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free/or 0.5$ |
| m83186 | The set of solutions to a system of linear inequalities Ax ≤ c is a convex set. |
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| m83187 | The set S = {x1, x2, . . . , xn} is affinely dependent if and only if the set {x2 - x1, x3 - x1, . . . , xn - x1} is linearly dependent.
Exercise 1.157 implies that the maximum number of affinely independent elements in an n dimensional space is n + 1. Moreover the maximum dimension of a proper affine subset is n. Analogous to exercise 1.133, we have the following alternative characterization of affine dependence. |
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| m83188 | The set S = {x1. x2. . . . , xn} is affinely dependent if and only if there exist numbers a1, a2, . . . , an, not all zero, such that
a1x1 + a2x2 +...........+ anxn = 0
with a1 + a2 +..........+ an = 0.
Analogous to a basis, every vector in the affine hull of a set has a unique representation as an affine combination of the elements of the set. |
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| m83191 | The space BL(X, Y) of all bounded linear functions from X to Y is a normed linear space, with norm
It is a Banach space (complete normed linear space) if Y is complete. The following proposition is an important result regarding bounded linear functions. |
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| m83194 | The system A is productive if and only if A-1 exists and is nonnegative. |
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| m83211 | The unit simplex in ℜ |
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| m83228 | There is no arbitrage if and only if there exist state prices. |
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| m83239 | To construct a suitable space for the application of a separation argument, consider the set of points
where eS is characteristic vector of the coalition S (example 3.19) and w(S) is its worth. Let A be the conic hull of A0, that is,
Let B be the interval
Clearly, A and B are convex and nonempty.
We assume that the game is balanced and construct a payoff in the core.
1. Show that A and B are disjoint if the game is balanced.
2. Consequently there exists a hyperplane that separates A and B. That is, there exists a nonzero vector (z, z0) ∈ ℜn × ℜ such that
for all y ∈ A and all ε > 0. Show that
a. (e∅, 0) ∈ A implies that c = 0.
b. (eN, w(N) ∈ A implies that z0 < 0. Without loss of generality, we can normalize so that z0 = -1.
3. Show that (36) implies that the payoff vector z satisfies the inequalities
Therefore z belongs to the core. |
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| m83240 | To prove corollary 2.4.1, let f: X → X be an increasing function on a complete lattice (X, ≿), and let E be the set of fixed points of f. For any S ⊆ E define
S* = {x ∊ X : x ≿ s for every s ∊ S}
S* is the set of all upper bounds of S in X. Show that
1. S* is a complete sublattice.
2. f (S*) ⊆ S*.
3. Let g be the restriction of f to the sublattice S*. g has a least fixed point .
4. is the least upper bound of S in E.
5. E is a complete lattice. |
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| m83241 | To prove corollary 2.4.2, let S c E and s* = sup S.
1. For every x ∊ S there exists some zx ∊ p(s*) such that zx ≿ x.
2. Let z* = sup zx. Then
a. z* ≾ s*
b. z* ∊ |
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| m83242 | To prove lemma 1.1, assume, to the contrary, that for every c > 0 there exists x ∈ lin{x1; x2; . . . , xn} such that
where x = a1x1 + a2x2 + • • • +anxn. Show that this implies that
1. There exists a sequence (x ) with |x ||→0
2. There exists a subsequence converging to some x ∈ lin {x1,x2,...xn}
3. x ≠ 0 contradicting the conclusion that ||x || → 0
This contradiction proves the existence of a constant c > 0 such that ||x|| ≥ c(|α1| + |α2| + ... +|αn|) for every x ∈ lin S. |
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| m83243 | To show that each of the hypotheses of Brouwer s theorem is necessary, find examples of functions f : S → S with S ⊆ R that do not have fixed points, where
1. f is continuous and S is convex but not compact
2. f is continuous and S is compact but not convex
3. S is compact and convex but f is not continuous
The following proposition, which is equivalent to Brouwer s theorem, asserts that it is impossible to map the unit ball continuously on to its boundary. |
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| m83254 | Two inputs are said to complementary if their cross-partial derivative D2xixjf(x) is positive, since this means that increasing the quantity of one input increases the marginal productivity of the other. Show that if a production function of two inputs is linearly homogeneous and quasi concave, then the inputs are complementary. |
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| m83256 | Two norms ||x||a and ||x||b on a linear space are equivalent if there are positive numbers A and B such that for all x ∈ X,
The following exercise shows that there essentially only one finite dimensional normed linear space. |
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| m83257 | Two players i and j are substitutes in a game (N, w) if their contributions to all coalitions are identical, that is, if
W(S ∪ {i}) = w(S ∪ {j} for every S ⊆ N {i, j}
Verify that the Shapley value treats substitutes symmetrically, that is i j substitutes ⇒ ϕiw = ϕjw |
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| m83260 | Under what circumstances, if any, could the IS curve be horizontal? |
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| m83261 | Union and intersection are one way of generating new sets from old. Another way of generating new sets is by welding together sets of disparate objects into another set called their product. The product of two sets X and Y is the set of ordered pairs
X × Y = {(x, y): x ∈ X, y ∈ Y}
A familiar example is the coordinate plane R × R which is denoted R2 (figure 1.2). This correspondence between points in the plane and ordered pairs x; y of real numbers is the foundation of analytic geometry. Notice how the order matters. (1, 2) and (2, 1) are different elements of R2.
Figure 1.2
The coordinate plane R2
The product readily generalizes to many sets, so that
X1 × X2 ×. . . . . . . .× Xn = {(x1, X2, . . . , xn} : xi ∈ Xi}
is the set of all ordered lists of elements of Xi, and Rn = {(x1, X2, . . . , xn) : xi A Rg is the set of all ordered lists of n real numbers. An ordered list of n elements is called an n-tuple. Rn and its nonnegative subset Rn+ provide the domain of most economic quantities, such as commodity bundles and price lists. To remind ourselves when we are dealing with a product space, we will utilize boldface to distinguish the elements of a product space from the elements of the constituent sets, as in
x = (x1, x2, . . . , xn) ∈ X
where X = X1 × X2 ×. . . . . . . . . . . Xn. |
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| m83342 | Use proposition 2.3 to provide an alternative proof of theorem 2.2
Theorem 2.2
A continuous functional on a compact set achieves a maximum and a minimum.
Proof Let M = supx∊X f (x). There exists a sequence xn in X with f (xn) → M. Since X is compact, there exists a convergent subsequence xm → x* and f (xm) → M. However, since f is continuous, f (xm) → f (x*). We conclude that f (x*) = M. |
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| m83343 | Use proposition 3.14 directly to prove the Farkas lemma when X is a Hilbert space. |
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| m83363 | Use the Bolzano-Weierstrass theorem to show that R is complete.
The following proposition is regarded as the most important theorem in topology. We give a simplified version for the product of two metric spaces. By induction, it generalizes to any finite product. In fact the theorem is also true of an infinite product of compact spaces. |
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