Contents

(A) Compactness and Continuity: Preliminaries

(B) Weierstrass Maximum Theorem

(C) Upper and Lower Semicontinuity of Correspondences

(D) Berge's Theorem

Selected References

**(A) Compactness and Continuity: Preliminaries**

Let (X, r ) be a metric space, with X as the space and r as the associated metric. Then, in any given metric space (X, r ), the following things may be defined:

Neighborhood: N_{e}(x) = {y ﾎ X ｽ r (x, y) < e } is a "neighborhood" around x, i.e. N_{e}(x) is a set of points in X which are within metric distance e of x.

Limit: x ﾎ X is a "limit point" of a set E ﾌ X ifeveryneighborhood N_{e}(x) (i.e. for every e > 0) has some point y ﾎ E where y ｹ x.

Complement: E^{c}is the "complement" of E if any x ﾏE implies x ﾎ E^{c}

Closed: a set E ﾌ X is "closed" if every limit point of E is in E.

Interior: x is an "interior point" of E if there is some e > 0 such that N_{e}(x) ﾌ E

Open: a set E is "open" if every point x ﾎ E is an interior point.

Bounded: a set E is "bounded" if for every x ﾎ E, there is some real number m such that for any y ﾎ E, r (x, y) < m.

Closure: cl(E) = E ﾈ E｢ where E｢ is the set of limit points in E is defined as the "closure" of E (i.e. cl(E) is the smallest closed set containing E).

Theorem:(Closed/Open Complementarity) E is open iff E^{c}closed.

Proof: (i) Suppose E^{c} closed. Consider any x ﾎ E. As E^{c} closed, then x ﾏ E^{c}
and x not limit of E^{c}. Thus, it must be that there is an e
such that N_{e}(x) ﾇ E^{c}
= ﾆ . This is true for all x ﾎ E.
Thus, E open. (ii) Suppose E open. Consider x = lim E^{c}. Then, N_{e}(x) ﾇ E^{c} ｹ ﾆ for all e
> 0. Thus, x is *not* an interior point of E. Since E is open and x is not an
interior point of E, then it must be that x ﾎ E^{c}.
Thus, E^{c} is closed.ｧ

Continuity: Let ｦ :(X, r ) ｮ (X, r ) be a function. ｦ is "continuous on x" if, for any e > 0, there exists a d > 0 such that r ｢ (ｦ (x), ｦ (x_{0}_{0})) < e wherever r (x, x_{o}) < d . ｦ is a "continuous" function when it is continuous on all x_{0}ﾎ X.

Theorem: (Continuity) ｦ :Xｮ Y is continuous iff for every open subset G ﾌ Y, ｦ^{-1}(G) is open.

Proof: (i) Let ｦ be continuous function.
Choose x_{0} ﾎ X. Then consider an open subset G ﾌ Y where ｦ (x_{0}) ﾎ G. As G is open, then ｦ (x_{0})
is in the interior of G so that we can then define N_{e}[ｦ (x_{0})] ﾌ G, i.e. an open e -ball around ｦ (x_{0}). By
continuity of ｦ at x_{0}, we know that there is a d > 0 such that ｦ (N_{d}(x_{0})) ﾌ N_{e}[ｦ (x_{0})] ﾌ G, thus N_{d}(x_{0}) ﾌ ｦ ^{-1}(G). As Nd (x_{0}) is an open d -ball
within ｦ ^{-1}(G), then x_{0} is an interior
point. Thus ｦ ^{-1}(G) is open.N_{e}

(ii) Suppose ｦ ^{-1}(G) is open for
every open set G ﾌ Y. As G is open, we can define an open
neighborhood N_{e} [ｦ (x_{0})]
ﾌ G for any ｦ (x_{0}) ﾎ G. As ｦ ^{-1}(G) is open in X,
then ｦ ^{-1}(N_{e}[ｦ (x_{0})]) is open ball in X which contains x_{0}.
Thus, there is a neighborhood N_{d}(x_{0}) ﾌ ｦ ^{-1}(N_{d} [ｦ (x_{0})]). Thus, ｦ (N_{d}(x_{0})) ﾌ N_{e} (ｦ
(x_{0})). Thus ｦ is continuous in X.ｧ

Open Cover: let {G_{a}} be a collection of open subsets of X such that E ﾌ ﾈ_{a}G_{a}, then {G_{a}} is an "open cover" of E.

Compactness: E is "compact" ifeveryopen cover of E has a finite subcover.

Thus, we approach one of the more important important theorems regarding a
special type of metric space - the Euclidian space R^{n}:

Theorem: (Compactness Theorems) Let E be a set in R^{n}. Then the following are equivalent:

(a) E is closed and bounded (Heine-Borel)

(b) E is compact

(c) Every infinite subset of E has a limit point in E (Bolzano-Weierstrass)

Proof: We want to prove (a) ﾞ (b) ﾞ (c) ﾞ (a). For this we need the following two lemmas:

Lemma: (Nested Set Property) If {I_{n}} is a sequence of closed intervals in R such that I_{n}ﾉ I_{n+1}(decreasing sequence of sets), then ﾇ^{･}_{n=1}I_{n}ｹ ﾆ .

Proof: I_{n} = [a_{n}, b_{n}], thus we have a
collapsing sequence of intervals. Let E = {a_{1}, a_{2}, ... } be the set
of lower bounds. Obviously, E is non-empty and bounded above by b_{1} (the upper
bound of the largest interval). Let x = sup E, i.e. the sup of the lower bounds. If m and
n are positive integers, then a_{n} ｣ a_{m+n}
< b_{m+n} < b_{m}, thus x < b_{m} for each m ﾎ 1 2, .... But, as x = sup E, then x ｳ
a_{m}. Thus, x ﾎ [a_{m}, b_{m}) for
each m = 1, 2, .... or, simply, x ﾎ I_{m} for m = 1,
2, ... or x ﾎ ﾇ^{･}_{n=1}I_{n}
ｹ ﾆ . Q.E.D.

Lemma: (Nested Boxes)Let {B_{k}} be a sequence of "boxes" in R^{n}such that B_{k}ﾉ B_{k+1}, n = 1, 2, ... Then ﾇ^{･}_{k=1}B_{k}ｹ ﾆ .

Proof: If B is a box, then B = {x ﾎ R^{n}
ｽ a_{i} ｣ x_{i} ｣ b_{i} for i = 1, ..., n}. Thus, box B_{k} = {x ﾎ R^{n} ｽ a_{ki} ｣ x ｣ b_{ki} for i = 1, .., n.}
and {B_{k}} is a sequence of collapsing boxes. Let interval B_{ki} = [a_{ki},
b_{ki}]. Consider sequence {B_{ki}} where B_{ki} ﾉ B_{(k+1)i} ...., i.e. a sequence of collapsing closed
intervals. By previous lemma, ﾇ^{･}_{k=1}
B_{ki} ｹ ﾆ , i.e. there is
a x_{i} such that x_{i} ﾎ B_{ki} for
all k = 1, ..., n. Of course, this is true for all i, thus x = [x_{1}, ...,x_{i},
... x_{n}] ﾎ B_{k} for all k = 1, .., n, i.e.
x ﾎ ﾇ^{･}_{k=1} B_{k}.
Q.E.D

(a) ﾞ (b) (**Heine-Borel Theorem**): If E
is bounded, we can enclose E in box B, i.e. E ﾌ B = {x ﾎ R^{n} ｽ a_{i} ｣ x_{i} ｣ b_{i}, i = 1,
.., n}. If B is compact, then as E is a closed subset of a compact set, is also compact.
Thus, we claim B is compact. Proof: If not, then {G_{a}
} is an open cover for B, but there is no finite sub-cover. Now, consider the following:
for every i, B_{i} = {x ﾎ Rｽ
a_{i} ｣ x_{i} ｣ b_{i}}.
Define d = ﾖ [・/font>
^{n}_{i=1} (b_{i} - a_{i})] so, for all x, y ﾎ B, r (x, y) ｣
d . Now, consider c_{i} = (a_{i} + b_{i})/2,
i.e. c_{i} bisects every interval B_{i} into two intervals, [a_{i},
c_{i}] and [c_{i}, b_{i}]. Thus, {c_{i}} divides B into 2^{n}
boxes, call then Q_{j} ﾌ R^{n} where ﾈ _{j}Q_{j} = B.

Now, recall that {G_{a} }is an open
cover for B, thus there must be at least one Q_{j} that is not covered by any
finite subcollection of {G_{a}} (otherwise B is
compact). Call Q_{j} = B_{1}. Thus, r (x, y) ｣ d /2 for any x, y ﾎ
B_{1}. Subdivide B_{1} again by bisection, e.g. there is a d_{i} =
(a_{i} + c_{i})/2, etc. Thus, we subdivide B_{1} into 2^{n}_{
}boxes J_{j} whose union ﾈ _{j}J_{j}
= B_{1}. So, define B_{2} as the "uncovered set" where B_{2}
ﾌ B_{1}. Obviously, r (x,
y) ｣ d /4 for any x, y ﾎ B_{2}. Continuing the process of subdividing eternally,
then we obtain a sequence of boxes B ﾉ B_{1} ﾉ B_{2} .... where any B_{k} in this sequence is not
covered by any finite subcollection of {G_{a}}.
Obviously, for any x, y ﾎ B_{k}, r
(x, y) ｣ d /2^{k}.

However, by previous lemma, there is an x ﾎ R^{n} such that x ﾎ B_{k}
for all k = 1, 2, ...., i.e. $ x ﾎ ﾇ^{･}_{k=1} B_{k}. But x ﾎ B and {G_{a} } is an open cover
of B, thus there is some a such that x ﾎ
G_{a} . At this point, define N_{e}
(x) = {y ﾎ R^{n} ｽ r (x, y) < e } as an open e -neighborhood of x. But, as G_{a}
is open, then for any e > 0, there is a y such that y ﾎ N_{e} (x). But as {B_{k}}
is an eternal subdivision, we can fit a box inside of N_{e}
(x), i.e. there is some B_{k} (k sufficiently large) such that B_{k} ﾌ N_{e} (x), which implies that r (x, y) > d /2^{k}. But then,
when this box is fitted, we need divide no further. Thus, for any B_{k}, there is
some G_{a} such that B_{k} ﾌ
G_{a} . Thus we have a finite subcollection of {G_{a} } that covers B. Thus B is compact. As E is a closed subset
of B, then E is compact. Q.E.D.

(b) ﾞ (c): Suppose E ﾌ
R^{n} is compact. Consider M ﾌ E where M is an
infinite subset of E. If M has no limit in E, then for each x ﾎ
E, we would have N_{e} (x) with at most a finite number
of elements in M. However, as M is infinite, thus no finite collection of open
neighborhoods{N_{e} (x)} can cover it. Thus M is not
compact. However, without limit points, M is closed. As M ﾌ E
is a closed subset but not compact, then E cannot be compact. A contradiction. Thus M must
have a limit point in E. Q.E.D.

(c) ﾞ (a): we use "not (a)" ﾞ "not (c)" to prove this. Take two cases. Case 1: suppose
E is not bounded. Then, there is points x_{k} ﾎ E such
that r (0, x_{k}) > k for all k = 1, 2, .... The set
S of such points x_{k} is infinite and has no limit in R^{n}. Thus, (c) is
not true. Case 2: suppose E is not closed. Then consider point x_{0} ﾎ R^{n} such that x_{0} ﾏ
E but is nonetheless a limit point of E. Thus, there is a sequence {x_{n}} ｮ x_{0 }as n ｮ ･ and x_{n} ﾎ E. Thus, {x_{n}}
is an infinite subset of E without limit point in E. Thus, (c) not true. Q.E.D.

Thus equivalence (a) ﾞ (b) ﾞ (c) ﾞ (a) is proved.ｧ

The usefulness of this theorem can hardly be overstated - but the simplest, in our case, is that in Euclidian space, if a set X is closed and bounded, then we can claim it is compact.

Finally, consider the following theorem:

Theorem: (Compact Range of Continuous Functions)X is a compact metric space, Y is a metric space and let ｦ : X ｮ Y is a continuous mapping. Then ｦ (X) is a compact set.

Proof: Let {G_{a} } be an open cover of
ｦ (X). As ｦ is continuous, then ｦ ^{-1}(G_{a}) is an open subset of X for all a
. Thus, ﾈ_{a}{ｦ^{-1}(G_{a} )} is an open cover of X. But X is compact. Thus, there is a
finite subcover, i.e. indices a _{1}, ..., a _{n} such that X ﾌ ﾈ a _{i}^{n}{ｦ ^{-1}(G_{a} _{i})}.
Since, by continuity, ｦ (ｦ ^{-1}(E)
ﾌ E for all E ﾌ Y, then ｦ (X) ﾌ ﾈ a _{i}^{n} ｦ {ｦ ^{-1}(Ga _{i})} or ｦ (X) ﾌ ﾈ a _{i}^{n}{G_{a} _{i}}.
Thus, we have a finite subcover of ｦ (X). Thus, ｦ (X) is compact.ｧ

**(B) The Weierstrass Maximum Theorem**

We want to prove the Weierstrass Theorem, i.e. that a continuous, real-valued function over a compact set attains a maximum and a minimum. For this we need the following lemma:

Lemma: Let E be a non-empty bounded subset of R. Let s = sup E and t = inf E. Then s, t ﾎ cl(E).

Proof: If s ﾎ E, then s ﾎ
cl(E). Suppose s ﾏ E, then for every e
> 0, there is a z ﾎ E such that s - e
< z < s, otherwise s - e would be an upper bound for E.
But, then, for all e > 0, there is a z ﾎ
N_{e} (s), thus s is a limit point of E. Thus, s ﾎ cl(E). We can do the same for t = inf E.ｧ

Now we can turn to the Weierstrass Maximum Theorem (Th. 1.7.4’ in Debreu, 1959):

Theorem:(Weierstrass) let ｦ : X ｮ R be a continuous real-valued function on a non-empty compact metric space X. Let M = sup_{xﾎ X}ｦ (x) and m = inf_{xﾎ X}ｦ (x). Then, there is a point x^{M}and a point x^{m}in X such that ｦ (x^{M}) = M and ｦ (x^{m}) = m, i.e. a continuous function ｦ (ｷ) attains a maximum and a minimum over a compact metric space.

Proof: X is compact. Thus, by the earlier theorem, ｦ(X) is compact. By Heine-Borel, ｦ (X) is closed and bounded. As ｦ (X) is a closed and bounded set of real numbers, then ｦ (X) = clｦ (X) and thus, by the previous lemma, contains its own supremum and infimum, i.e. sup ｦ (X) ﾎ ｦ (X) and inf ｦ (X) ﾎ ｦ (X).ｧ

**(C) Upper and Lower Semicontinuity of
Correspondences**

What can we say about cases when we are not dealing with single-valued
functions but rather correspondences (i.e. set-valued functions)? Thus, a correspondence j : X ｮ Y maps to a set and not a point,
e.g. X ﾍ R^{m} and Y ﾍ R^{n}.
How do we define the "continuity" for a correspondence? A different approach
must be followed.

Upper Semicontinuity: (Def: 1.8.d. in Debreu, 1959) a correspondence j : X ｮ Y (e.g. X ﾍ R^{m}, Y ﾍ R^{n}) is "upper semicontinuous at x" if whenever the following three conditions are met for any two sequences {x_{n}} and {y_{n}}:

(i) x

_{n}ｮ x

(ii) y_{n}ﾎ j (x_{n})

(iii) y_{n}ｮ y

thenit is also true that (iv) y ﾎ j (x). If j is "upper semicontinuous at x" for all x ﾎ X, then it is "upper semicontinuous".

Intuitively, we can see this criteria graphically in Figure 1: given a
sequence {y_{n}} in the graph of the correspondence which approaches some endpoint
y, does this endpoint y lie in the graph of j (x)? If so, then
the correspondence is "upper semicontinuous". We can see this in the figure
below where any sequence {y_{n}} drawn in the graph of the correspondence will
approximate a point (e.g. y) within the graph of the correspondence.

Figure 1- Upper Semicontinuous Correspondence

We now turn to an interesting theorem:

Theorem: (Closed Graph)Let j : X ｮ Y be a correspondence, where X, Y ﾍ R^{n}are compact. Then, j is upper semicontinuous if and only if the graph of j is closed, i.e. G_{j}={(x, y) ﾎ X ｴ Y ｽ y ﾎ j (x)} is a closed set in R^{n}ｴ R^{n}_{.}

Proof: As this is an "if and only if" statement, we need to prove it both ways.

(i) usc ﾞ closed graph: Let j be upper semicontinuous. Then for all y^{n} in the
convergent sequence y^{n} ｮ y, it is true that y^{n}
ﾎ j (x^{n}) " x^{n} ｮ x. Thus, by the
definition of the graph of j , it is true that (x^{n},
y^{n}) ﾎ G_{j} "
x^{n}, y^{n} in their respective convergent sequences. As (x^{n},
y^{n}) converges to (x, y), by the definition of u.s.c., then (x, y) is a limit
point of the graph of j . Is (x, y) ﾎ
G_{j} ? Again, yes, as by the definition of u.s.c., y ﾎ j (x). Thus every convergent sequence
(x^{n}, y^{n}) in G_{j} has its limit
in G_{j} . Thus G_{j}
is closed.

(ii) closed graph ﾞ usc: If G_{j} is closed, then every convergent sequence (x^{n}, y^{n})
ﾎ G_{j} has its limit (x,
y) in G_{j} . This implies that for all sequences x^{n}
ｮ x, y^{n} ﾎ j (x^{n}), y^{n} ｮ y,
then y ﾎ j (x), i.e. it cannot be
that the first three properties are fulfilled and the last (y ﾎ
j (x)) is not for that would mean that (x, y) ﾏ G_{j} and thus G_{j} would not be closed. Thus j is
upper semicontinuous at x. Since x is arbitrary, then this is true for any x ﾎ X. Thus, j is upper semicontinuous.ｧ

*Note*: the assumption in this proof that the range of the
correspondence j is compact can be seen to be necessary. A
counterexample can be constructed of a closed graph which is not u.s.c. if the range fails
to be compact. Consider the following correspondence: j (x) =
{1/x} for x > 0 and j (x) = {0} for x = 0. There is indeed a
closed graph, but it is not u.s.c. at x = 0. Also, for the first part of the proof, that
we must assume that the domain X is closed - or else j (.) may
not be defined at x.

The sum and Cartesian product of a series upper semicontinuous correspondences is also upper semicontinuous.

Theorem: (Sum of usc) Let {j_{i}}^{k}_{i=1 }be a series of upper semicontinuous correspondences where j_{i}: X ｮ Y. Then, the sum correspondence j : X ｮ Y (i.e. j = ・/font>^{k}_{i=1 }j_{ i}) is also upper semicontinuous.

Proof: Let us have convergent sequence{x_{n}} in X where x_{n}
ｮ x and consider a sequence {y_{n}} in Y where y_{n}
ﾎ j (x_{n}) and where j (x_{n}) = ・/font> ^{k}_{i=1j }(x_{n}) and y_{n} = ・/font>
^{k}_{i=1}y_{n}^{i} with y^{i}_{n} ﾎ j _{i}(x_{n}). As j _{i} is upper semicontinuous, then there is a y^{i}
ﾎ j _{i}(x) such that the
sequence {y_{n}^{i}} converges to it, i.e. y_{n}^{i} ｮ y^{i}. As this is true for all i = 1, .., k, then lim_{nｮ ･ }・/font> _{i=1}^{k}
y_{n}^{i} = ・/font> _{i=1}^{k} y^{i}
and ・/font> _{i=1}^{k} y^{i} ﾎ ・/font> _{i=1}^{k} j _{i}(x) = j (x). Defining y = ・/font> _{i=1}^{k} y^{i}, then lim_{nｮ ･ }y_{n} = y and y ﾎ j (x). Thus j
is upper semicontinuous.ｧ

The next theorem is practically the same:

Theorem: (Product of usc) Let {j_{i}}^{k}_{i=1 }be a series of upper semicontinuous correspondences where j_{i}: X ｮ Y. Then, the product correspondence j : X ｮ Y (i.e. j = ﾕ^{k}_{i=1 }j_{ i}= j_{1}ｴ j_{2}ｴ ... ｴ j_{k}) is also upper semicontinuous.

Proof: Let us have convergent sequence{x_{n}} in X where x_{n}
ｮ x and consider a sequence {y_{n}} in Y where y_{n}
ﾎ j (x_{n}) and with j (x_{n}) = ﾕ ^{k}_{i=1j }(x_{n}) and y_{n} = [y_{n}^{1},
y_{n}^{2}, .., y_{n}^{k}] with y^{i}_{n} ﾎ j _{i}(x_{n}). As j _{i} is upper semicontinuous, then there is a y^{i}
ﾎ j _{i}(x) such that the
sequence {y_{n}^{i}} converges to it, i.e. y_{n}^{i} ｮ y^{i}. As this is true for all i = 1, .., k, then lim_{nｮ ･ }[y_{n}^{1}, y_{n}^{2}_{,
...,} y_{n}^{k}] = [y^{1}, y^{2}, ..., y^{k}]
and [y^{1}, y^{2}, .., y^{k}] ﾎ j _{1}(x)ｴ j
_{2}(x)ｴ ..ｴ j _{k}(x) = ﾕ _{i=1}^{k}
j _{i}(x). Defining y = [y^{1}, y^{2},
.., y^{k}] and recalling that ﾕ _{i=1}^{k}
j _{i}(x) = j (x), then lim_{nｮ ･ }y_{n} = y and y ﾎ j (x). Thus j
is upper semicontinuous.ｧ

Let us now turn to the analogous concept of lower semicontinuity:

Lower Semicontinuity: (Def: 1.8.e in Debreu, 1959) a correspondence j : X ｮ Y (e.g. X ﾍ R^{m}, Y ﾍ R^{n}) is "lower semicontinuous at x" if whenever the following three conditions are met for any two sequences {x_{n}} and {y_{n}}:

(i) x

_{n}ｮ x

(ii) y_{n}ﾎ j (x_{n})

(iii) y ﾎ j (x)

thenit is also true that (iv) y_{n}ｮ y. If j is "lower semicontinuous at x" for all x ﾎ X, then it is "lower semicontinuous".

Intuitively, we can see this criteria graphically again, this time in
Figure 2: given an endpoint y ﾎ j
(x) in the graph, is there a sequence {y_{n}} in the graph that approaches it? If
so, then the correspondence is "lower semicontinuous". In the earlier Figure 1,
we do *not* have a lower semicontinuous function at x as the point y ﾎ j (x) cannot be approximated from the
right by a sequence of points in the graph. However, in Figure 2, the correspondence is
indeed lower semicontinuous as any endpoint y in j (x) is
approximated from either side by a sequence {y_{n}} in the graph. However, the
graph below is *not* upper semicontinuous as, for instance, we can imagine a sequence
in the graph approximating y｢ yet y｢
ﾏ j (x).

Figure 2- Lower Semicontinuous Correspondence

Finally, we should note that the sum and product of a series of a lower
upper semicontinuous correspondences is also lower semicontinuous, i.e. correspondence j : X ｮ Y where j
= ・/font> _{i=1}^{k}j^{
}_{i} or j = ﾕ _{i=1}^{k}j^{ }_{i} are lower semicontinuous if the series {j _{i}}_{i=1}^{k} where j
_{i}:X ｮ Y is also lower semicontinuous. The proof is
analogous to that of the upper semicontinuous case.

Finally we turn to continuity:

Continuity(Def. 1.8.f in Debreu, 1959): a correspondence j : R^{m}ｮ R^{n}is "continuous" if it is both uppersemicontinuous and lower semicontinuous at all x ﾎ R^{m}.

Thus, in the previous figures, the correspondences are *not*
continuous as the first is upper semicontinuous (but not lower semi-continuous) while the
second is lower semicontinuous (but not upper semi-continuous).

The following theorem (Berge's Theorem, i.e. Th. 1.8.4 in Debreu, 1959) is very useful. Let ｦ : R^{n} ｴ R^{m} ｮ R and j : R^{m} ｮ R^{n} (i.e. ｦ is a function
and j a correspondence). Then let:

f(t) = max ｦ (x, t) s.t. x ﾎ j (x)

F(t) = argmax ｦ (x, t) s.t. x ﾎ j (x).

thus, f(t) is a function and F(t) is a correspondence (for "argmax" read "the argument (x) that maximizes ｦ (x, t)"). Then the following holds:

Theorem: (Berge) If ｦ (x, t) is a jointly continuous function and j (x) is both an upper semicontinuous and lower semicontinuous correspondence (i.e. a continuous correspondence), then: (i) f(t) is a continuous function and (ii) F(t) is an upper semicontinuous correspondence.

Proof: Omitted. See Hildenbrand (1974: p.30) or Border (1985: p.64)

Kim C. Border (1985) *Fixed Point Theorems with Applications to Economics and Game
Theory*. Cambridge, UK: Cambridge University Press.

Gerard Debreu (1959) *Theory of Value: An axiomatic analysis of economic equilibrium*.
New York: Wiley.

Werner Hildenbrand (1974) *Core and Equilibria of a Large Economy*. Princeton:
Princeton University Press.

Hukukane Nikaido (1968) *Convex Structures and Economic Theory*. New York:
Academic Press.

Back | Top | Next |