c
structure Λ and each type t ∈ T , there is a type tc ∈ T c with δ c (tc ) = δ t .
c
Definition 2.7 says that the structure Λ is finitely terminal if, for each t type that occurs in
c
some structure and each m, there is a type tcm in Λ whose hierarchy agrees with t up to level m.
c
Definition 2.8 says that the structure Λ is terminal if, for each t type that occurs in some structure,
c
there is a type tc in Λ with the same hierarchy of beliefs as t . Appendix D (Proposition D1)
establishes that, if Λ is terminal, then for any hierarchy of beliefs consistent with “coherency and
common belief of coherency” there is a type in Λ associated with this hierarchy of beliefs.
3
Main Theorem
Theorem 3.1 Fix a complete type structure
Λ = X a , X b ; (T a , S (T a )) , T b , S T b ; λa , λb .
(i) If Λ is analytic, then Λ is finitely terminal.
(ii) If Λ is compact and continuous, then Λ is terminal.
To prove Theorem 3.1, it will be useful to define the notion of an X a , X b -belief structure.
These will correspond to the hierarchies consistent with “coherency and common belief of coherency”
mentioned in Section 1.1.
10
First, we will show that any type structure can be embedded into the belief structure. Then
we will show that, for any analytic complete type structure and any element of the belief structure,
viz. h = (µ1 , µ2 , . . .), there is a type in the complete structure whose hierarchy of beliefs coincides
with h up to level m. If, in addition, the structure is compact and continuous, for any element of
the belief structure, viz. h = (µ1 , µ2 , . . .), there is a type of the complete structure whose hierarchy
of beliefs coincides with h.
3.1
Full Set of Coherent Hierarchies
This section inductively constructs the set of all hierarchies of belief that are coherent, assign probability 1 to coherent hierarchies of beliefs, etc.
k
Fix measurable spaces Ω1 , . . . , Ωm and a sequence of measures (µ1 , . . . , µm ) ∈ m
k=1 M( n=1 Ωn ).
Write margΩn µk for the marginal of µk on Ωn . If margΩn µk = µn , for all 1 k m and all n k,
say the sequence (µ1 , . . . , µm ) is coherent.
c
c
Inductively define sets Xm
, Ymc , Hm
as follows: Set X1c = X c and Y1c = H1c = M (X1c ). Assuming
c c
d
c
the sets are defined for m, set Xm+1
= X1c × Hm
, Ym+1
= Ymc × M Xm+1
, and
c c
c
c
Hm+1
= Hm
× M Xm+1
: margXm
∩ µ1 , . . . , µm+1 ∈ Ym+1
.
c µm+1 = µm
c c
to be well-defined we need that, for any measure µm+1 ∈ M Xm+1
For Hm+1
, margXm
c µm+1 is
indeed well-defined. But this is so since
c
Xm+1
d
= X1c × Hm
d
d
⊆ X1c × Hm−1
× M Xm
d
c
= Xm
× M Xm
.
c
Notice that, if µ1 , . . . , µm+1 ∈ Hm+1
, then (µ1 , . . . , µn ) ∈ Hnc for all n m. So, the hierarchy of
belief up to level m is coherent.
d
c
Several facts: First, a standard inductive argument gives that Xm+1
⊆ X1c × m
n=1 M Xn and
c
c
c
Hm
⊆ Ymc for all m. Second, for each m, the sets Xm
, Ymc are Polish and Hm
is a closed subset of a
a
b
c
Polish space (and, so, Polish). See Lemma B2. When X , X are both compact, each of Xm
, Ymc ,
c
and Hm
is compact metrizable, for all m. (See Remark B1.)
Definition 3.1 Let
Ha
Hb
a
a
= {(µ1 , µ2 , . . .) ∈ ∞
m=1 M (Xm ) : for each m, (µ1 , . . . , µm ) ∈ Hm }
b
∞
b
=
(µ1 , µ2 , . . .) ∈ m=1 M Xm : for each m, (µ1 , . . . , µm ) ∈ Hm
.
The set H a × H b is called the X a , X b -based belief structure.
11
The X a , X b -based belief structure consists of all hierarchies of beliefs associated with “coa
b
herency and common belief of coherency.” In line with this terminology, refer to the sets Hm
× Hm
as the mth -level X a , X b -based belief structure.
The construction of the X a , X b -based belief structure is akin to the canonical constructions in
Mertens-Zamir [23, 1985] and Brandenburger-Dekel [7, 1993]. The X a , X b -based belief structure
is a closed subset of a Polish space and, so, Polish. Similarly, when X a , X b are both compact, it is
compact metrizable. (See Lemma B3 for these results.)
3.2
Embeddings
a
b
This subsection establishes that each mth -level X a , X b -based belief structure, viz. Hm
× Hm
,
m
c
a
b
a
b
can be embedded into n=1 M (Zn ). And, similarly, the X , X -based belief structure H × H
c
a
b
can be embedded into ∞
m=1 M (Zm ). That is, H × H is topologically equivalent to a subset of
∞
c
m=1 M (Zm ).
c
c
c
c
For this, begin by defining maps ζ cm : Xm
→ Zm
and η cm : Hm
→ m
n=1 M (Zn ) inductively: Let
c
c
c
ζ 1 , ηc1 be the identity maps. Assuming ζ m , ηcm are defined, take ζ m+1 and ηcm+1 so that
ζ cm+1 xc1 , hdm = ζ c1 (xc1 ) , ηdm hdm
η cm+1 hcm , µcm+1 = (ηcm (hcm ) , ζ cm+1 (µcm+1 )).
Note that
ζ cm+1 xc1 , µd1 , . . . , µdm−1 , µdm = (ζ cm xc1 , µd1 , . . . , µdm , ζ dm µdm )
ηcm (µc1 , . . . , µcm ) = (µc1 , ζ c2 (µc2 ) , . . . , ζ cm (µcm )).
See Lemma B4.
c
c
c
Lemma 3.1 For each m, the maps ζ cm : Xm
→ Zm
and ηcm : Hm
→
Now define η c : H c →
∞
c
m=1 M (Zm )
m
c
n=1 M (Zn )
are embeddings.3
such that, for each (µ1 , µ2 , . . .) ∈ H c ,
η c (µ1 , µ2 , . . .) = (µ1 , ζ c2 (µc2 ) , . . .).
Notice that, by Lemma B4, if (µ1 , ζ c2 (µc2 ) , . . . , ζ cm (µcm )) is an initial segment of ηc (µ1 , µ2 , . . .) of
length m then
ηcm (µ1 , . . . , µm ) = (µ1 , ζ c2 (µc2 ) , . . . , ζ cm (µcm )).
We will repeatedly make use of this fact. Now:
Lemma 3.2 The map ηc : H c →
3 Proofs
∞
c
m=1 M (Zm )
is an embedding.
not found in the main text can be found in the appendices.
12
3.3
From Type to Belief Structures
Section 2 showed that types induce hierarchies of beliefs, via the mapping δ c . In this section,
we will see that δ c (T c ) can be embedded in the belief structure, via the inverse of ηc . That is,
δ c (T c ) ⊆ ηc (H c ). Begin with a finite version of this result:
Lemma 3.3 For each m:
c
(i) for any xc , td ∈ X c × T d , ρcm xc , td ⊆ ζ cm (Xm
);
c
(ii) for any tc ∈ T c , δ cm (tc ) (ζ cm (Xm
)) = 1;
c
(iii) for any tc ∈ T c , (δ c1 (tc ) , . . . , δ cm (tc )) ∈ ηcm (Hm
).
Part (iii) is the result of interest. It says that the mth -level hierarchies of beliefs induced by
some type is topologically equivalent to some mth -level hierarchy of beliefs in the belief structure.
To establish this, we will need some preliminary results. First, types induce coherent hierarchies
of beliefs:
c
c
c
c
Lemma 3.4 For each m, margZm
c δ m+1 (t ) = δ m (t ).
c
Proof. Fix an event E in Zm
and notice that
c −1 d d ρm+1
E × M Zm
= [(ρcm )−1 (E)] ∩ [X c × (δ dm )−1 M Zm
]
= (ρcm )−1 (E) .
From this,
d −1 d δ cm+1 (tc ) E × M Zm
= λc (tc ) ρcm+1
E × M Zm
= λc (tc ) ((ρcm )−1 (E))
= δ cm (tc ) (E) ,
as required.
Corollary 3.1 For each n m, margZnc δ cm (tc ) = δ cn (tc ).
Next:
Lemma 3.5 Let Ω, Φ be Polish spaces and let f : Ω → Φ be a continuous injective map. If
ν ∈ M (Φ) with ν (f (Ω)) = 1, then there exists some µ ∈ M (Ω) such that ν is the image measure
of µ under f.
13
Lemma 3.5 can be given an independent proof by the Lusin-Souslin Theorem [16, 1995; Theorem
15.1]. But since it will be a specific case of Lemma 3.8 below, the proof is omitted.
Now we can prove the stated result:
Proof of Lemma 3.3. By induction on m.
m = 1: Part (i) is immediate from the fact that X1c = Z1c and ζ c1 is the identity map. Part
(ii) follows, then, from the fact that ζ c1 (X1c ) = Z1c . Part (iii) follows again from the fact that
M (X1c ) = M (Z1c ) and ηc1 is the identity map.
m ≥ 2: Assume that the result holds for m. For Part (i), fix xc , td ∈ X c × T d . By Lemma
A1,
ρcm+1 xc , td = (xc , δ d1 (td ), . . . , δ dm (td )).
d
with
So, using Part (iii) of the induction hypothesis, there is some hdm ∈ Hm
By construction, then,
ρcm+1 xc , td = xc , η dm hdm .
ζ cm+1 xc , hdm = ζ c1 (xc ) , ηdm hdm
= ρcm+1 xc , td ,
as required.
It follows from Part (i) applied to (m + 1) that
So
−1 c
c X c × T d ⊆ ρcm+1
ζ m+1 Xm+1
⊆ X c × T d.
c c
δ cm+1 (tc ) ζ cm+1 Xm+1
= λc (tc ) ((ρcm+1 )−1 (ζ cm+1 (Xm+1
)))
c c
c
d
= λ (t ) X × T
= 1,
establishing Part (ii).
c
Finally, by the induction hypothesis, there is some hcm ∈ Hm
with (δ c1 (tc ) , . . . , δ cm (tc )) =
c ηcm (hcm ). Write hcm = (µc1 , . . . , µcm ). It suffices to show that there is some µcm+1 ∈ M Xm+1
with
c c
c
c
c
c
ζ m+1 µm+1 = δ m+1 (t ) and margXm
c µm+1 = µm .
c c
First, we construct a measure µm+1 ∈ M Xm+1
with ζ cm+1 µcm+1 = δ cm+1 (tc ). To see this,
c c
c
note that Xm+1
and Zm+1
are Polish. Also, Part (ii) gives that δ cm+1 (tc ) ζ cm+1 Xm+1
= 1. By
c c
c
c
c
Lemma 3.5, there is a measure, viz. µm+1 , with ζ m+1 µm+1 = δ m+1 (t ).
14
c
c
c
Now we show that, for this measure µcm+1 , margXm
Fix an event E in Xm
and
c µm+1 = µm .
d c
note that G = E × M Xm ∩ Xm+1 is Borel. By the Lusin-Souslin Theorem [16, 1995; Theorem
c
c
15.1], ζ cm+1 (G) is Borel. (This uses the fact that Xm+1
, Zm+1
are Polish and ζ cm+1 is injective
continuous.) Then
d c
∩ Xm+1
,
δ cm+1 (tc ) ζ cm+1 (G) = µcm+1 E × M Xm
since ζ cm+1 is injective. Also, by Lemma B4,
d
.
ζ cm+1 (G) ⊆ ζ cm (E) × M Zm
So, by Part (ii) applied to m + 1,
d δ cm+1 (tc ) ζ cm+1 (G) = δ cm+1 (tc ) ζ cm (E) × M Zm
.
So, putting the above together with Lemma 3.4, we have
d ζ cm+1 µcm+1 (G) = δ cm+1 (tc ) ζ cm (E) × M Zm
= δ cm (tc ) (ζ cm (E))
= ζ cm (µcm ) (E) ,
where the last line follows from Lemma B4.
Lemma 3.6 For each type tc ∈ T c , δ c (tc ) ∈ η c (H c ).
Proof. Fix a type tc ∈ T c . Inductively construct a sequence (µ1 , µ2 , . . .) ∈ H c as follows: Choose
µ1 ∈ H1c so that δ c1 (tc ) = ζ c1 (µc1 ) = µ1 . Assuming that we have found measures (µ1 , . . . , µm ) ∈
c
Hm
with
(δ c1 (tc ) , . . . , δ cm (tc )) = (µ1 , ζ c2 (µc2 ) , . . . , ζ cm (µcm )),
c c
Lemma 3.3(iii) says that we can find a measure µm+1 ∈ M Xm+1
with µ1 , . . . , µm , µm+1 ∈ Hm+1
and
c c
δ 1 (t ) , δ c2 (tc ) , . . . , δ cm (tc ) , δ cm+1 (tc ) = (µ1 , ζ c2 (µc2 ) , . . . , ζ cm (µcm ) , ζ cm+1 µcm+1 ).
So, inductively, we identify a sequence (µ1 , µ2 , . . .) ∈ H c with δ cm (tc ) = ζ cm (µcm ) for each m. This
establishes the result.
3.4
From Belief to Complete Structures
This section relates complete type structures to belief structures. In particular, we begin by establishing conditions under which the mth -level belief structure is topologically equivalent to a complete
type structure. We then go on to show conditions under which the belief structure is topologically
15
equivalent to a complete structure. As discussed in Section 1.3, these results appear to require
additional topological structure of the type sets T a and T b .
We begin by stating one of the main results of this section:
Proposition 3.1 Fix an complete analytic type structure. For each m:
c
(i) ρcm X c × T d = ζ cm (Xm
);
c
(ii) for each ηcm (µ1 , . . . , µm ) = (̟1 , . . . , ̟m ), ̟m (ζ cm (Xm
)) = 1;
c
(iii) for each (µ1 , . . . , µm ) ∈ Hm
, there exists a type tc such that ηcm (µ1 , . . . , µm ) = (δ c1 (tc ) , . . . , δ cm (tc )).
Part (iii) says that, for an analytic complete structure Λ, each element of the mth -level belief
structure is topologically equivalent to some mth -level hierarchy of beliefs associated with a type tc
in Λ.
Showing this requires some preliminary results. First, coherency is preserved under the map
c
ηm .
c
Lemma 3.7 Fix µ1 , . . . , µm+1 ∈ Hm+1
. Then, for each n m, margZnc ζ cm+1 µm+1 = ζ cn (µn ).
Proof. Fix some n m and some event G in Znc . Then
ζ cn (µn ) (G) = µn ((ζ cn )−1 (G)).
We know that margXnc µm+1 = µn , so that
c
µn ((ζ cn )−1 (G)) = µm+1 ({(ν 1 , . . . , ν m+1 ) ∈ Hm+1
: (ν 1 , . . . , ν n ) ∈ (ζ cn )−1 (G)})
d
= µm ((ζ cm+1 )−1 (G × m+1
k=n+1 M(Zk )))
d
= ζ cm+1 (µm+1 )(G × m+1
k=n+1 M(Zk ))
= margZnc ζ cm+1 µm+1 (G) ,
as required.
c
Next, as discussed in Section 1.3, given a measure on µm on Zm
, we will want to guarantee that
c
d
we can find a measure ν m on X × T so that µm is the image measure of ν m under ρcm . This is
given by the following result due to Lubin [18, 1974]. (Again, see also Landers-Rogge [17, 1974]
and Yershov [26, 1974].)
Lemma 3.8 (Lubin) Let Ω, Φ be analytic subsets of a Polish space and let f : Ω → Φ be a measurable map. If ν ∈ M (Φ) with ν (f (Ω)) = 1, then there exists some µ ∈ M (Ω) such that ν is the
image measure of µ under f .
16
We now turn to proving Lemma 3.1.
Proof of Proposition 3.1. By induction on m.
m = 1: By Lemma 3.3, ρc1 X c × T d ⊆ ζ c1 (X1c ). Conversely, fix some xc ∈ X c = X1c and note
that, for any td ∈ T d , ρc1 xc , td = xc = ζ c1 (xc ), establishing Part (i). Part (ii) is immediate from
the fact that ζ c1 (X1c ) = Z1c . For Part (iii), fix some µ1 ∈ H1c = M (X c ). By Lemma 3.8, there exists
a measure ν 1 ∈ M X c × T d such that µ1 is the image measure of ν 1 under ρc1 . By completeness,
there is a type tc with λc (tc ) = ν 1 . Since δ c1 (tc ) is the image measure of λc (tc ) under ρc1 , the result
is established.
m ≥ 2: Assume that the result holds for m. We will show that it also holds for m + 1.
c Begin with Part (i). By Lemma 3.3, ρcm+1 X c × T d ⊆ ζ cm+1 Xm+1
. Fix some ζ cm+1 (xc1 , µd1 , . . .
, µdm ). By Part (iii) of the induction hypothesis, there exists a type td such that
η dm (µd1 , . . . , µdm ) = (δ d1 (td ), . . . , δ dm (td )).
So, ρcm+1 xc1 , td = (xc1 , δ d1 (td ), . . . , δ dm (td )) as required.
Now turn to Part (ii). Recall from Lemma B4 that ̟m+1 =ζ cm+1 µm+1 . Certainly
so that
c −1 c
c c
ζ m+1
ζ m+1 Xm+1
= Xm+1
,
c −1 c
c ̟m+1 ζ cm+1 Xm+1
= µm+1 ζ cm+1
ζ m+1 Xm+1
=1
as required.
c
For Part (iii), fix some µ1 , . . . , µm+1 ∈ Hm+1
with ηcm+1 µ1 , . . . , µm+1 = (̟1 , . . . , ̟m+1 ).
First, notice that there exists a measure ν m+1 ∈ M X c × T d such that ̟m+1 is the image mea
sure of ν m+1 under ρcm+1 : This follows from Lemma 3.8 and the fact that ̟m+1 ρcm+1 X c × T d =
1 (Parts (i)-(ii) of this Lemma).
Now, notice that, for each n m, margZnc ̟m+1 is the image measure of ν m+1 under ρcn . To
see this, fix an event E in Znc . Then
d margZnc ̟m+1 (E) = ̟m+1 E × m
k=n+1 M Zk
−1 d = ν m+1 ρcm+1
E× m
k=n+1 M Zk
= ν m+1 (ρcn )−1 (E) ,
as required.
By completeness, there exists a type tc such that λc (tc ) = ν m+1 . Since δ cm+1 = ρcm+1 · λc , we
also have that δ cm+1 (tc ) = ̟m+1 . Moreover, by Lemma 3.7, ̟n = margZnc ̟m+1 for all n m.
Using this and the fact that, for all n, δ cn = ρcn · λc , we also have that δ cn (tc ) = ̟n for all n m.
17
This establishes the result.
Now:
Proposition 3.2 Fix a complete compact continuous type structure. For each (µ1 , µ2 , . . .) ∈ H c ,
there exists a type tc ∈ T c such that η c (µ1 , µ2 , . . .) = (δ c1 (tc ) , δ c2 (tc ) , . . .).
Proof. Fix a type structure. Also fix some hc = (µ1 , µ2 , . . .) ∈ H c . Write hc |m for the initial
segment of hc of length m, i.e., hc |m = (µ1 , . . . , µm ). Let T c [hc |m] be the set of types tc ∈ T c with
ηcm (hc |m) = (δ c1 (tc ) , . . . , δ cm (tc )), i.e., the set of types whose hierarchy induces an initial segment
of hc |m.
By Proposition 3.1, the sets T c [hc |m] are each non-empty. These sets are decreasing. Moreover,
they are closed. To see this last fact, note that each set {hc |m} is closed. Since ηcm is an embedding,
it follows that ηcm ({hc |m}) is closed. Now, using the fact that λc is continuous, each δ cn , where
n m, is continuous. (This is Lemma A4.) Then the map tc → (δ c1 (tc ) , . . . , δ cm (tc )) is continuous
(see Munkres [24, 1975; Theorem 8.5]), establishing that T c [hc |m] is closed.
We have established that T c [hc |1] , T c [hc |2] , . . . is a decreasing sequence of non-empty closed
sets in a compact space. It follows that m T c [hc |m] = ∅, establishing the result.
3.5
Proof of Main Theorem
Now, Theorem 3.1 is a Corollary of the results of Sections 3.3-3.4:
Proof of Theorem 3.1. Part (i) follows from Lemma 3.6 and Proposition 3.1. Part (ii) follows
from Lemma 3.6 and Proposition 3.2.
4
When a Terminal Structure is Complete
Section 3 provides conditions under which a complete structure is finitely terminal or terminal. This
section addresses the converse: Is a terminal structure complete? Section 1.4 provided an example
of a terminal structure that is not complete. There, we argued that the difficulty lies in the fact
that the terminal structure is redundant, i.e., has two types associated with the same hierarchy of
beliefs. As such, we will be interested in non-redundant type structures (Mertens-Zamir [23, 1985]):
Definition 4.1 An X a , X b -based type structure, viz.
Λ = X a , X b ; (T a , T (T a )) , T b , T T b ; λa , λb ,
is non-redundant if the maps δ a , δ b are injective.
18
We will see that:
Theorem 4.1 Fix a non-redundant Borel type structure Λ. If Λ is terminal, then it is complete.
Now we give the idea of the proof. Doing so will illuminate the role of a Borel type structure.
Suppose Λ is a structure under which δ b is injective (so no two types of Bob induce the same
hierarchy of beliefs). Further, suppose that the structure is not complete. That is, there is a
measure µa on X a × T b such that each λa (ta ) = µa . We want to argue that the hierarchy of beliefs
induced by µa cannot be associated with any type of Ann. If so, then Λ is not terminal. (See
Proposition D1.)
To see this, define a map
ρa : X a × T b → X a ×
∞
m=1 M
so that ρa xa , tb = xa , δ b tb . This induces the map
b
Zm
b ρa : M X a × T b → M X a × ∞
m=1 M Zm
so that ρa (µa ) is the image measure of µa ∈ M X a × T b under ρa . Suppose the map ρa is
injective: Then, for each type ta in Λ, λa (ta ) and µa induce different beliefs and X a and Bob’s
hierarchies of beliefs. (See Lemma C4.) It will then follow that each λa (ta ) induces a different
hierarchy of beliefs than µa , as required.
Tb
Hb
F = (ρa)-1(ρa(F))
ρa injective
(ρa(F))
Xa
Xa
Figure 4.1
But is the map ρa injective? Since no two types of Bob induce the same hierarchy of beliefs,
δ —and so ρa —must be injective. Now refer to Figure 4.1. Fix two distinct measures ν a and
̟a on X a × T b . There is some event, viz. F , with ν a (F ) = ̟a (F ). Since ρa is injective,
F = (ρa )−1 (ρa (F )), so that
b
ν a ((ρa )−1 (ρa (F ))) = ̟a ((ρa )−1 (ρa (F ))).
19
b
a
Supposing ρa (F ) is an event in X a × ∞
m=1 M Zm , it follows that the image measure of ν under
ρa must assign to the event ρa (F ) and different probability than the image measure of ̟a under ρa
assigns to this event, i.e., that ρa (ν a ) = ρa (̟a ), as required.
To show that ρa (F ) is an event: Notice that the map ρa is injective and measurable, mapping
a metrizable space into a Polish space. Using a result due to Purves [20], ρa (F ) is measurable
provided T a is a Borel subset of a Polish space. This is why we require that Λ be Borel.
5
Application to Game Theory
The results here shed light on epistemic conditions for the iteratively undominated (IU) strategies—
specifically, on the role of compactness and continuity in these conditions. In doing so, they suggest
new epistemic conditions for the IU strategies. We begin by reviewing the known conditions and
subsequently turn to the new conditions.
Fix a two-player finite strategic-form game S a , S b , πa , π b , where S a , S b are the finite strategy
sets and πa , π b are payoff functions of Ann and Bob, respectively. Extend πc to M(S c ) × M(S d )
in the usual way, i.e. π c (σc , σd ) = (sc ,sd )∈S c ×S d σ c (sc )σd (sd )π c (sc , sd ).
Definition 5.1 Fix Qa × Qb ⊆ S a × S b . A strategy sc ∈ Qc is strongly dominated with respect
to Qc × Qd if there exists σ c ∈ M(S c ) with σ c (Qc ) = 1 and πc (σ c , sd ) > πc (sc , sd ) for every sc ∈ Qc .
Otherwise, say sc is undominated with respect to Qc × Qd . If sc is undominated with respect
to S c × S d , simply say that sc is undominated.
We have the usual equivalence:
Lemma 5.1 A strategy sc ∈ Qc is undominated with respect to Qc × Qd if and only if there exists
σd ∈ M(S d ), with σd (Qd ) = 1 and πc (sc , σd ) ≥ πc (rc , σd ) for every rc ∈ Qc .
Definition 5.2 Set S0c = S c and define inductively
c
c
c
d
Sm+1
= {sc ∈ Sm
: sc is undominated with respect to Sm
× Sm
}.
c
A strategy sc ∈ Sm
is called m-undominated. A strategy sc ∈
undominated (IU).
∞
c
m=0 Sm
is called iteratively
c
c
Note that there is an M such that ∞
m=0 Sm = SM = ∅. It is well known that the IU strategy
c
d
set is a Pearce best-response set [19, 1984], i.e., each sc ∈ SM
is undominated given S c × SM
.
Now turn to defining the terms rationality, belief of rationality, etc. To do so, we will want to
append to the game an S b , S a -based type structure
Λ = S b , S a ; (T a , T (T a )) , T b , T T b ; λa , λb .
20
After all, a strategy of player a may be rational if she holds certain beliefs about what Bob will do,
and irrational if she holds other beliefs. Similarly, to talk about whether or not player a believes
player b is rational, we must first state what player a believes about b’s strategies and beliefs. And
so on. Under an S b , S a -based type structure
λa
λb
: T a → M Sb × T a
: T b → M Sa × T b ,
so that each type is mapped into a measure on the strategies and types of the other player, giving
us the required information.
Rationality and belief are defined relative to this fixed structure Λ:
Definition 5.3 A strategy sc ∈ Qc is optimal under σ d ∈ M(S d ) if πc (sc , σd ) ≥ πc (rc , σd ) for
every rc ∈ S c . Say (sc , tc ) ∈ S c × T c is rational if sc is optimal under margS d λc (tc ).
Rationality is a property of a strategy-type pair: A strategy may be rational given certain types,
and irrational given other types.
Definition 5.4 Say a set E ⊆ S d × T d is believed under µ ∈ M S d × T d if E is Borel and
µ (E) = 1. Say a type tc believes E if E is believed under λc (tc ).
Let
B c (E) = {tc ∈ T c : E is believed under λc (tc )}.
c
Now, for each m define Rm
inductively. Take Rc0 = S c × T c , R1c to be the set of all rational pairs
(sc , tc ), and for all m ≥ 1
c
d
∩ [S c × B c (Rm
)].
Rcm+1 = Rm
Definition 5.5 If (sa , ta , sb , tb ) ∈ Ram+1 × Rbm+1 , say there is rationality and mth-order belief
∞
a
b
of rationality (RmBR) at this state. If (sa , ta , sb , tb ) ∈ ∞
m=1 Rm ×
m=1 Rm , say there is
rationality and common belief of rationality (RCBR) at this state.
We can now state the characterization of the iteratively undominated strategies:
Proposition 5.1 Fix a game and an associated complete S b , S a -based type structure Λ.
b
a
b
(i) If T a and T b are Polish, then projSa Ram × projS b Rm
= Sm
× Sm
for each m.
∞
a
b
a
b
(ii) If Λ is compact and continuous, then projS a [ ∞
m=1 Rm ] × projS b
m=1 Rm = SM × SM .
Part (i) says that in a complete Polish structure, the strategies consistent with rationality and
mth -order belief of rationality are the strategies that survive (m + 1) rounds of eliminating dominated
strategies. Part (ii) says that if, in addition, the structure is compact and continuous, then the
strategies consistent with RCBR are exactly the IU strategies.
21
By now, Proposition 5.1 is well known. In particular, it is related to results in BrandenburgerDekel [6, 1987] and Tan-Werlang [25, 1988]. It is a special case of Proposition 6 in BattigalliSiniscalchi [3, 2002].
We now turn to the requirements for Proposition 5.1. This next example shows the role of
completeness in Part (i) of the Proposition.
Example 5.1 Consider the game in Figure 5.1 and note that the entire strategy set is IU.
L
R
U
2, 2
0, 0
M
0, 2
2, 0
D
1, 0
1, 2
Figure 5.1
Append to the game the type structure in Figure 5.2.
Here, there is one rational strategy type pair for Ann, viz. (U, ta ), and two rational strategy type
pairs for Bob, viz. L, tb and R, ub . Also, λa (ta ) assigns probability 1 to L, tb , a rational
strategy-type pair of Bob, and so believes R1b . Similarly, λb tb believes R1a . But λb ub assigns
probability 1 to (D, ta ), an irrational strategy-type pair of Ann, and so doesn’t believe that Ann is
rational.
λa(ta)
L
tb
1
ub
0
λb(tb)
ta
U
1
U
0
M
0
M
0
D
0
D
1
λb(ub)
ta
R
0
0
Figure 5.2
22
In sum:
projS a R1a × projS b R1b
= {U} × S b
S1a × S b = S a × S b ,
and, for all m ≥ 2,
a
b
× projS b Rm
projS a Rm
= {U} × {L}
a
Sm
× Sb = Sa × Sb.
In Example 5.1, the strategies M, D of Ann are inconsistent with rationality despite the fact
that they are undominated. The reason is that there are too few types of Ann. For instance, there
is no type of Ann that assigns probability 12 to each of L and R, the only belief under which D
could be a best response. This suggests that we expand the type set of Ann so that λa can map
Ann’s type set onto all possible beliefs on S b × T b . But even this is not enough. Not only do we
want M and D to be consistent with rationality, but we also want these strategies to be consistent
with rationality when Ann believes “Bob is rational and believes I am rational.” Ann must assign
(i) positive probability to R and (ii) probability 1 to the strategy-type pairs of Bob consistent with
rationality and belief of rationality, i.e., Rb2 . But, the strategy R is inconsistent with R2b . So, we
need to expand Bob’s type set too, so that there is a strategy-type pair so that R ∈ projSb R2b . And
so on. Proceeding along these lines, we mimic the rationale for looking at a complete type structure.
(Refer back to Section 1.2.)
Why the need for compactness and continuity in Proposition 5.1? The next example gives the
mathematical reason. Then, we turn to the conceptual reason suggested by this paper.
Example 5.2 Consider the example in Figure 5.3 below, where the entire strategy set is IU.
L
R
U
1, 1
0, 0
D
0, 0
0, 0
Figure 5.3
a
a
Append to the game a complete S b , S a -based type structure. In this structure, Rm
\Rm+1
= ∅ and
b
b
a
Rm \Rm+1 = ∅ for all m. We argue by induction that, for each m, there are types tm ∈ T a and
a
a
b
tbm ∈ T b with (D, tam ) ∈ Rm
\Rm+1
and R, tbm ∈ Rbm \Rm+1
. Begin with m = 0 and note that,
a a
a
a
by completeness, there is a type t0 ∈ T with λ (t0 ) {L} × T b = 1. So (D, ta0 ) ∈ R0a \Ra1 . And
23
b
similarly for R. Assume that the result holds for m, i.e., is a type such that R, tbm ∈ Rbm \Rm+1
.
a a
a
a
b
a
a
By completeness, there is a type tm+1 ∈ T with λ tm+1 R, tm = 1. Certainly D, tm+1 ∈ R1 .
b
Using the fact that tbm ∈ Rm
, it follows by induction that D, tam+1 ∈ Ram+1 . But D, tam+1 ∈
/
a
Rm+2
, since tbm ∈
/ Rbm+1 . And similarly for R.
a
may
The import of Example 5.2 is that, for a given game, in any complete structure the sets Rm
shrink forever. This is so, even though the IU strategies stop shrinking at some finite level M.
Specifically, in this example, at each level m, there is some type tam ∈ T a , such that (D, tam ) is
a
contained in Ram and not in Rm+1
. This fits with part (i) of Proposition 5.1, which says that, for
a
a
each m, there is some type tm ∈ T a with (D, tam ) ∈ Rm
.
Ta
Ta[3]
Ta[2]
Ta[1]
U
D
Sa
Figure 5.4
The question is whether there is some type ta in the complete structure such that (D, ta ) ∈
a
a
a
a
a
a
m Rm . Refer to Figure 5.4, where we write T [m] for the set of types t ∈ T with (D, t ) ∈ Rm .
a
This is a decreasing sequence of sets, each of which is non-empty. When each T [m] is closed
and S a × T a is compact, then they have a non-empty intersection—as desired. A continuous type
structure guarantees the former and a compact structure guarantees the latter.
But there is another route: Notice that D is a best response for Ann if she assigns positive
probability 1 to R. And similarly for the strategy R of Bob. So, if Ann believes that Bob plays
R and assigns probability 1 to Ann playing D, then playing D is rational and she believes Bob is
rational. And so on. More formally, let γ denote a Dirac measure and consider
a
γ 1 (R) , γ a2 R, γ b1 (D) , . . . ,
i.e., the associated hierarchy of beliefs for Ann where she assigns probability 1 to (1) R, to (2) R and
Bob assigning probability 1 to D, and so on. If there is a type, viz. ta , in the complete structure
that is associated with this hierarchy of beliefs, then (D, ta ) will be contained in m Ram . Such a
24
type will exist if the structure is terminal. Theorem 3.1(ii) says that when a complete structure is
compact and continuous, it is also terminal. This seems to be why the argument above required
that the structure be compact and continuous.
This second route suggests that the epistemic conditions for IU given in the literature may be
indirect. In particular, this second route makes use of the fact that the type structure is terminal
(guaranteed by completeness, continuity, and compactness) but does not directly reference the fact
the structure is complete. More generally:
Proposition 5.2 Fix a game and an associated terminal S b , S a -based type structure Λ.
a
b
(i) For each m, projS a Ram × projS b Rbm = Sm
× Sm
.
(ii) Moreover, projS a
∞
m=1
Ram × projS b
∞
b
m=1 Rm
a
b
= SM
× SM
.
For the idea of the proof, refer to Figure 5.5.
Ta
SaM
βa
Sa
Sa
SaM
SaM
Figure 5.5
a
b
Fix a game and the IU strategies SM
× SM
. We can construct a type structure Λ with the following
a
b
a
b
property: The types sets are copies of the IU strategies, i.e., T × T = SM
× SM
, and the diagonal
a
b
∞
∞
a
a
b
b
of SM × SM (resp. SM × SM ) is contained in m=1 Rm (resp.
(See, for instance,
m=1 Rm ).
Brandenburger-Friedenberg-Keisler [9, 2004].)
a
a
Now consider a terminal structure Λ. For every type t ∈ T , we can find some type in T a , resp.
b
a
a
b
β a t , with the same hierarchy of beliefs as t . And similarly for types t ∈ T . An inductive
a a ∞
a a a ∞
a
argument establishes that if s , t ∈ m=1 Rm , then s , β t
∈ m=1 Ram (see Lemma E4).
So, with this,
a
b
SM
× SM
⊆
∞
a
m=1 Rm
a
projS a ∞
m=1 Rm
= projS a
25
×
∞
b
m=1 Rm
b
projS b ∞
m=1 Rm .
× projSb
Moreover, a standard argument establishes that
projSa
as required.
6
∞
a
m=1 Rm
× projS b
∞
b
m=1 Rm
a
b
⊆ SM
× SM
,
Discussion
This section discusses some technical and conceptual aspects of the paper.
a. Large Type Structures: The literature has a number of notions of a “large” type structure.
We have established a relationship between two such type structures: Terminal type structures and
complete type structures. The former goes back to Böge-Eisele [4, 1979]. The latter is more recent,
due to Brandenburger [5, 2003].
There is another notion of a “large type structure,” namely type structures satisfying a universal
property. Here is roughly the idea.
Suppose the analyst constructs a belief structure that contains all hierarchies that are coherent,
that assign probability 1 to coherent hierarchies, etc. The first level of these hierarchies reflects
players’ beliefs about the underlying state of uncertainty. The second level reflects players’ beliefs
both about the underlying state of uncertainty and other players’ beliefs, and so on. Does this
reasoning ever stop? That is, does there exist an ordinal α so that the hierarchies of beliefs up to
level α determine all subsequent beliefs? A structure is said to satisfy the universal property if such
an ordinal exists.
Mertens-Zamir [23, 1985], Brandenburger-Dekel [7, 1993] and Heifetz [13, 1993] each address the
question of whether a universal structure exists. In particular, each of these papers constructs a
“canonical type structure” and shows that this type structure is universal. See point b. below and
Appendix D.
By definition, a universal structure is terminal. As such, Mertens-Zamir [23, 1985], BrandenburgerDekel [7, 1993] and Heifetz [13, 1993] each (also) construct a terminal structure. But a terminal
structure need not be universal. For instance, taken together, Heifetz-Samet [14, 1998]-[15, 1999],
give an example of a terminal structure that is not universal.
b. The Canonical Construction of a Large Type Structure: The constructions of a canonical type structure (e.g., as in Mertens-Zamir [23, 1985], Brandenburger-Dekel [7, 1993], Heifetz [13,
1993] etc.) are terminal, universal, and complete. To see this, refer to Appendix D. These papers
build a belief structure that contains all hierarchies of beliefs that are coherent, assign probability
1 to coherent beliefs, etc., i.e., akin to H a × H b here. They then show that all the beliefs up to
level ω 0 (the first infinite ordinal) determine the belief at ω 0 . With this, they take the type sets to
26
be the hierarchies associated with H a × H b and (using universality) show that this naturally gives
bijective continuous maps λa , λb . Thus, their constructions satisfy all three properties. Moreover, by construction, any two types induce distinct hierarchies of beliefs—that is, the structure is
non-redundant.
This raises the question: Is there a complete structure that is distinct from the canonical construction of a large type structure? Such a structure does indeed exist. Mertens-Sorin-Zamir [22,
page 193, Remark 2; 1993] gave the first such example. The following example is similar in spirit.4
Let X a = X b = {x}. Then, the canonical construction of a large type structure involves one
type for each player, each of which assigns probability 1 to x, to “x and the other player assigning
probability 1 to x,” etc. But there is another complete structure. In particular, take T a , T b to
be the Cantor space and note that there exists a continuous surjective map λa (resp. λb ) from T a
(resp. T b ) to M {x} × T b (resp. M ({x} × T a )). (See Kechris [16, Theorem 4.18; 1995].) This
gives a complete structure that is not canonical. In particular, all types in T a = {0, 1}N induce the
same hierarchy of beliefs, while the canonical construction is non-redundant.
This example might suggest that any complete structure that differs from the canonical structure
differs by redundancy—it may have too many types. But, Sections 1.3 and 3 suggest that this may
not be the case—that a complete type structure may not “contain” all hierarchies of beliefs in H a × H b .
That said, at this time, such an example is lacking.
c. Existence of a Non-Compact Terminal Structure: If X a , X b are compact, the canonical
structure is both compact and continuous. This raises the question: In this case, does there exist
a complete structure that lacks one of these properties? The answer is yes and can be found
in Brandenburger-Friedenberg-Keisler [9, Appendix G; 2004]: Take T a = T b to be copies of the
Baire space. Then there exists a continuous surjection from T a (resp. T b ) onto M X a × T b
b
(resp. M X × T a ). (See Kechris [16, Theorem 7.9; 1995].) This gives a non-compact complete
structure. We do not know if this structure is terminal.
On the other hand, suppose that X a is Polish but not compact. Theorem 3.1 says that a complete
structure is terminal if T a , T b are compact and λa , λb are continuous. But if λa : T a → M X a × T b
is continuous and T a is compact, then M X a × T b must also be compact—which cannot be. This
says that there is no complete, compact, and continuous type structure. However, this is not to
say that there is no complete and terminal structure—in fact, Brandenburger-Dekel [7, 1993] show
that one does exist. In their construction, λa , λb are continuous, but T a , T b are only Polish (not
compact).
d. Definition of a Terminal Structure: The terminal property considered here can be
thought of as requiring that there be a hierarchy morphism from every type structure Λ to a terminal
4 The example follows Proposition 5.2 in Brandenburger-Friedenberg-Keisler [8, 2004] and Theorem 10.4 in
Brandenburger-Keisler [10, 2006].
27
a
structure Λ: Say that there is a hierarchy morphism from Λ to Λ if there are maps β a : T → T a
a
a
b
and β b : T → T b with δ a · β a = δ and δ b · β b = δ .
a
Another idea found in the literature is that of a type morphism: Fix maps β a : T → T a and
b
a : X b × T a → X b × T a and β
b : X a × T b → X a × T b so that β
a = id × β a
β b : T → T b and construct β
b
b
a
= id × β b . Say β a , β b are type morphisms if λa · β a = β
· λa and λb · β b = β
· λb .
and β
If the type sets are Polish: Any type morphism is also a hierarchy morphism. But the converse
need not hold when the structure Λ contains two types that induce the same hierarchy of beliefs. If
the type structure Λ is non-redundant, any hierarchy morphism is also a type morphism.5
Conceptually, hierarchy morphisms are the objects of interest and, often, type morphisms are
used because they only implicitly make reference to the hierarchies of beliefs. Given that the former
are of interest, it is used in the definition of terminality.
e. Definition of a Type Structure: In the definition of a type structure, we took T a , T b to be
metrizable. The reason we did so is to insure that types induce hierarchies of beliefs. Specifically,
this condition guarantees that the maps ρam ,ρbm are measurable: If f : Ω → Φ where Ω is metrizable,
Φ is Polish, and f is measurable, then f : M (Ω) → M (Φ) is also measurable. Of course, this
conclusion may obtain even when Ω is not metrizable, i.e., for some particular σ-algebra (as in
Heifetz-Samet [14, 1998]).
f. Epistemic Conditions for IU: Proposition 5.1 states that the strategies consistent with
RmBR in a complete Polish structure are the strategies that survive (m + 1) rounds of elimination
of dominated strategies. Why require T a , T b to be Polish? The reason is that this is sufficient to
a
b
ensure that the sets Rm
, Rm
are Borel. Actually, a weaker condition will do—namely, if T a , T b are
a
b
separable metrizable, then the sets Rm
, Rm
are Borel.6
Why, then, doesn’t Proposition 5.2 require that T a , T b be separable metrizable? The reason
is that, for the specific case when the type structure is terminal, we can independently show that
the sets Ram , Rbm are measurable. Since the set of hierarchies is Polish, the sets associated with
a
b
, Rm
in hierarchy space will be Polish. Then use the measurability of the maps ρam , ρbm to get
Rm
a
b
that Rm
, Rm
must be measurable. (See Lemma E3.)
Appendix A
Results for Section 2
Lemma A1 For each m:
d
c
(i) Zm+1
= Z1c × m
n=1 M Zn ;
5 This
result is shown in joint work with Martin Meier.
steps are necessary: First, the set Rc1 is Borel. This can be shown using the Portmanteau Theorem—to show
a stronger statement—that Rc1 is closed. This is one reason to require metrizability. The other reason is to show
that the set of measures that assign probability 1 to R1d is Borel. The standard argument uses the metrizability and
separability of S d × T d .
6 Two
28
(ii) ρcm+1 xc , td = xc , δ d1 td , . . . , δ dm td .
Proof. For m = 1 both parts are immediate. Assume that the result holds for m ≥ 2. Then
c
Zm+2
Also,
d c
= Zm+1
× M Zm+1
d d
= Z1c × m
n=1 M Zn × M Zm+1 .
ρcm+2 xc , td =
=
establishing the result.
ρcm+1 xc , td , δ dm+1 td
xc , δ d1 td , . . . , δ dm td , δ dm+1 td ,
Lemma A2 Let (Ω, T (Ω)) be metrizable and (Φ, T (Φ)) be Polish. If f : Ω → Φ is measurable
(resp. continuous) then f : M (Ω) → M (Φ) is measurable (resp. continuous).
Proof. The case where f is continuous is found in Aliprantis-Border [2, 1999; Theorem 14.14]. We
treat the case where f is measurable.
First note that, since (Φ, T (Φ)) is Polish, B (M (Φ)) is generated by all the sets of the form
{ν ∈ M (Φ) : ν (E) ∈ K} ,
ranging over E Borel in Φ and K measurable in [0, 1] (see Kechris [16, 1995; Theorem 17.24]). It
suffices to show that each set
f −1 ({ν ∈ M (Φ) : ν (E) ∈ K})
is in B (M (Φ)). If so, then f is measurable. (See Aliprantis-Border [2, 1999; Corollary 4.23].)
Note that
µ ∈ M (Ω) : f (µ) (E) ∈ K
=
µ ∈ M (Ω) : µ f −1 (E) ∈ K .
f −1 ({ν ∈ M (Φ) : ν (E) ∈ K}) =
Since f is measurable, f −1 (E) is Borel in Ω. Now, by Lemma A2 in Aliprantis-Border [2, 1999]
f −1 ({ν ∈ M (Φ) : ν (E) ∈ K})
is in B (M (Φ)).
Lemma A3 Let I be (at most) a countable collection of integers 1, 2, . . .. Let (Ω, T (Ω)) be a
topological space and, for each i ∈ I, let (Φi , T (Φi )) be a Polish space. Fix measurable maps
29
fi : Ω → Φi and define f : Ω →
Φi so that
i
f (ω) = (f1 (ω) , f2 (ω) , . . .)
for all ω ∈ Ω. Then f is measurable.
Proof. Let E be a Borel set on
so that f −1 (E) is measurable.
i
Φi . Then f −1 (E) =
−1
i fi (E).
Each fi−1 (E) is measurable,
Lemma A4 The maps ρcm and δ cm are measurable. Moreover, if the structure is continuous, the
maps ρcm and δ cm are continuous.
Proof. First note that ρc1 = projX c , and so is certainly continuous. So, by Lemma A2, ρc1
is continuous. Also using this Lemma, we have that δ c1 is measurable (and continuous if λc is
continuous).
Now assume that the result holds for m. By the induction hypothesis and Lemma A3, ρcm+1 is
measurable. (When ρcm , δ dm are continuous, Theorem 8.5 in Chapter 2 of Munkres [24, 1975] says
that ρcm+1 is continuous.) Now, by Lemma A2, ρcm+1 is measurable (resp. continuous when ρcm+1
is continuos). So, each δ cm+1 is measurable (and continuous when λc is continuous).
Lemma A5 The map δ c is measurable. Moreover, δ c is continuous if the structure is continuous.
Proof. Measurability is from Lemmas A3 and A4. Continuity is from Theorem 8.5 in Chapter 2
of Munkres [24, 1975] and Lemma A4.
Appendix B
Proofs for Section 3
Lemma B1 Let Ω1 , Ω2 , . . . and Φ1 , Φ2 , . . . be Polish space where Φm is measurable in
Then
{(µ1 , µ2 , . . .) ∈ ∞
m=1 M (Φm ) : for all m and all n m, margΦn µm = µn }
∞
is closed in m=1 M (Φm ).
n· m Ωn .
Proof. Fix a sequence µk1 , µk2 , . . . in ∞
M (Φm ) with margΦn µkm = µkn for all m and all n m.
k k
m=1
Fix also (µ1 , µ2 , . . .) with µ1 , µ2 , . . . → (µ1 , µ2 , . . .). We’d like to show that margΦn µm = µn for
all m and all n m. To show this, fix some m and note that it suffices to show margΦn µkm →
margΦn µm whenever n m. If so, then µkn → margΦn µm whenever n m. Since each µkn → µn ,
it follows that µn = margΦn µm for all n m, as required.
Now, we turn to show that margΦn µkm → margΦn µm . By the Portmanteau Theorem, it suffices
to show that
lim inf margΦn µkn (U ) ≥ margΦn µm (U ) ,
30
for any open set U in Φn . Fix such an open set and note that U × m
l=n+1 Ωl ∩ Φm is open in
the relative topology. Using the fact that µkm → µm and the Portmanteau Theorem,
lim inf µkm
Since
U× m
U× m
l=n+1 Ωl ∩ Φm ≥ µm
l=n+1 Ωl ∩ Φm .
margΦn µkm (U ) = µkm
for each k, and
margΦn µm (U ) = µm
we have established the result.
U× m
l=n+1 Ωl ∩ Φm
m
U × l=n+1 Ωl ∩ Φm ,
Lemma B2 For each c and each m,
c
(i) Xm
, Ymc are Polish;
c
is a closed subset of Ymc and so Polish.
(ii) Hm
Proof. By induction on m. For m = 1, the result is immediate. Assume that the result holds for
c
c
m. Then it is immediate that Xm+1
and Y c
are Polish. By the induction hypothesis, Hm
is a
c m+1
c
c
c
c
closed subset of Ym , so that Hm × M Xm+1 is a closed subset of Ym+1 . As such, to show Hm+1
c
is a closed subset of Ym+1
, it suffices to show that
c
: margXm
µ1 , . . . , µm+1 ∈ Ym+1
c µm+1 = µm
is closed. This follows from Lemma B1.
Remark B1 The proof of Lemma B2 can readily be amended to cover the following case. Suppose
c
c
X a , X b are compact metrizable. Then, for each m, Xm
, Ymc , and Hm
are compact metrizable.
Lemma B3 The set H c is a closed subset of a Polish space and so Polish. If X a , X b are compact
metrizable, then H c is a closed subset of a compact metrizable space and so compact metrizable.
c
Proof. First, note that H c is a subset of m M (Xm
), a Polish space. Lemma B1 says that H c is
c
c
closed in m M (Xm
). If X a , X b are compact metrizable, then m M (Xm
) is compact metrizable,
c
establishing that H is compact metrizable.
Lemma B4 For each m ≥ 1,
(i) ηcm (µ1 , . . . , µm ) = (µ1 , ζ c2 (µ2 ) , . . . , ζ cm (µm ));
(ii) ζ cm+1 xc1 , µd1 , . . . , µdm−1 , µdm = (ζ cm xc1 , µd1 , . . . , µdm , ζ dm µdm ).
31
Proof. By induction on m.
m = 1: Since ηc1 is the identity map, part (i) is immediate. Now note that that since ζ d1 is
the identity map, so is ζ d1 , and so ζ d1 = η d1 . This establishes that ζ c2 xc1 , µd1 = (ζ c1 (xc1 ) , ζ d1 µd1 )
establishing part (ii).
m ≥ 2: Assume that the result holds for m. Then part (i) follows immediately from the
induction hypothesis. Now
ζ cm+2 xc1 , µd1 , . . . , µdm , µdm+1 = ζ c1 (xc1 ) , ηdm+1 µd1 , . . . , µdm , µdm+1
=
ζ c1 (xc1 ) , η dm µd1 , . . . , µdm , ζ dm+1 µdm+1
=
ζ cm+1 xc1 , µd1 , . . . , µdm , ζ dm+1 µdm+1 ,
where the second line follows from part (i) established for (m + 1). This establishes part (ii).
Lemma B5 Let I = {1, 2, . . .} be (at most) a countable set of integers. For each i, let fi : Ωi → Φi
be an embedding. Define f : i∈I Ωi → i∈I Φi so that f (ω 1 , ω2 , . . .) = (f1 (ω 1 ) , f2 (ω 2 ) . . .).
Then f is also an embedding.
Proof. Let gi : Ωi → fi (Ωi ) be such that gi (ω i ) = fi (ω i ). We have that each gi is a homeomor
phism. Define f as in the statement of the Lemma and g : i Ωi → f ( i Φi ) so that g (ω) = f (ω)
for all ω ∈ i Ωi . We will show that g is also a homeomorphism. The injectivity of g is immediate
from the injectivity of each of the maps gi . The surjectivity of g is immediate.
To show that g is continuous, fix closed sets Ci in Φi where Ci = Φi for all but finitely many i.
These sets form a basis for i∈I Φi in the product topology, so it suffices to show that g−1
i∈I Ci
is closed. Let J be the set of i with Ci = Φi . Then
g −1
i∈I
Ci
=
=
i∈J
gi−1 (Ci ) ×
i∈J
gi−1 (Ci ) ×
i∈I\J
i∈I\J
gi−1 (Ci )
Ωi .
Since each gi is continuous, then each gi−1 (Ci ) is closed. It follows that g −1
i∈I Ci is indeed
closed, as required.
To show that g is closed, fix closed sets Fi in Ωi , where Fi = Ωi for all but finitely many i.
Again, these sets form a basis for i∈I Ωi in the product topology, so that it suffices to show that
g
i∈I Fi is closed. Let J be the subset of i with Fi = Ωi . Then
g
i∈I
Fi
=
=
i∈J
gi (Fi ) ×
i∈J
gi (Fi ) ×
i∈I\J
gi (Fi )
i∈I\J
Ωi ,
where the last line follows from the fact that each gi is surjective. Since each gi (Fi ) is closed, it
follows that g
i∈I Fi is closed.
32
Proof of Lemma 3.1. By induction on m. For m = 1, the result is immediate since ζ c1 and
ηc1 are the identity maps. Assume that the result holds for m. We will show that it also holds
for m + 1. By the induction hypothesis, ζ c1 and ηdm are embeddings. Since ζ cm+1 is the product
of ζ c1 and ηdm , Lemma B5 in Appendix B gives that ζ cm+1 is an embedding. Similarly, ηcm+1 is the
product of ηcm and ζ cm+1 . The former is an embedding by the induction hypothesis. Moreover, the
induction hypothesis gives that ζ cm+1 is an embedding so that ζ cm+1 is an embedding (Kechris [16,
1995; Exercise 17.28]). Again, by Lemma B5, η cm+1 is an embedding.
Proof of Lemma 3.2. First, note that ηc is injective since each of the maps ηcm is injective:
Fixing (µ1 , µ2 , . . .) = (̟1 , ̟2 , . . .) in H c , we can find some initial segment of these sequences that
are distinct, i.e., some m with (µ1 , . . . , µm ) = (̟1 , . . . , ̟m ). Then by Lemma 3.1, ηcm (µ1 , . . . , µm ) =
ηcm (̟1 , . . . , ̟m ) so that ηc (µ1 , µ2 , . . .) = η c (̟ 1 , ̟2 , . . .).
c
Now turn to showing the continuity of ηc . For each m, fix closed sets Cm in M (Zm
) with
c
Cm = M (Zm ) for all but finitely many m. Since these sets form a basis for the product topology,
it suffices to show that (η c )−1 ( m Cm ) is closed. Note, there is some M such that
c −1 M
c
c
( m=1 Cn ) × ∞
(ηc )−1 ( ∞
m=1 Cm ) = [(η M )
m=M+1 M (Zm )] ∩ H .
c
By Lemma 3.1 (ηcm )−1 ( M
It follows that
n=1 Cn ) is closed and, by Lemma B3, H is closed.
∞
c −1
(η ) ( m=1 Cm ) is closed.
c
Next, we show that ηc is a closed map. For this, fix closed sets Fm in M (Xm
) with Fm =
c
M (Xm ) for all but finitely many m. Since these sets form the basis for the product topology for
∞
∞
c
c
c
m=1 M (Xm ), it suffices to show that η (H ∩ ( m=1 Fm )) is closed. Note that there exists some
M with
M
∞
c
c
c
c
ηc (H c ∩ ∞
m=1 Fm ) = [η M (HM ∩
m=1 Fm ) ×
m=M+1 M (Xm )] ∩ H .
c
By Lemma B2, HM
is closed. So, again using Lemmas 3.1 and B3, ηc (H c ∩ ∞
m=1 Fm ) must be
closed, as required.
Appendix C
Results for Section 4
In this section, we prove Theorem 4.1. Fix a type structure Λ and define maps ρa , ρb as in the text,
i.e., for each player c, let
d
ρc : X c × T d → Z1c × ∞
m=1 M Zm
be a map with ρc xc , td = xc , δ d td . Note, if Λ is non-redundant, then ρc is injective. Also
note that ρc is measurable since it is the product of two measurable maps.
Lemma C1 Let Ω be a Borel subset of a Polish space and let Φ be Polish. If the map f : Ω → Φ
is injective and measurable, then the map f : M (Ω) → M (Φ) is also injective.
33
Proof. Suppose f (µ) = f (ν). Fix some event E in B (Ω). Then, by Purves’ Theorem [20, 1966],
f (E) is an event in Φ. So,
µ (E) = µ f −1 (f (E)) = ν f −1 (f (E)) = ν (E) ,
where the first and third equalities follow from the fact that f is injective and the second from the
fact that f (µ) = f (ν). As such, µ = ν.
Corollary C1 Fix a non-redundant and Borel structure Λ. The maps ρa and ρb are injective.
Another fact about the map ρc :
Lemma C2 For any event Em in B Z1c ×
(ρc )−1 (Em ×
Proof. We have that
(ρc )−1 (Em ×
∞
n=1
n=m+1 M
d
n=m+1 M Zn ) =
∞
m−1
M Znd ,
d
Zn ) = (ρcm )−1 (Em ) .
c d
x , t : xc , δ d1 td , . . . , δ dm−1 td ∈ Em
= (ρcm )−1 (Em ) .
c
Now define a map ϕc : M X c × T d → m M (Zm
) so that
ϕc (µ) = (ρc1 (µ) , ρc2 (µ) , . . .).
We now state a relationship between the maps ρc and ϕc . As usual, say ρc (µ) extends ϕc (µ) =
d
c
c
(ρc1 (µ) , ρc2 (µ) , . . .) to Z1c × ∞
m=1 M Zm if (i) ρ1 (µ) = margZ1c ρ (µ) and (ii) for each m ≥ 1,
ρcm+1 (µ) = margZ c ×m M(Z d ) ρc (µ).
1
n=1
n
d
Lemma C3 For each µ ∈ M X c × T d , ρc (µ) extends ϕc (µ) to Z1c × ∞
m=1 M Zm .
Proof. Fix some µ ∈ M X c × T d . First, note that, for any event E1 in Z1c ,
−1
ρc1 (µ) (E1 ) = µ (ρc1 ) (E1 )
= µ E1 × T d
= µ (ρc )−1 (E1 )
= margZ1c ρc (µ) (E1 ) ,
34
establishing that ρc1 (µ) = margZ1c ρc (µ).
d
m
n=1 M Zn . Then
Similarly, fix some m ≥ 1 and some Em+1 in Z1c ×
c −1
ρm+1
(Em+1 )
= µ (ρc )−1 (Em+1 )
= margZ c ×m M(Z d ) ρc (µ) (Em+1 ) ,
ρcm+1 (µ) (Em+1 ) = µ
1
n=1
n
where the second line follows from Lemma C2.
Lemma C4 Given a non-redundant Borel structure, the map ϕc is injective.
Proof. Fix a non-redundant structure Λ and some µ, ν ∈ M (X c × T c ) with µ = ν. By Corollary
d
C1, ρc (µ) = ρc (ν). Now, note that each Z1c and M Zm
is Polish. So, applying Lemma C3 and
c
c
Kolmogorov’s Extension Theorem, ϕ (µ) = ϕ (ν). (If they were equal, then the fact that ρc (µ) is
not the same as ρc (ν) would contradict that there is a unique extension.)
Proof of Theorem 4.1. Fix a non-redundant, Borel, terminal X a , X b -based structure
Λ = X a , X b ; (T a , T (T a )) , T b , T T b ; λa , λb .
Suppose, contra hypothesis, that the structure is not complete. That is, there is a measure µ ∈
M X a × T b where λa (ta ) = µ for all types ta ∈ T a . By injectivity of ϕa (Lemma C4), for all
types ta ∈ T a , we then have that
δ a (ta ) = ϕa (λa (ta )) = ϕa (µ) .
We will now show that there is an X a , X b -based type structure
a
a
Λ = X a, X b; T , T
T
a
b
b
a
, (T , T (T )); λ , λ
b
and a type ta∗ with δ (ta∗ ) = ϕa (µ), contradicting that Λ is terminal. (Of course, an analogous
argument holds for b.)
a
a
Take T = T a ∪ {ta∗ } with T a ∩ {ta∗ } = ∅. Let T T
so that it is generated by T (T a ) ∪ {ta∗ }
a
(and so both T (T a ) and {ta∗ } can be viewed as the topology induced by T T ). It follows that
a
T T
is separable metrizable. (Separability is immediate, metrizability follows from Engelking
b
a
b
[11, Proposition 2.24 and Theorem 4.2.1; 1977].) Take (T , T (T )) = (T b , T (T b )). Choose λ (ta ) =
a
b
λa (ta ) for all ta ∈ T a and λ (ta∗ ) = µ. Then, for any event E a in M(X a × T ),
a −1
λ
a
(E ) =
(λa )−1 (E a )
(λa )−1 (E a ) ∪ {ta∗ }
35
if µ ∈
/ Ea
if µ ∈ E a .
a
Since λa is measurable, λ is also measurable. Take
λ
b b t
b
F b = λb tb F b ∩ X b × T b
a
for all F b in X b × T .
b
It is immediate that λ is a probability measure and also that λ is measurable. (The latter follows
a
from Kechris [16, Exercise 17.28] and the fact that T is separable metrizable.) It is immediate
a
b that, for any type ta ∈ T a and tb ∈ T b , δ (ta ) = δ a (ta ) and δ tb = δ b tb . From this, it is
immediate that
a
δ (ta∗ ) = ϕa (µ) = ϕa (µ) ,
as required.
Appendix D
The Canonical Type Structure
This appendix provides a canonical construction of an X a , X b -based type structure from the
a b
X , X -based belief structure H a × H b . This is very similar to the constructions in MertensZamir [23, 1985] and Brandenburger-Dekel [7, 1993]. The interest in doing this here is that we’d
like to know that, up to an embedding, a terminal structure contains all hierarchies of beliefs available
in the belief structure. Specifically:
Proposition D1 Fix a terminal X a , X b -based type structure
Λ = X a , X b ; (T a , T (T a )), (T b , T (T b )); λa , λb .
If (µc1 , µc2 , . . .) ∈ H c , then there is a type tc ∈ T c with δ c (tc ) = ηc (µc1 , µc2 , . . .).
c
a
Begin by defining type sets so that T c = ηc (H c ) ⊆ ∞
m=1 M (Zm ) and T (T ) is the topology
∞
c
relative to m=1 M (Zm
). Since ηc is an embedding, ηc (H c ) is a closed subset of a Polish space
and so Polish. By the Kolmogorov Extension Theorem, there is an injective map
d c
λ̂ : T c → M Z1c × ∞
m=1 M Zm
c
such that λ̂ (tc ) extends tc to Z1c ×
∞
m=1 M
d
Zm , i.e., if tc = (µc1 , µc2 , . . .) then
c
margZ1c λ̂ (tc ) = µc1
and
margZ c ×m
1
c
n=1
c
d)
M(Zn
Lemma D1 The map λ̂ is closed and continuous.
36
λ̂ (tc ) = µcm+1 .
c
Proof. First, we show that λ̂ is a closed map. To see this, it will be convenient to introduce
c . By the Kolmogorov Extension Theorem, we can choose
another map λ
c : ∞ M (Z c ) → M Z c × ∞ M Z d
λ
m
1
m
m=1
m=1
c (µ , µ , . . .) extends (µ , µ , . . .) to M Z c × ∞ M Z d . In
such that it is injective and λ
1
2
1
2
1
m
m=1
c
c restricted to ηc (H c ). So, to show that λ̂c
fact, since the extension is uniquely determined, λ̂ is λ
c maps closed sets to closed sets.
maps closed sets to closed sets, it suffices to show that λ
∞
c
Fix a closed set C ⊆ m=1 M (Zm
). We will show that that, given a sequence of measures
c
n
n
c (C). Since λ
c is injective, each
ν contained in λ (C) with ν → ν, ν is also contained in λ
c )−1 (λ
c (C)) = C. So, it suffices to show that (λ
c )−1 (ν n ) → (λ
c )−1 (ν). (Note that this is well(λ
c
c
)−1 (ν n ) , (λ
)−1 (ν) is a singleton.) If so, since C is closed (λ
c )−1 (ν) ∈ C
defined since each of (λ
c (C) as required.
and so ν ∈ λ
c )−1 (ν n ) = (µn1 , µn2 , . . .) and (λ
c )−1 (ν) = (µ1 , µ2 , . . .). Fix an set Um where (i) if m = 1
Write (λ
m−1
M Zkd . Then
then Um is open in Z1c and (ii) if m ≥ 2 then Um is open in Z1c × k=1
d µnm (Um ) = ν n Um × ∞
k=m M Zk
d µm (Um ) = ν Um × ∞
.
k=m M Zk
(D1)
(D2)
Since ν n → ν, by the Portmanteau Theorem,
d d lim inf ν n Um × ∞
≥ ν Um × ∞
.
k=m M Zk
k=m M Zk
Now, using Equations D1-D2,
lim inf µnm (Um ) ≥ ν (Um ) .
Since this must hold for an arbitrary open set Um , it follows that µm → µ for all m, so that
(µn1 , µn2 , . . .) → (µ1 , µ2 , . . .) as required.
d c −1
Fix a closed set D in M Z1c × ∞
We will show that λ̂
(D) is closed.
m=1 M Zm .
c −1
To show this, it suffices to show that if (µn1 , µn2 , . . .) is a sequence contained in λ̂
(D) with
−1
c
c
c
c
(µn1 , µn2 , . . .) → (µ1 , µ2 , . . .) then (µ1 , µ2 , . . .) ∈ λ̂
(D). Note that, since λ̂ is closed, λ̂ ((λ̂ )−1 (D))
c
c
must be closed. So, λ̂ (µn1 , µn2 , . . .) → λ̂ (µ1 , µ2 , . . .), as required.
d with
There is also an embedding θ c : M Z1c × T d → M Z1c × ∞
m=1 M Zm
d θc M Z1c × T d = ν ∈ M Z1c × ∞
: ν Z1c × T d = 1 .
m=1 M Zm
See Kechris [16, 1995; Exercise 17.28].
c d Lemma D2 If µc1 , . . . , µcm+1 ∈ ηcm+1 Hm+1
then µcm+1 X c × ηdm Hm
= 1.
37
Proof. Fix some ηcm+1 ν c1 , . . . , ν cm+1 = µc1 , . . . , µcm+1 . Then
d −1 c
d µcm+1 X c × ηdm Hm
= µcm+1 ζ cm+1 ζ cm+1
X × ηdm Hm
−1 c
d = ν cm+1 ζ cm+1
X × η dm Hm
d
= ν cm+1 X c × Hm
= 1,
as required.
Lemma D3 For each m,
c
) = projm
ηcm (Hm
k=1
Proof. It is immediate that
projm
k=1
M(Zkc )
M(Zkc )
ηc (H c ) .
c
ηc (H c ) ⊆ ηcm (Hm
).
To show the opposite inclusion, fix some ηcm (ν c1 , . . . , ν cm ) = (µc1 , . . . , µcm ). It suffices to show that
c c c
there is some pair of measures ν cm+1 ∈ M Xm+1
and µcm+1 ∈ M Zm+1
such that margXm
c ν m+1 =
c
c
c
c
ν m and ζ m+1 ν m+1 = µm+1 . If so, a standard induction argument establishes that there is some
ηc ν c1 , . . . , ν cm , ν cm+1 , . . . = µc1 , . . . , µcm , µcm+1 , . . . .
c c
c
(This can be
To see this, first find a measure ν cm+1 ∈ M Xm+1
with margXm
c ν m+1 = ν m .
c
c
done: Consider the projection map from Xm+1 to Xm . Then, by Lemma 3.8, there is a measure
ν cm+1 such that ν cm+1 is the image measure of ν cm under the projection map. It follows that
c
c
c
c
margXm
c ν m+1 = ν m .) Finally, note that we can simply take µm+1 to be the image measure of ν m+1
under ζ cm+1 . This establishes the result.
c
Lemma D4 λ̂ (T c ) ⊆ θc M Z1c × T d .
c
Proof. Fix tc = (µc1 , µc2 , . . .) ∈ ηc (H c ). It suffices to show that λ̂ (tc ) Z1c × ηd H d = 1. To see
c
this, note that λ̂ (tc ) extends tc = (µc1 , µc2 , . . .). So, for all m,
µcm+1 (Z1c × projm
k=1
M(Zkd )
d
η d (H d )) = µcm+1 (Z1c × ηdm (Hm
))
= 1,
c
where the first equality follows from Lemma D3 and the second from Lemma D2. So, since λ̂ (tc )
c
extends tc = (µc1 , µc2 , . . .), it follows that λ̂ (tc ) Z1c × ηd H d = 1, as required.
Proof of Proposition D1.
Let
Λ = X a , X b ; (T a , T (T a )), (T b , T (T b )); λa , λb
38
c
λ̂ (tc ) . To
c
see that λc is well defined first notice
that λ̂ (tc ) is a single point contained in θc M Z1c × T d
c
c
−1
−1
λ̂ (tc ) is non-empty. Since θc is injective, (θc )
λ̂ (tc ) must be a
(Lemma D4). So, (θc )
single point.
It remains to show that λc is measurable. In fact, we will show that it is continuous. To see
this, fix a closed set C in X c × T d . Then
be such that (T a , T (T a )) and (T b , T (T b )) are given as above. Let λc (tc ) = (θ c )
(λc )
−1
(C) =
=
=
tc ∈ T c : (θc )
−1
c
−1
c
λ̂ (tc ) ∈ C
tc ∈ T c : λ̂ (tc ) ∈ θc (C)
c −1
λ̂
(θc (C)) .
c
Since θc is an embedding, θc (C) is closed. Now, use the fact that λ̂ is continuous (Lemma D1), to
get that (λc )−1 (C) is closed, as required.
Appendix E
Proofs for Section 5
The first result characterizes rationality (Definition 5.3):
Lemma E1 The strategy-type pair (sc , tc ) ∈ S c × T c is rational if and only if it is optimal under
δ c1 (tc ).
Proof. This follows from the fact that δ c1 (tc ) = margS d λc (tc ): For any event E in S d ,
δ c1 (tc ) (E) = λc (tc ) ((ρc1 )−1 (E)) = λc (tc ) E × T d ,
as required.
The next result is standard and so the proof is omitted.
Friedenberg-Keisler [9, 2004].)
(See, for instance, Brandenburger-
Lemma E2 Fix a Polish space Ω and some E is closed in Ω. Then
{µ ∈ M (Ω) : µ (E) = 1}
is closed in M (Ω).
Lemma E3 Fix a terminal S b , S a -based type structure structure Λ. For each m ≥ 1:
(i) if (sc , tc ) ∈ Rcm and (δ c1 (tc ) , . . . , δ cm (tc )) = (δ c1 (uc ) , . . . , δ cm (uc )) for some type uc , then
(sc , uc ) ∈ Rcm ;
−1 d
(ii) Rcm = ρdm+1
ρm+1 (Rcm ) ;
39
(iii) the sets ρcm+1 Rdm are closed;
(iv) the sets Rcm are Borel.
Proof. By induction on m.
m = 1: Part (i) follows immediately from Lemma E1. Turn to part (ii). It is immediate that
−1 d c −1 d c c
R1 ⊆ ρd2
ρ2 (R1 ) . Fix (sc , tc ) ∈ ρd2
ρ2 (R1 ) . Then there is some (sc , uc ) ∈ Rc1 with
c c
c
d
c c
d
c c
c
ρ2 (s , t ) = ρ2 (s , u ). Then δ 1 (t ) = δ 1 (u ) so that, by part (i), (sc , tc ) ∈ R1c , as required. For
part (iii) notice that
ρc2 R1d = sd ∈S d F sd ,
where
F sd = {µd1 ∈ M (S c ) : sd is optimal under µd1 }.
This follows from Proposition D1, which uses the fact that the structure Λ is terminal. Now,
using the proof of Lemma A3 in Brandenburger-Friedenberg-Keisler [9, 2004], we have that F sd
is closed, so that ρc2 R1d is the finite union of closed sets and so closed. This establishes Part (iii).
−1 d c For part (iv), use the fact that ρd2 (R1c ) is measurable (part (iii)), R1c = ρd2
ρ2 (R1 ) (part (i)),
d
c
and ρ2 is measurable (Lemma A4), to get that R1 is measurable.
c
m ≥ 2: Begin with part (i). Fix some (sc , tc ) ∈ Rm+1
and some type uc with
c c
δ 1 (t ) , . . . , δ cm+1 (tc ) = δ c1 (uc ) , . . . , δ cm+1 (uc ) .
c
By Part (i) of the induction hypothesis, (sc , uc ) ∈ Rm
. As such, we only need show that uc believes
d
Rm
.
−1 c
d d
By part (ii) of the induction hypothesis, Rm
= ρcm+1
ρm+1 Rm
. Moreover, by part (iv)
d
is Borel. So,
of the induction hypothesis, Rm
−1 c
λc (uc ) Rdm = λc (uc ) ρcm+1
ρm+1 Rdm
−1 c
d = λc (tc ) ρcm+1
ρm+1 Rm
= λc (tc ) Rdm
= 1.
d
With this, uc ∈ B c Rm
, so that (sc , uc ) ∈ Rcm+1 .
−1 d
c c
Next, turn to part (ii): Certainly Rm+1
⊆ ρdm+2
ρm+2 Rm+1
. Fix some (sc , tc ) ∈
d −1 d
c c c
c
c
ρm+2
ρm+2 Rm+1 . Then there exists some (s , u ) ∈ Rm+1 with ρm+2 (sc , tc ) = ρcm+2 (sc , uc ).
It follows that
c c
δ 1 (t ) , . . . , δ cm+1 (tc ) = δ c1 (uc ) , . . . , δ cm+1 (uc ) ,
c
so that, by part (i) applied to m + 1, (sc , tc ) ∈ Rm+1
.
40
Now part (iii): To show this, it will suffice to show that
d d
d d
ρcm+2 Rm+1
= ρcm+1 Rm
× {µdm+1 ∈ M Zm+1
: µm+1 believes ρdm+1 (Rcm )}
c
d
d
∩ sd , µd1 , . . . , µdm+1 ∈ Zm+2
: margZm
.
d µm+1 = µm
(E1)
d must be closed: By the induction hypothesis and Lemma E2,
If so, then ρcm+2 Rm+1
d d
c
ρcm+1 Rdm × {µdm+1 ∈ M Zm+1
: µm+1 believes ρdm+1 (Rm
)}
is closed. By Lemma B1,
d d
c
d
d
s , µ1 , . . . , µdm+1 ∈ Zm+2
: margZm
d µm+1 = µm
is closed. It follows that ρcm+2 Rdm+1 is the intersection of two closed sets and so closed.
We now turn to showing Equation E1. Begin by showing that the left-hand side is contained in
d
d
. Then sd , td ∈ Rm
,
the right-hand side. For this, fix a point ρcm+2 sd , td where sd , td ∈ Rm+1
c
d d
c
d
d
c
so that ρm+1 s , t ∈ ρm+1 Rm . Also, using the induction hypothesis, ρm+1 (Rm ) is closed. So,
since td believes Rcm ,
c
λd td (ρdm+1 )−1 (ρdm+1 (Rcm )) = λd td (Rm
) = 1,
so that δ dm+1 td believes ρdm+1 (Rcm ). (This uses part (ii) of the induction hypothesis.) The fact
d
d
that margZm
= δ dm td follows from Lemma 3.4.
d δ m+1 t
Now we establish that the right-hand side is contained in the left-hand side. To show this,
c
d
so that (a) there is a type td ∈ T d with sd , td ∈ Rm
and
fix some sd , µd1 , . . . , µdm+1 ∈ Zm+2
c
d d
d
d
d
d
d
c
d
d
ρm+1 s , t = s , µ1 , . . . , µm , (b) µm+1 believes ρm+1 (Rm ), and (c) margZm
As
d µm+1 = µm .
d
d
an implication of (a), (c), and Lemma 3.4, we have that margZnd µm+1 = µn for all n m.
d
Notice that there is some ν d1 , . . . , ν dm+1 ∈ Hm+1
so that ηdm+1 ν d1 , . . . , ν dm+1 = µd1 , . . . , µdm+1 .
To see this, begin by recalling that
η dm ν d1 , . . . , ν dm+1 = (ζ d1 (ν 1 ) , . . . , ζ dm+1 (ν m+1 )).
d d (This is Lemma B4.) Given ζ dm+1 and µdm+1 ∈ M Zm+1
, there exists a measure ν dm+1 ∈ M Xm+1
so that µdm+1 is the image of ν dm+1 under ζ dm+1 (see Lemma 3.5). Fix ν dm+1 and take ν dn =
margXnd ν dm+1 for all n m. It suffices to show that µdn is the image of ν dn under ζ dn for each n m.
41
For this
m+1
µdn (En ) = µdm+1 En × k=n+1 M Zkd
−1
m+1
= ν dm+1
ζ dm+1
En × k=n+1 M Zkd
d
(sc , µc1 , . . . , µcm ) ∈ Xm+1
: ζ dm+1 (sc , µc1 , . . . , µcm ) ∈ En ×
d
= ν dm+1 (sc , µc1 , . . . , µcm ) ∈ Xm+1
: ζ dn sc , µc1 , . . . , µcn−1 ∈ En
−1
= ν dn
ζ dn
(En ) ,
= ν dm+1
m+1
k=n+1 M
Zkd
where the first (resp. last) line follows from the fact that µdn (resp. ν dn ) is the marginal of µdm+1
(resp. ν dm+1 ) on Znd (resp. Xnd ), the second line follows from the fact that ν dm+1 is the image measure
of µdm+1 under ζ dm+1 , and the fourth line follows from Lemma B4.
d
Now fix ν d1 , . . . , ν dm+1 ∈ Hm+1
with ηdm+1 ν d1 , . . . , ν dm+1 = µd1 , . . . , µdm+1 . By Proposition
D1, there is a type ud with ρcm+2 sd , ud = sd , µd1 , . . . , µdm+1 . It suffices to show that sd , ud ∈
d d
: If so sd , µd1 , . . . , µdm+1 ∈ ρcm+2 Rm+1
Rm+1
.
d
. Also, by part (iv) of the
Using part (i) of the induction hypothesis and (a), sd , ud ∈ Rm
c
induction hypothesis, Rm is Borel. So
c
d −1 d
c
λd ud (Rm
) = λ d ud
ρm+1
ρm+1 (Rm
)
= δ dm+1 ud ρdm+1 (Rcm )
c
= µdm+1 ρdm+1 (Rm
)
= 1,
where the first line follows from part (ii) of the induction hypothesis, and the last line follows from
d
(b). Given this, sd , ud ∈ Rm+1
, as required.
c Finally, turn to part (iv). Use the fact that ρdm+2 Rm+1
is measurable (part (iii) established
d −1 d
c c
already), Rm+1 = ρm+2
ρm+2 Rm+1 (part (i) established already), and ρdm+2 is measurable
c
(Lemma A4), to get that Rm+1
is measurable.
In what follows, we will fix a terminal S b , S a -based type structure structure
Λ = S b , S a ; (T a , T (T a )), (T b , T (T b )); λa , λb
and also an arbitrary S b , S a -type structure
a
b
a
a
b
b
Λ = S b , S a ; (T , T (T )), (T , T (T )); λ , λ .
42
We will use upper-bars to indicate all sets and maps associated with the structure Λ. We also have
hierarchy morphisms β a , β b , i.e., maps
a
βa
: T → Ta
βb
: T → Tb
b
a a
a with δ a β a (t ) = δ t and similarly for b.
Lemma E4 For each m:
c c
c
c
∈ Rm
;
(i) if sc , t ∈ Rm then sc , β c t
c
−1
c
(ii) Rm ⊆ ρdm+1
ρdm+1 (Rm
) .
Proof. By induction on m.
c c
c
c
m = 1: First, fix sc , t ∈ R1 and note, by Lemma E1, sc is optimal under δ 1 t . Since
c c c c c
δ 1 t = δ c1 β c t , sc is optimal under δ c1 β c t . By Lemma E1, sc , β c t
∈ R1c establishing
Part (i). Part (ii) is immediate.
m ≥ 2: Assume that the result holds for m and we will show that it also holds for (m + 1).
c c
c
c
For part (i), fix sc , t ∈ Rm+1 and note that, by the induction hypothesis, sc , β c t
.
∈ Rm
c c
d
d
As such, it suffices to show that β t believes Rm . By Lemma E3(iv), Rm is Borel. Part (iii) of
d
that same Lemma says that ρcm+1 Rm
is closed. So
c d c c −1 c
d λc β c t
R m = λc β c t
ρm+1
ρm+1 Rm
c c
d = δ cm+1 β c t
ρm+1 Rm
c d c
= δ m+1 t ρcm+1 Rm
−1 c
c c c
= λ t
ρm+1
ρm+1 Rdm
c c
d
≥ λ t (Rm )
= 1,
where the fifth line follows from part (ii) of the induction hypothesis.
Part (ii) is immediate, completing the proof.
Corollary E1 For any type structure Λ and any terminal structure Λ,
projS a
a
m Rm
× projS b
Lemma E5 Given a type structure Λ,
a
b
m Rm
⊆ projS a
b
a
b
(i) for each m, projS a Rm × projSb Rm ⊆ Sm
× Sm
;
43
a
m Rm
× projS b
b
m Rm .
(ii) projS a
a
m Rm
× projS b
b
m Rm
a
b
⊆ SM
× SM
.
c
Proof. We will prove (i) by induction on m. Fix (sc , tc ) ∈ R1 and note that, by Lemma E1, sc is
optimal under δ c1 (tc ) ∈ M S d so that sc is undominated and contained in S1c . Assume that the
c
d
result holds for m and fix (sc , tc ) ∈ Rm+1 . Then λc (tc ) (Rm ) = 1 and so
d
margS d λc (tc ) (projS d Rm ) = 1.
d
= 1.
By the induction hypothesis, margS d λc (tc ) Sm
c c
c
c
margS d λ (t ), and so s ∈ Sm+1 .
For part (ii) use part (i), i.e.,
projS a
a
m Rm
× projS b
b
m Rm
c
Since (sc , tc ) ∈ R1 , sc is optimal under
a
b
⊆ projS a RM × projS b RM
a
b
⊆ SM
× SM
,
as required.
Proof of Proposition 5.2. By Theorem 6.1 in Brandenburger-Friedenberg-Keisler [9, 2004], there
is an S b , S a -type structure Λ with
a
b
a
b
projS a Rm × projS b Rm = Sm
× Sm
and
projS a
So, by Lemma E4,
a
m Rm
× projS b
b
m Rm
for all m ≥ 1
a
b
= SM
× SM
.
a
b
a
b
Sm
× Sm
⊆ projS a Ram × projS b Rbm = Sm
× Sm
for all m ≥ 1
and, similarly, by Corollary E1,
a
b
SM
× SM
⊆ projS a
Now both parts follow from Lemma E5.
a
m Rm
44
× projS b
b
m Rm .
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