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BEBR
FACULTY WORKING
PAPER NO. 1329
FE8
On
Complete Regularity of Spaces of Economic Agents
Endowed with the Order Topology
Ai AH Khan
Ye Neng Sim
College of
Bureau
Commerce and Business
of Economic
University of Illinois,
Administration
and Business Research
Urbana-Champaign
BEBR
FACULTY WORKING PAPER HO.
1329
College of Commerce and Business Administration
University of Illinois at Urbana-Champaign
February 1987
On Complete Regularity of Spaces of Economic Agents
Endowed with the Order Topology
Ali Khan, Professor
Department of Economics
M.
Ye Neng Sun
Department of Mathematics
On Complete Regularity of Spaces of Economic Agents
Endowed with the Order Topologyt
by
M.
Ali Khan* and Ye Neng Sun**
January 1987
Abstract
We show that the order topology of Chichilnisky on closed
subsets of a compact Hausdorff space is completely regular and finer
than the topology of set convergence of Choquet and Kuratowski.
:
t
This research was supported, in part, by a N.S.F. grant to the
University of Illinois. The second author would like to acknowledge the constant help and encouragement of Peter Loeb during the
course of his studies at Illinois.
*
Department of Economics, University of Illiois,
Street, Champaign, IL 61820.
1206 South Sixth
** Department of Mathematics, University of Illinois,
Street, Urbana, IL 61801.
1409 West Green
1.
Introduction
In
G.
[1],
Chichilnisky introduced
a
topology on the space of eco-
nomic agents that is sensitive to a measure of the graph of an agent's
preference and which
Choquet
is
finer than the topology of set convergence of
and Kuratowski
[2]
[7]
as well as
the topology induced by the
Hausdorff metric, both in the setting of a compact metric space of
Chichilnisky termed this the order topology and derived
commodities.
its basic properties.
In
[5,6], the authors investigate the existence of Cournot-Nash
equilibria of games as formalized by Mas-Colell [8].
In this work,
an
agent's preference depends on the distribution of actions taken by all
the other agents and hence an agent is formalized as a continuous
function from a compact Hausdorff space to the space of preferences
defined on
a
compact Hausdorff space.
This leads one to consider a
space of economic agents in the presence of externalities and to endow
this space with a suitable topology.
As argued
in
[5,6], two proper-
ties which are crucial for the existence and upper hemicontinuity of
Cournot-Nash equilibria are complete regularity and compactness of
this topology.
Complete regularity hinges on the complete regularity
of the range space and compactness results are generalizations of the
Ascoli-Arzela theorem and hence require a uniform structure on the
range space.
Given that every completely regular space is unifor-
mizable, the compactness property also hinges on complete regularity
of the range space.
This motivation led to the principal result presented in this
paper.
We show that in the case of a compact Hausdorff space of com-
modities, the order topology proposed by Chichilnisky is completely
-2-
We also show that it is finer than the topology of set con-
regular.
This result simply drops the metric
vergence and hence Hausdorff.
assumption from Theorem
Section
tion
2.
of
[1].
presents the notation and the basic definitions and Sec-
2
Section
the results.
3
1
devoted to the proofs.
is
4
Notation and Preliminary Definitions
We shall work with a topological space (X,t).
For any A
C
X,
let
A° denote the interior of A,
denote the closure of
A
A
c
A,
denote the complement of A,
all of these operations relative to (X,x).
note that (A
refers to the interior of the closure of the
)°
complement of A whereas (A
complement of
reader
By way of clarification,
)°
the closure of the interior of the
is
For undefined notions of point-set topology,
A.
the
referred to [9].
is
Following
[1]
we
,
consider topologies on a subset of the set of
Accordingly, let
closed subsets of (X,x).
3=
C
{A
(A
X:
3")
= A}
For any A, B in j, we shall say
A > B if and only if B
Let t° be the topology on
[B,.]
for all B
any A
,
A
e
{A
=
^.
e
generated by
A > B};
:
A°.
[-,B]
a
=
sub-base consisting of
{A
e *J:
B > A}
We shall refer to x° as the order topology on J.
in "^with A
[A,, A
J
j
C
]
2
e
(A
>
A
e3:
let
,
A
>
2
A > A
}.
For
-3-
Let
be the topology on
t
[K]
(A
E
-J
generated by
a
sub-base consisting of
e^: AOKM); <G>E (AeJ:
for all compact sets K in X and open sets G in X.
x
HG
A
is
*
}
referred to as
the topology of set-convergence and also termed the Kuratowski topology
for details and additional references.
see Section III in [3]
3.
;
The Results
We can now present the two principal results of this paper.
Theorem
1 .
Let (X,t) be a compact Hausdorff space.
topology t° on
Remark
1 .
$
Then the order
is
finer than the topology of set convergence
1
drops the metric hypothesis in the corres-
Theorem
ponding statement in Theorem
1
of
In this
[1].
t
cm
theorem, Chichil-
nisky also makes the statement that "if X is metric separable and
locally compact, and X denotes its compactif ication
,
then the order
topology is finer than the topology of closed (set) convergence" when
defined as an analogue of
v7"in X.
It
is
clear that the assumptions of
X being metric separable and locally compact can be replaced by the
requirement that X be
Remark
2
.
It
a
completely regular T -space.
should be noted that for X a unit square in R
2
,
Chichi-
lnisky shows that the order topology is strictly finer than the topology of set convergence.
Next, we present
Theorem
2.
Let (X,t) be a compact Hausdorff space.
completely regular.
Then
t°
on
j
is
j".
.
-4-
4.
Proofs
We begin with the following
Lemma
Let
1 .
E,
generated by
E
and each x
U,
Proof
Then the toplogy
X.
finer than that generated by
is
there exists U
e
E
E'
such that x
for each U'
if
C
e U
e
E'
U'
Straightforward.
:
Lemma
e
denote two sub-basis in
E'
2
For any topological space (X,t),
.
C
(i) for any A
^
X,
{
C
X,
(Aj r\ A
2
and
= A° r\ A°
>
(A^A^
= A
\J
(ii) for any closed set A, A° = (A°)°
(iii) for any open set B, B = (B)°.
Proof
:
for example,
See,
[9,
Theorems 3.7 and 3.11]
For (ii), A° C (A°) implies A
C
For (iii),
(B)°
given openness of
5"
A,
C
A.
C (B)°.
B
Hence
implies (B)°
B
Z.
B,
"A*
Moreover, A° C A implies,
(A°)°.
A"
given closedness of
(
C
5") °
B.
for (i).
C
A°.
Moreover, B C (B) implies,
Hence B C (B)° and we are done.
We can now present a
Proof of Theorem
Then A C K
1 .
Pick, a
sub-basic set
and by regularity (see,
there exists an open set U such that
A
C
U
C
U
C
K°.
[K]
for example
and let A
[9,
e
[K].
Lemma 43.3]),
A^
A
.
-5*
Now let V = U.
Then by Lemma
C
[-,V]
C
= (U)°
implies U
"u
C
is open,
Since
"J
e
,
well-defined
a
e
CUC
A = (A
Hence B r> K =
K°.
75
and
")
(
5")
r\ G * 0.
and B
e
[K].
Let A
e
<G>.
Let x
e
.
,
N(x)
OA°
A°aG
Thus
* 0.
Since (A
* 0.
[9,
V
e
zf
and hence [V,»]
e
The first claim is obvious since V =
C
or that A > V or that A
A°
e
Since G
C
V = W
C
C
B°
A° r\ G, we have
C
[V,
•]
and that
WCA°AGCA
BHG^W
is
a
Let V =
[V,»]
C
and hence
and hence B
t
e
(BAG).
Then
Since
<G> and we are done.
Q
Lemma
(A°)
3
C
e
.
(i)
3.
2
For any A and B
we shall need the following lemmata.
i_n
x°.
<G>.
The proof is now complete.
For a proof of Theorem
W.
[V, •].
Hence (V C\ (A°nG)) C
B.
G.
Theorem 14.3],
For the second claim, pick any arbitrary element B in [V,»],
B > V which means V
* 0.
a well-defined sub-basic set of
is
All that remains to be shown is that A
V
Then
r\ G)
there exists a nonempty open set W such that W C W C (A°DG).
,
V°
Then A (~\G
((A°) r\ G)
nonempty open set, by regularity, see, for example,
2
C
any arbitrary element B in [»,V].
pick,
there exist a neighborhood of x, N(x), such that N(x)
x e (k°)
By Lemma
[»,V]
(U)° = V° and hence A
Next, we consider a sub-basic set <G>.
Since A
is
[• ,V]
e
For the second claim,
BCV°
[«,V]
[K].
For the first claim, U
leading to A
jr and hence
e
All that remains to be shown is that A
sub-basic set of t°.
and that
V
2
3"
»
(AU B)
e
J.
(ii) For any A e +,
:
-6-
Certainly (AUB) is
Proof
C (AUB).
implies (AUB)°
(AUB)° and hence
C (AUB) implies
On the other hand, A
C (AUB)°.
A*"
Hence (AUB)° C (AUB) which
closed set.
a
Since A
3"
e
,
A =
"A
5"
C
A°
and we can
Similarly, B C (AUB)° and hence (AUB) C (AUB)
conclude A C (AUB)°.
This completes the proof of (i).
As regards (ii), we claim that for any subset A of X,
(A°)
(a)
C
For (a), x
U(x)
x,
C\
A
=
(A°)
e
(A
and
<=> x
*
C
=
((A°) )°.
<=> x i A° <=> for all open neighborhoods U(x) of
)
C
(A°V
(b)
(A°).
e
For (b)
C
x
,
<=> x k (A 5 ") <=>
(A°~)
e
C
there exists an open neighborhood U(x) of x such that U(x)
(A°)
<=> x
c
e
((A°) )°.
Now A
e
3
implies A = (A75").
C
Therefore, ((A°) )° = (A
C
= A° =
)°
C
Hence, A° =
= ((A°)
(A°~)
Hence (A°)
(A°)°.
C
e
C
5
)°
=
C
(A )°.
and we are
done.
Lemma
Proof
4
.
{
[A, ,A„
]
:
A. ,A
?
e
J
A
',
?
>
A,
is
}
base for x° £n_
a
The proof is a consequence of Theorem 2.2 in [4, Chapter III]
:
according to which we have to show that for any open set G in
any A
e
G,
there exists
Pick any G
e
x°
[B_,
and any A
B
]
e
such that A
G.
basic sets, certainly there exist A
=
1,
...,
a e
,
A
m such that
(
r\
1=1
u.,.])n( ni-.A
j=l
e
[B
,
B
]
x°
,
and
CG.
Since G is an arbitrary union of
basic sets and since every basic set is
j
J".
])
a
finite intersection of sub-
where
i
=
1,
...,
n and
.
.
-7-
m
n
.
AJ
O
=
Let A
- (A°) and B
B
,
=
.
By Lemma 2, B
A..
[J
O
e
.
i=l
j=l
By an induction argument based on Lemma 3(i),
We have to show that
implies B
e
-J
.
B„
>
B.
ej
A.
n
We show that A
e
[B
Since A
B ].
,
Z
e
1
[A
(~\
,
which
A > A
•],
i
i
i=1
n
means
C
A,
for all
A"
i
=
....
1,
This implies
n.
i
\
C
A.
I
ji
i=
A° which
*
l
m
means
C
B
and hence A > B
A°
On the other hand, A
.
e
(~\
j=
[«,A
]
l
m
implies A > A which means A C (A )° and hence A C ( (~\ (A )°) which
m
,
j=l
equals ( f\ A )° by an induction argument based on Lemma 2(i).
But
j
=l
m
)° = A° which equals B° as shown in the paragraph above.
f\ A
(
j
=l
B
Putting the two parts of the argument together, we can conclude
> A.
that B
and that A
B
>
[B
e
Next, we show that [B
arbitrary B
m
[B_,
e
B
J
(A )° and hence B
j
=
Hence
].
C
B
,
B ].
,
Now
]
B
J
(A )°
C
G.
>
B
Towards this end, pick an
means B C B° which implies B C A° =
for all
j
=
..., m.
1,
Hence AJ > B and
l
hence B
e
m
C\
(
j
=
[ •
,A ]).
On the other hand,
B >
means
B
B
l
C
B°
which
l
n
implies
I
J
A.
C
C
B° and hence A.
B°
for all
i
=
1,
...,
n.
This
n
means that B
> A
and hence B
e
(
C\
i
e
latter set is a subset of G,
,
Putting the two parts
•]).
m
n
together, we obtain that B
[A
=l
f\ [k. , •]) (\( C\
1
i=l
j=l
we are done.
(
[',£?]).
Since the
D
Lemma
a
5
.
Consider the map
homeomorphism.
Q:
j
+J"
such that Q(A) = (A°)
.
Then Q is
-8-
Proof
Lemma
:
3
(
ii
)
establishes that Q is well-defined.
the proof of Lemma 3(ii), we showed
((A°)
C
(0 C
=
)°
AeJ,
If
.
From (b) in
for any subset A of X,
that
this implies (((A°)
C
)°)
C
=
Hence (Q»Q)(A) = A for any A ejand we can state that Q
((A)
C
= A.
an iden-
is
«Q
C
)
tity mapping on Jf.
Next, we show that Q is
establish that B
Q(A)
open set U such that Q(U) C [B
B.
e
<7
by Lemma 3(ii),
,
Q(B„), (B°)
(B?)°
U.
A
e
U implies
).
(B°)°
[B.
,
B
].
e
J-
.
We have to show that Q(A)
e
C
(B°)
C
,
For this, we first
C
A° which equals
B
].
>
(B°)
A >
C
.
e
Let U - [(B°)°,
Since B
[B
,
B
]
((A°)
C
)°
<=>
We have to find an
Since
(B°)°].
implies Q(B
B
>
and hence U is well-defined.
(B°)
>
A
Q((B°)
<=>
B°
Now pick any basic set
Q(B).
>
C
A <=> A
>
continuous mapping.
a
>
)
Pick any arbitrary
i.e.,
B
Q(A) > B
>
From this we obtain Q((B°)
Since Q-Q is an identity, this implies that B„
>
C
)
>
But
.
Q(A) >
Q(A) > B
,
and we are done.
a
Lemma
6
Let
.
X + R be a continuous
ty:
function and let
and by f(0) = Inf
defined by f(A) = Sup ijj(x) for A t
xeA
xeX
f is continuous with al endowed with the topology t°.
Proof
:
Note that [»,0]
empty set.
tinuity of
Hence
f
at
f
is
3A = 0,
continuous at
tinuous at
A.
j-
+ R be
Then
ij;(x).
and we need only to show con-
an arbitrary nonempty element A of
then A = A°.
:
an open set in t° and consists only of the
is
Let 3A denote the boundary of A.
If
f
Since A > A,
Since A
e
j",
[A, A]
{A
}
We therefore consider the case
there exists an open set U
it
Pick any
is
a
and hence
8A * 0.
such that for all x
P
J.
e U
P
e >
0.
closed set.
f
For all
is
p
cone
3A,
.
,
-9-
-
|i|i(x)
i|/(p)
<
|
e.
there exists by [A, VII. 2. 2(2)]
By regularity of X,
such
an open set W
that
CW
W
e
p
P
= W
Now let V
C U
.
P
P
Hence by Lemma
.
2 V
or.
e
P
P
P
Since 3A is a closed set of compact space, it is compact; see, for
example
[4,
Hence there exists a finite sub-
III. 4. 10 and XI. 1.4(3)].
n
cover {W
of
}
"l
Let A ? = AL>( I) V
3A,
M
1-1
based on Lemma 4(i), A_
j
e
We can now claim that A
By an induction argument
).
P
l
.
>
A and that A
implies f(B)
> B
f(A) +
<_
n
To see the first claim, note that A = (A°
e.
n
C A°U
O
(
•
i
i=l
3A)
U
C A
v°
P
C (AU(
)
•
))° = A°
J V
*
P
i=l
l
ej
such that A„
>
B.
This means B C A^
Thus, suppose B
certainly f(0) = Inf ij>(x) _< f(A) + e.
xeX
corresponding to any xeB, we pick x' e A such that
x'
= x if
= p.
p
)
P-
•
i
B = 0,
Now,
W
i
i=l
i
For the second
and we are done.
I
•
l
M
(
•
claim, pick any arbitrary B
If
U
n
* 6.
x e A
otherwise,
i
where
i
is
chosen to be such that x
e V
Since x
•
p
t A and
x
i
such that x
e
V
.
p
there is more than one index, pick any one.
us to claim
_<
ip(x')
+
e
<_
Sup
yeA
i|;(y)
+
A°
2
certainly there exists at least one index
ij^(x)
e
i
e.
If
i
This construction enables
-10-
This implies
ij;(x)
f(A) +
_<
and hence f(B) - Sup
e
i^(x) _<
f(A) +
We
e.
xeB
are done.
consider
Next,
Hausdorff and
A such that
= Sup
ij>(x).
Since A is compact,
xeA
is continuous, certainly there exists such a p.
Again,
i|>
by continuity of
e
p
\\>(
p)
we can assert the existence of an open set W
i|>,
such
P
that
p
and for all x
W
e
W
e
P
,
P
*(p)
Since W p| A
*(x) + e.
<
^
(3
and since A = A°
we can argue just as in the fourth
,
paragraph of the proof of Theorem
AA°)
Since (W
to assert
1,
an open nonempty set,
is
that W
f\
A°
# 0.
by regularity of X (see,
for
P
example,
[4,
VIII. 2. 2(3)
]
we can assert the existence of a nonempty
)
such that
open set V
P
(3
C
* V
Now let
A,
1
= V
C
V
(W
P
P
.
f\
A°).
P
Hence by Lemma
1
We can now claim that A
>
The first claim is obvious.
B > A,
i|)(
p) _<
A,
2
p
iKp') + £> an d hence
4
^
and that B
A
.
>
implies f(A)
A
_<
f(B) +
For the second claim, pick any arbitrary
C
which means that A
e
Now pick any
B°.
i|>(p)
S.
^"P
*K X )
+
p'
£•
from A
.
We have
This implies that
xeB
f(A) < f(B) + e.
By putting the two parts of the argument together, we can conclude
that for any arbitrary B
|f(B) - f(A)|
e
< e.
[A
,
A
],
e.
-11-
Since [A.,
an open neighborhood of A,
Is
A,,]
the proof is complete.
a
Lemma
Let
7.
X + R be a continuous function and let g: J* * R be
i|»:
^(x) _for A t X, and by g(X) = Inf
X £a.
defined by g(A) = Sup
/ a
xe(A
is
o \
C
i|»(x)
Then g
.
)
continuous.
Proof
g =
Note that for any A
:
f
g(A) = f((A°)
,
C
Since Q is a homeoraorphism by Lemma
Q.
•
3"
e
)
= f(Q(A)).
5
and
f
is
Hence
continuous
by Lemma 6, g is continuous.
n
We are now ready for a
Proof of Theorem
2.
and such that A
i
Pick A n £^and F C
j
[A
,
A
i F,
CF
]
A
c
e
F
Since F
.
are open sets in
{x}.
Since x° is finer than
j
,
e
c
[A
is
,
A
by Lemma 5,
open,
on
t
on
j
Proposition 3.2]),
Now note that [-,0] and
].
and since
{0} and
JX
t
[A
,
A
{0} and
is Hausdorff
(see,
are closed sets in
}
Thus, without any loss of generality, we may consider A
If not, we simply consider the open set
and
there exist
are respectively equal to the sets
[X,«]
[3,
t°
We begin with the case where A n t
F.
and such that A
for example,
closed in
that takes the value zero on A and the value
unity on each element of
Since A
F is
We have to show that there exists a con-
F.
tinuous function on
J where
]
- ({0}
j"
and A
*
U{x}).
.
t X.
Since,
by Lemma 5, the base for t° consists of sets of the form [B,C], we can
find
[A
,
A
]
such that A
t
and A
t X
and such that
[A
,
A
]
CF
.
-12-
Since A
[4,
and A
*
certainly A
* X,
[A
z
A
,
];
see for example,
III.2.2].
C
Since A
A°
and since A
(A°)
* X,
*
0.
Hence we have two
n
and (A°)
disjoint nonempty closed sets A
.
Given that X is compact
Hausdorff and hence normal, we can appeal to Urysohn's Lemma
VII.4.1.] to assert the existence of a continuous function
[0,1]
[4,
X +
:
ij>
such that
^(A
-
)
C
10
Since
A°
A,
{0} and
^((AJ)
=
)
{l }.
10
and since
A,
and A° * X, we have two disjoint
J
*
n
nonempty closed sets A
and (A°)
By a second appeal to Urysohn's
.
Lemma, we can assert the existence of a continuous function
i|>
:
such that
X + [0,1]
^(Aj)
We now define
f:
=
{1} and
\j +
f(A) = Sup
((A°)
t|-
C
)
2
[0,1]
and g:
\p
(x)
for A *
iJj
(x)
for A -
=
{0}.
cr *
[0,1]
such that for any A
xeA
-
Inf
xeX
g(A) =
Sup
>J;
xe(A°)
=
Inf
C
i|>«(x)
(x)
for A * X
2
for A =
X.
xeX
By Lemmata
+ [0,1]
6
and 7,
f
and g are continuous on
such that h(A) = Max (f(A),g(A)}.
>7'.
Now define
h:
e
-
j
-13-
Since *,CA
=
)
{O} and
((A°)°) = {o}, certainly h(A
i|>
consider any A eJfor which h(A) <
ticular A £
g(A) <
g(A) <
=
and hence A
=
e
=
1.
An
implies f(A)
<
and
1
Z.
A
C
O
(x) < 1 which implies (A°)
Sup
C ^l
xe(A°)
We have g(0) = 1 and f(X) =
which means A > A .
A
and hence h(0) = h(X) =
A
1
1
implies for all A
1
Now
implies for all A * 0, Sup ty (x) < 1 which means
xeA
and hence A C A° which means A > A.
On the other hand
But f(A) <
1.
A H(A°)
= 0.
We shall show that this par-
1.
note that h(A) <
To see this,
F.
)
1.
t X,
The case A =
or A = X was ruled out.
Therefore we have shown that for all A £ [A
h(A)
All that remains is the consideration of the cases where A_ =
or
,
].
Since A
e
F implies k £
Suppose A
= X.
closed set in J*.
A * 0.
=
[A
,
A ],
and that
Then define f(A
£ F
)
we obtain h(A) =
where F is
= f(0)
=
,
as
,
1.
before, a
and f(A) =
1
Since {0} = [«,0] is both an open and closed set in
The case for A n = X is virtually identical.
continuous.
define f(A
)
= F(X)
=
and f(A) =
1
the openness and closedness of [X,»]
A
for all
j
,
f
is
Simply
for all A t X and again appeal to
in
j
l
1
Hence
],
[A
to establish continuity.
We thus have our desired function in all cases and the proof is
finished.
D
A,
-14-
REFERENCES
[1]
Chichilnisky "Spaces of Economic Agents," Journal of Economic Theory
15, (1977) 160-173.
G.
,
,
Choquet,
55-112.
"Convergences," Ann. Univ. de Grenoble
23
(1947-48)
[2]
G.
[3]
G. Salinetti and R. Wets, "Convergence of Functions:
S. Dolecki
Equi-Semicontinui ty ," Transactions of the American Mathematical
Society 276 (1983) 409-429.
,
,
,
[4]
J.
Dugundji, Topology
[5]
M.
Ali Khan,
,
Allyn and Bacon Inc., Boston,
1966.
"On a Variant of a Theorem of Schmeidler," B.E.B.R.
Revised in December 1986.
1306.
Working Paper No.
Ali Khan and Y. Sun, "On a Reformulation of Cournot-Nash
Equilibria," University of Illinois, mimeo.
[6]
M.
[7]
K.
Kuratowski
[8]
A.
Mas-Colell, "On
,
tical Economics
[9]
S.
D/108A
Topology
,
13
a
,
Academic Press, New York, 1966.
Theorem of Schmeidler," Journal of Mathema(1984) 201-206.
Willard, General Topology, Addison-Wesley
,
New York,
1970.
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