The respective inverses are:

𝜑⁻¹₂ 𝜑⁺₂ 𝜑⁻₂

{𝑡1} ∅ ∅ {𝑠1}

{𝑡₂} ∅ ∅ {𝑠₁, 𝑠₂} {𝑡1, 𝑡2} {𝑠1} {𝑠1} {𝑠1, 𝑠2} {𝑡2, 𝑡3} {𝑠2} {𝑠2} {𝑠1, 𝑠2} {𝑡₁, 𝑡₂, 𝑡₃} ∅ {𝑠₁, 𝑠₂} {𝑠₁, 𝑠₂}

2.100 Let 𝑇 be an open interval meeting 𝜑(1), that is 𝜑(1) ∩ 𝑇 ∕= ∅.Since 𝜑(1) = {1}, we must have 1 ∈ 𝑇 and therefore 𝜑(𝑥) ∩ 𝑇 ∕= ∅ for every 𝑥 ∈ 𝑋.Therefore 𝜑 is lhc at 𝑥 = 1.On the other hand, the open interval 𝑇 = (1/2, 3/2) contains 𝜑(1) but it does not contain 𝜑(𝑥) for any 𝑥 > 1.Therefore, 𝜑 is not uhc at 𝑥 = 1.

2.101 Choose any open set 𝑇 ⊆ 𝑌 and 𝑥 ∈ 𝑋.Since 𝜑(𝑥) = 𝐾 = 𝜑(𝑥^′) for every 𝑥, 𝑥^′∈ 𝑋

∙ 𝜑(𝑥) ⊆ 𝑇 if and only if 𝜑(𝑥^′) ⊆ 𝑇 for every 𝑥, 𝑥^′ ∈ 𝑋

∙ 𝜑(𝑥) ∩ 𝑇 ∕= ∅ if and only if 𝜑(𝑥^′) ∩ 𝑇 ∕= ∅ for every 𝑥, 𝑥^′ ∈ 𝑋.

Consequently, 𝜑 is both uhc and lhc at all 𝑥 ∈ 𝑋.

2.102 First assume that the 𝜑 is uhc.Let 𝑇 be any open subset in 𝑌 and 𝑆 = 𝜑⁺(𝑇 ).

If 𝑆 = ∅, it is open.Otherwise, choose 𝑥0 ∈ 𝑆 so that 𝜑(𝑥0) ⊆ 𝑇 .Since 𝜑 is uhc, there exists a neighborhood 𝑆(𝑥0) such that 𝜑(𝑥) ⊆ 𝑇 for every 𝑥 ∈ 𝑆(𝑥0).That is, 𝑆(𝑥0) ⊆ 𝜑⁺(𝑇 ) = 𝑆.This establishes that for every 𝑥0∈ 𝑆 there exist a neighborhood 𝑆(𝑥0) contained in 𝑆.That is, 𝑆 is open in 𝑋.

Conversely, assume that the upper inverse of every open set in 𝑌 is open in 𝑋.Choose some 𝑥0∈ 𝑋 and let 𝑇 be an open set containing 𝜑(𝑥0).Let 𝑆 = 𝜑⁺(𝑇 ). 𝑆 is an open set containing 𝑥0.That is, 𝑆 is a neighborhood of 𝑥0 with 𝜑(𝑥) ⊆ 𝑇 for every 𝑥 ∈ 𝑆.

Since the choice of 𝑥₀was arbitrary, we conclude that 𝜑 is uhc.

The lhc case is analogous.

2.103 Assume 𝜑 is uhc and 𝑇 be any closed set in 𝑌 .By Exercise 2.97 𝜑⁻(𝑇 ) =[

𝜑⁺(𝑇^𝑐)]

𝑇^𝑐 is open.By the previous exercise, 𝜑⁺(𝑇^𝑐) is open which implies that 𝜑⁻(𝑇 ) is closed.

Conversely, assume 𝜑⁻(𝑇 ) is closed for every closed set 𝑇 .Let 𝑇 be an open subset of 𝑌 so that 𝑇^𝑐 is closed.Again by Exercise 2.97,

𝜑⁺(𝑇 ) =[

𝜑⁻(𝑇^𝑐)]

By assumption 𝜑⁻(𝑇^𝑐) is closed and therefore 𝜑⁺(𝑇 ) is open.By the previous exercise, 𝜑 is uhc.

The lhc case is analogous.

2.104 Assume that 𝜑 is uhc at 𝑥₀.We first show that (𝑦^𝑛) is bounded and hence has a convergent subsequence.Since 𝜑(𝑥₀) is compact, there exists a bounded open set 𝑇 containing 𝜑(𝑥₀).Since 𝜑 is uhc, there exists a neighborhood 𝑆 of 𝑥₀such that 𝜑(𝑥) ⊆ 𝑇 for 𝑥 ∈ 𝑆.Since 𝑥^𝑛 → 𝑥₀, there exists some 𝑁 such that 𝑥^𝑛 ∈ 𝑆 for every 𝑛 ≥ 𝑁.

Consequently, 𝜑(𝑥^𝑛) ⊆ 𝑇 for every 𝑛 ≥ 𝑁 and therefore 𝑦^𝑛 ∈ 𝑇 for every 𝑛 ≥ 𝑁.

The sequence 𝑦^𝑛 is bounded and hence has a convergent subsequence 𝑦^𝑚→ 𝑦₀. To complete the proof, we have to show that 𝑦₀ ∈ 𝜑(𝑥₀).Assume not, assume that 𝑦0 /∈ 𝜑(𝑥0).Then, there exists an open set 𝑇 containing 𝜑(𝑥0) such that 𝑦0 /∈ 𝑇 (Exercise 1.93). Since 𝜑 is uhc, there exists 𝑁 such that 𝜑(𝑥^𝑛) ⊆ 𝑇 for every 𝑛 ≥ 𝑁.

This implies that 𝑦^𝑚∈ 𝑇 for every 𝑚 ≥ 𝑁.Since 𝑦^𝑚→ 𝑦0, we conclude that 𝑦0∈ 𝑇 , contradicting the specification of 𝑇 .

Conversely, suppose that for every sequence 𝑥^𝑛 → 𝑥0, 𝑦^𝑛 ∈ 𝜑(𝑥^𝑛), there is a subse-quence of 𝑦^𝑚 → 𝑦0 ∈ 𝜑(𝑥0).Suppose that 𝜑 is not uhc at 𝑥0.That is, there exists an open set 𝑇 ⊇ 𝜑(𝑥0) such that every neighborhood contains some 𝑥 with 𝜑(𝑥) ∕⊆ 𝑇 . From the sequence of neighborhoods 𝐵_1/𝑛(𝑥₀), we can construct a sequence 𝑥^𝑛 → 𝑥 and 𝑦^𝑛 ∈ 𝜑(𝑥^𝑛) but 𝑦^𝑛 /∈ 𝑇 .Such a sequence cannot have a subsequence which con-verges to 𝑦⁰∈ 𝜑(𝑥₎, contradicting the hypothesis.We conclude that 𝜑 must be uhc at 𝑥₀.

2.105 Assume that 𝜑 is lhc.Let 𝑥^𝑛 be a sequence converging to 𝑥₀ and 𝑦₀ ∈ 𝜑(𝑥₀).

Consider the sequence of open balls 𝐵_1/𝑚(𝑦0), 𝑚 = 1, 2, . . . .Note that every 𝐵_1/𝑚(𝑦0) meets 𝜑(𝑥0).Since 𝜑 is lhc, there exists a sequence (𝑆^𝑚) of neighborhoods of 𝑥0 such that 𝜑(𝑥) ∩ 𝐵_1/𝑚∕= ∅ for every 𝑥 ∈ 𝑆^𝑚.Since 𝑥^𝑛→ 𝑥, for every 𝑚, there exists some 𝑁𝑚 such that 𝑥^𝑛 ∈ 𝑆𝑚 for every 𝑛 ≥ 𝑁𝑚.Without loss of generality, we can assume that 𝑁1 < 𝑁2 < 𝑁3. . . .We can now construct the desired sequence 𝑦^𝑛.For each 𝑛 = 1, 2, . . . , choose 𝑦^𝑛 in the set 𝜑(𝑥^𝑛) ∩ 𝐵_1/mwhere 𝑁𝑚≤ 𝑛 ≤ 𝑁𝑚+1since

𝑛 ≥ 𝑁𝑚 =⇒ 𝑥^𝑛 ∈ 𝑆𝑚 =⇒ 𝜑(𝑥^𝑛) ∩ 𝐵_1/𝑚∕= ∅ Since 𝑦^𝑛∈ 𝐵_1/m(𝑦0), the sequence (𝑦^𝑛) converges to 𝑦0and 𝑛 → ∞.

Conversely, assume that 𝜑 is not lhc at 𝑥0, that is there exists an open set 𝑇 with 𝑇 ∩𝜑(𝑥0) ∕= ∅ such that every neighborhood 𝑆 ∋ 𝑥0contains some 𝑥 with 𝜑(𝑥)∩𝑇 = ∅.

Therefore, there exists a sequence 𝑥^𝑛 → 𝑥 with 𝜑(𝑥)∩𝑇 = ∅.Choose any 𝑦0∈ 𝜑(𝑥0)∩𝑇 . By assumption, there exists a sequence 𝑦^𝑛→ 𝑦 with 𝑦^𝑛∈ 𝜑(𝑥^𝑛).Since 𝑇 is open and 𝑦₀∈ 𝑇 , there exists some 𝑁 such that 𝑦^𝑛∈ 𝑇 for all 𝑛 ≥ 𝑁, for which 𝜑(𝑦^𝑛) ∩ 𝑇 ∕= ∅.

This contradiction establishes that 𝜑 is lhc at 𝑥₀.

2.106 1.Assume 𝜑 is closed.For any 𝑥 ∈ 𝑋, let (𝑦^𝑛) be a sequence in 𝜑(𝑥).Since 𝜑 is closed, 𝑦^𝑛 → 𝑦 ∈ 𝜑(𝑥).Therefore 𝜑(𝑥) is closed.

2.Assume 𝜑 is closed-valued and uhc.Choose any (𝑥, 𝑦) /∈ graph(𝜑).Since 𝜑(𝑥) is closed, there exist disjoint open sets 𝑇1and 𝑇2in 𝑌 such that 𝑦 ∈ 𝑇1and 𝜑(𝑥) ⊆ 𝑇₂ (Exercise 1.93). Since 𝜑 is uhc, 𝜑⁺(𝑇₂) is a neighborhood of 𝑥.Therefore 𝜑⁺(𝑇₂) × 𝑇₁ is a neighborhood of (𝑥, 𝑦) disjoint from graph(𝜑).Therefore the complement of graph(𝜑) is open, which implies that graph(𝜑) is closed.

3.Since 𝜑 is closed and 𝑌 compact, 𝜑 is compact-valued.Let (𝑥^𝑛) → 𝑥 be a sequence in 𝑋 and (𝑦^𝑛) a sequence in 𝑌 with 𝑦^𝑛 ∈ 𝜑(𝑥^𝑛).Since 𝑌 is compact, there exists a subsequence 𝑦^𝑚 → 𝑦.Since 𝜑 is closed, 𝑦 ∈ 𝜑(𝑥).Therefore, by Exercise 2.104, 𝜑 is uhc.

2.107 Assume 𝜑 is closed-valued and uhc.Then 𝜑 is closed (Exercise 2.106). Con-versely, if 𝜑 is closed, then 𝜑(𝑥) is closed for every 𝑥 (Exercise 2.106). If 𝑌 is compact, then 𝜑 is compact-valued (Exercise 1.110). By Exercise 2.104, 𝜑 is uhc.

2.108 𝜑1 is closed-valued (Exercise 2.106). Similarly, 𝜑2 is closed-valued (Proposition 1.1). Therefore, for every 𝑥 ∈ 𝑋, 𝜑(𝑥) = 𝜑1(𝑥) ∩ 𝜑2(𝑥) is closed (Exercise 1.85) and hence compact (Exercise 1.110). Hence 𝜑 is compact-valued.

Now, for any 𝑥0∈ 𝑋, let 𝑇 be an open neighborhood of 𝜑(𝑥0).We need to show that there is a neighborhood 𝑆 of 𝑥₀ such that 𝜑(𝑆) ⊆ 𝑇 .

Case 1 𝑇 ⊇ 𝜑₂(𝑥₀): Since 𝜑₂ is uhc, there exists a neighborhood such that 𝑆 ∋ 𝑥₀ such that 𝜑₂(𝑆) ⊆ 𝑇 which implies that 𝜑(𝑆) ⊆ 𝜑₂(𝑆) ⊆ 𝑇

Case 2 𝑇 ∕⊇ 𝜑₂(𝑥₀): Let 𝐾 = 𝜑₂(𝑥₀) ∖ 𝑇 ∕= ∅.For every 𝑦 ∈ 𝐾, there exist neighbor-hoods 𝑆_𝑦(𝑥₀) and 𝑇 (𝑦) such that 𝜑₁(𝑆_𝑦(𝑥₀))∩𝑇 (𝑦) = ∅ (Exercise 1.93). The sets 𝑇 (𝑦) constitute an open covering of 𝐾.Since 𝐾 is compact, there exists a finite subcover, that is there exists a finite number of elements 𝑦1, 𝑦2, . . . 𝑦𝑛 such that

𝐾 ⊆∪^𝑛

𝑖=1

𝑇 (𝑦𝑖)

Let 𝑇 (𝐾) denote∪_𝑛

𝑖=1𝑇 (𝑦𝑖).Note that 𝑇 ∪𝑇 (𝐾) is an open set containing 𝜑2(𝑥0).

Since 𝜑₂ is uhc, there exists a neighborhood 𝑆^′(𝑥₀) such that 𝜑₂(𝑆^′(𝑥₀)) ⊆ 𝑇 ∪ 𝑇 (𝐾).Let

𝑆(𝑥₀) =∩^𝑛

𝑖=1

𝑆_𝑦𝑖(𝑥₀) ∩ 𝑆^′(𝑥₀)

𝑆(𝑥0) is an open neighborhood of 𝑥0for which

𝜑1(𝑆(𝑥0)) ∩ 𝑇 (𝐾) = ∅ and 𝜑2(𝑆(𝑥0)) ⊆ 𝑇 ∪ 𝑇 (𝐾) from which we conclude that

𝜑(𝑆(𝑥0)) = 𝜑1(𝑆(𝑥0)) ∩ 𝜑2(𝑆(𝑥0)) ⊆ 𝑇 2.109 1.Let x ∈ 𝑋(p, 𝑚) ∩ 𝑇 .Then x ∈ 𝑋(p, 𝑚) and ∑_𝑛

𝑖=1𝑝𝑖𝑥𝑖 ≤ 𝑚.Since 𝑇 is open, there exists 𝛼 < 1 such that ˜x = 𝛼x ∈ 𝑇 and

∑𝑛 𝑖=1

𝑝𝑖˜𝑥𝑖= 𝛼

∑𝑛 𝑖=1

𝑝𝑖𝑥𝑖<

∑𝑛 𝑖=1

𝑝𝑖𝑥𝑖≤ 𝑚

2.(a) Suppose that 𝑋(p, 𝑚) is not lhc.Then for every neighborhood 𝑆 of (p, 𝑚), there exists (p^′, 𝑚^′) ∈ 𝑆 such that 𝑋(p^′, 𝑚^′) ∩ 𝑇 = ∅.In particular, for every open ball 𝐵𝑛(p, 𝑚), there exists a point (p^𝑛, 𝑚^𝑛) ∈ 𝐵𝑛(p, 𝑚) such that 𝑋(p^𝑛, 𝑚^𝑛) ∩ 𝑇 = ∅.((p^𝑛, 𝑚^𝑛)) is the required sequence.

(b) By construction, ∥p^𝑛− p∥ < 1/𝑛 → 0 which implies that 𝑝^𝑛_𝑖 → 𝑝_𝑖for every 𝑖.Therefore (Exercise 1.202)

∑𝑝^𝑛_𝑖˜𝑥_𝑖→∑

𝑝_𝑖˜𝑥_𝑖 < 𝑚 and 𝑚^𝑛→ 𝑚 and therefore there exists 𝑁 such that

∑𝑝^𝑁_𝑖 ˜𝑥𝑖< 𝑚^𝑁

which implies that

˜x ∈ 𝑋(p^𝑁, 𝑚^𝑁)

(c) Also by construction 𝑋(p^𝑁, 𝑚^𝑁) ∩ 𝑇 = ∅ which implies 𝑋(p^𝑁, 𝑚^𝑁) ⊆ 𝑇^𝑐 and therefore

˜x ∈ 𝑋(p^𝑛, 𝑚^𝑛) =⇒ ˜x /∈ 𝑇

The assumption that 𝑋(p, 𝑚) is not lhc at (p, 𝑚) implies that ˜x /∈ 𝑇 , contra-dicting the conclusion in part 1 that ˜x ∈ 𝑇 .

3.This contradiction establishes that (p, 𝑚) is lhc at (p, 𝑚).Since the choice of (p, 𝑚) was arbitrary, we conclude that the budget correspondence 𝑋(p, 𝑚) is lhc for all (p, 𝑚) ∈ 𝑃 (assuming 𝑋 = ℜ^𝑛₊).

4.In the previous example (Example 2.89), we have shown that 𝑋(p, 𝑚) is uhc.

Hence, the budget correspondence is continuous for all (p, 𝑚) such that 𝑚 >

inf_x∈𝑋∑_𝑚

𝑖=1𝑝_𝑖𝑥_𝑖.

2.110 We give two alternative proofs.

Proof 1 Let 𝒞 = {𝑆} be an open cover of 𝜑(𝐾).For every 𝑥 ∈ 𝐾, 𝜑(𝑥) ⊆ 𝜑(𝐾) is compact and hence can be covered by a finite number of the sets 𝑆 ∈ 𝒞.Let 𝑆𝑥 denote the union of the finite cover of 𝜑(𝑥).Since 𝜑 is uhc, every 𝜑⁺(𝑆𝑥) is open in 𝑋.Therefore { 𝜑⁺(𝑆𝑥) : 𝑥 ∈ 𝐾 } is an open covering of 𝐾.If 𝐾 is compact, it contains an finite covering { 𝜑⁺(𝑆𝑥1), 𝜑⁺(𝑆𝑥2), . . . , 𝜑⁺(𝑆𝑥𝑛) }.The sets 𝑆𝑥1, 𝑆𝑥2, . . . , 𝑆𝑥𝑛 are a finite subcovering of 𝜑(𝐾).

Proof 2 Let (𝑦^𝑛) be a sequence in 𝜑(𝐾).We have to show that (𝑦^𝑛) has a convergent subsequence with a limit in 𝜑(𝐾).For every 𝑦^𝑛, there is an 𝑥^𝑛 with 𝑦^𝑛 ∈ 𝜑(𝑥^𝑛).Since 𝐾 is compact, the sequence (𝑥^𝑛) has a convergent subsequence 𝑥^𝑚→ 𝑥 ∈ 𝐾.Since 𝜑 is uhc, the sequence (𝑦^𝑚) has a subsequence (𝑦^𝑝) which converges to 𝑦 ∈ 𝜑(𝑥) ⊆ 𝜑(𝐾).Hence the original sequence (𝑦^𝑛) has a convergent subsequence.

2.111 The sets 𝑋, 𝜑(𝑋), 𝜑²(𝑋), . . . form a sequence of nonempty compact sets.Since 𝜑(𝑋) ⊆ 𝑋, 𝜑²(𝑋) ⊆ 𝜑(𝑋) and so on, the sequence of sets 𝜑^𝑛𝑋 is decreasing.Let

𝐾 =

∩∞ 𝑛=1

𝜑^𝑛(𝑋)

By the nested intersection theorem (Exercise 1.117), 𝐾 ∕= ∅.Since 𝐾 ⊆ 𝜑^𝑛−1(𝑋), 𝜑(𝐾) ⊆ 𝜑^𝑛(𝑋) for every 𝑛, which implies that 𝜑(𝐾) ⊆ 𝐾.

To show that 𝐾 ⊆ 𝜑(𝐾), let 𝑦 ∈ 𝐾.For every 𝑛 there exists an 𝑥^𝑛 ∈ 𝜑^𝑛(𝑋) such that 𝑦 ∈ 𝜑(𝑥^𝑛).Since 𝑋 is compact, there exists a subsequence 𝑥^𝑚 → 𝑥0.Since 𝑥^𝑚∈ 𝜑^𝑚(𝑋) for every 𝑚, 𝑥0∈ 𝐾.The sequence (𝑥^𝑚, 𝑦) → (𝑥0, 𝑦).Since 𝜑 is closed (Exercise 2.107), 𝑦 ∈ 𝜑(𝑥0).Therefore 𝑦 ∈ 𝜑(𝐾) which implies that 𝐾 ⊆ 𝜑(𝐾).

2.112 𝜑(𝑥) is compact for every 𝑥 ∈ 𝑋 by Tychonoff’s theorem (Proposition 1.2).

Let 𝑥^𝑘 → 𝑥 be a sequence in 𝑋 and let 𝑦^𝑘 = (𝑦₁^𝑘, 𝑦₂^𝑘, . . . , 𝑦^𝑘_𝑛) with 𝑦_𝑖^𝑘 ∈ 𝜑(𝑥^𝑘) be a corresponding sequence of points in 𝑌 .For each 𝑦^𝑘_𝑖, 𝑖 = 1, 2, . . . , 𝑛, there exists a subsequence 𝑦_𝑖^𝑘′ → 𝑦𝑖with 𝑦𝑖∈ 𝜑𝑖(𝑥) (Exercise 2.104). Therefore 𝑦 = (𝑦1, 𝑦2, . . . , 𝑦𝑛) ∈ 𝜑(𝑥) which implies that 𝜑 is uhc.

2.113 Let 𝑣 ∈ 𝐶(𝑋).For every x ∈ 𝑋, the maximand 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) is a continuous function on a compact set 𝐺(𝑥).Therefore the supremum is attained, and max can replace sup in the definition of the operator 𝑇 (Theorem 2.2). 𝑇 𝑣 is the value function for the constrained optimization problem

𝑦∈𝐺(𝑥)max { 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) }

satisfying the requirements of the continuous maximum theorem (Theorem 2.3), which ensures that 𝑇 𝑣 is continuous on 𝑋.We have previously shown that 𝑇 𝑣 is bounded (Exercise 2.18). Therefore 𝑇 𝑣 ∈ 𝐶(𝑋).

2.114 1. 𝑆 has a least upper bound since 𝑋 is a complete lattice.Let 𝑠^∗ = sup 𝑆.

Then 𝑆^∗= ≿(𝑠^∗) is a complete sublattice of 𝑋 (Exercise 1.48).

2.For every 𝑠 ∈ 𝑆, 𝑠 ≾ 𝑠^∗ and since 𝑓 is increasing and 𝑠 is a fixed point 𝑠 = 𝑓(𝑠) ≾ 𝑓(𝑠^∗)

Therefore 𝑓(𝑠^∗) ∈ 𝑆^∗. (𝑓(𝑠^∗) is an upper bound of 𝑆).Again, since 𝑓 is increas-ing, this implies that 𝑓(𝑥) ≿ 𝑓(𝑠^∗) for every 𝑥 ∈ 𝑆^∗.Therefore 𝑓(𝑆^∗) ⊆ 𝑆^∗. 3.Let 𝑔 be the restriction of 𝑓 to the sublattice 𝑆^∗.Since 𝑓(𝑆^∗) ⊆ 𝑆^∗, 𝑔 is an

increasing function on a complete lattice.Applying Theorem 2.4, 𝑔 has a smallest fixed point ˜𝑥.

4.˜𝑥 is a fixed point of 𝑓, that is ˜𝑥 ∈ 𝐸.Furthermore, ˜𝑥 ∈ 𝑆^∗.Therefore ˜𝑥 is an upper bound for 𝑆 in 𝐸.Moreover, ˜𝑥 is the smallest fixed point of 𝑓 in 𝑆^∗. Therefore, ˜𝑥 is the least upper bound of 𝑆 in 𝐸.

5.By Exercise 1.47, this implies that 𝐸 is a complete lattice.

In Example 2.91, if 𝑆 = {(2, 1), (1, 2)}, 𝑆^∗= {(2, 2), (3, 2), (2, 3), (3, 3)} and ˜𝑥 = (3, 3).

2.115 1.For every 𝑥 ∈ 𝑀, there exists some 𝑦𝑥∈ 𝜑(𝑥) such that 𝑦𝑥≾ 𝑥.Moreover, since 𝜑 is increasing and ˜𝑥 ≾ 𝑥, there exists some 𝑧𝑥∈ 𝜑(˜𝑥) such that

𝑧_𝑥≾ 𝑦_𝑥≾ 𝑥 for every 𝑥 ∈ 𝑀 2.Let ˜𝑧 = inf{𝑧_𝑥}

(a) Since 𝑧_𝑥≾ 𝑥 for every 𝑥 ∈ 𝑀, ˜𝑧 = inf{𝑧_𝑥} ≾ inf{𝑥} = ˜𝑥.

(b) Since 𝜑(˜𝑥) is a complete sublattice of 𝑋, ˜𝑧 = inf{𝑧𝑥} ∈ 𝜑(˜𝑥).

3.Therefore, ˜𝑥 ∈ 𝑀.

4.Since ˜𝑧 ≾ ˜𝑥 and 𝜑 is increasing, there exists some 𝑦 ∈ 𝜑(˜𝑧) such that 𝑦 ≾ ˜𝑧 ∈ 𝜑(˜𝑥)

Hence ˜𝑧 ∈ 𝑀.

5.This implies that ˜𝑥 ≾ ˜𝑧.Therefore

˜𝑥 = ˜𝑧 ∈ 𝜑(˜𝑥)

˜𝑥 is a fixed point of 𝜑.

6.Since 𝐸 ⊆ 𝑀, ˜𝑥 = inf 𝑀 is the least fixed point of 𝜑.

2.116 1.Let 𝑆 ⊆ 𝐸 and 𝑠^∗ = sup 𝑆.For every 𝑥 ∈ 𝑆, 𝑥 ∈ 𝜑(𝑥).Since 𝜑 is increasing, there exists some 𝑧_𝑥∈ 𝜑(𝑠^∗) such that 𝑧_𝑥≿ 𝑥.

2.Let 𝑧^∗= sup 𝑧_𝑥.Then

(a) Since 𝑧_𝑥≿ 𝑥 for every 𝑥 ∈ 𝑆, 𝑧^∗= sup 𝑧_𝑥≿ sup 𝑥 = 𝑠^∗ (b) 𝑧^∗∈ 𝜑(𝑠^∗) since 𝜑(𝑠^∗) is a complete sublattice.

3.Define

𝑆^∗= { 𝑥 ∈ 𝑋 : 𝑥 ≿ 𝑠 for every 𝑠 ∈ 𝑆 }

𝑆^∗is the set of all upper bounds of 𝑆 in 𝑋.Then 𝑆^∗ is a complete lattice, since 𝑆^∗= ≿(𝑠^∗)

4.Let 𝜇: 𝑆^∗⇉ 𝑆^∗ be the correspondence

𝜇(𝑥) = 𝜑(𝑥) ∩ 𝜓(𝑥)

where 𝜓 : 𝑆^∗⇉ 𝑆^∗is the constant correspondence defined by 𝜓(𝑥) = 𝑆^∗for every 𝑥 ∈ 𝑆^∗.Then

(a) Since 𝜑 is increasing, for every 𝑥 ≿ 𝑠^∗, there exists some 𝑦𝑥 ∈ 𝜑(𝑥) such that 𝑦𝑥≿ 𝑠^∗.Therefore 𝜇(𝑥) ∕= ∅ for every 𝑥 ∈ 𝑆^∗.

(b) Both 𝜑(𝑥) and 𝜓(𝑥) are complete sublattices for every 𝑥 ∈ 𝑆^∗.Therefore 𝜇(𝑥) is a complete sublattice for every 𝑥 ∈ 𝑆^∗.

(c) Since both 𝜑 and 𝜓 are increasing on 𝑆^∗, 𝜇 is increasing on 𝑆^∗ (Exercise 2.47).

5.By the previous exercise, 𝜇 has a least fixed point ˜𝑥.

6.˜𝑥 ∈ 𝑆^∗ is an upper bound of 𝑆.Therefore ˜𝑥 is the least upper bound of 𝑆 in 𝐸.

7.By the previous exercise, 𝐸 has a least element.Since we have shown every subset 𝑆 ⊆ 𝐸 has a least upper bound, this establishes that 𝐸 is complete lattice (Exercise 1.47).

2.117 For any 𝑖, let a¹_−𝑖, a²_−𝑖∈ 𝐴−𝑖with a²_−𝑖≿ a¹_−𝑖.Let ¯𝑎¹_𝑖 = 𝑓(a¹_−𝑖) and ¯𝑎²_𝑖 = 𝑓(a²_−𝑖).

We want to show that ¯𝑎²_𝑖 ≿ ¯𝑎¹_𝑖.Since ¯𝑎¹_𝑖 ∈ 𝐵(a¹_−𝑖) and 𝐵(a−𝑖) is increasing, there exists some 𝑎𝑖∈ 𝐵(a²_−𝑖) such that 𝑎𝑖≿ ¯𝑎¹_𝑖.(Exercise 2.44).Therefore

sup 𝐵(a_−𝑖) = ¯𝑎²_𝑖 ≿ 𝑎_𝑖≿ ¯𝑎¹_𝑖 𝑓¯_𝑖 is increasing.

2.118 For any player 𝑖, their best response correspondence 𝐵_𝑖(a_−𝑖) is 1.increasing by the monotone maximum theorem (Theorem 2.1).

2.a complete sublattice of 𝐴𝑖 for every a−𝑖∈ 𝐴−𝑖(Corollary 2.1.1).

The joint best response correspondence

𝐵(a) = 𝐵1(a−1) × 𝐵2(a−2) × ⋅ ⋅ ⋅ × 𝐵𝑛(a−𝑛) is also

1.increasing (Exercise 2.46)

2.a complete sublattice of 𝐴 for every a ∈ 𝐴

Therefore, the best response correspondence 𝐵(a) satisfies the conditions of Zhou’s theorem, which implies that the set 𝐸 of fixed points of 𝐵 is a nonempty complete lattice. 𝐸 is precisely the set of Nash equilibria of the game.

2.119 In proving the theorem, we showed that

𝜌(𝑥^𝑛, 𝑥^𝑛+𝑚) ≤ 𝛽^𝑛

1 − 𝛽𝜌(𝑥⁰, 𝑥¹) for every 𝑚, 𝑛 ≥ 0.Letting 𝑚 → ∞, 𝑥^𝑛+𝑚→ 𝑥 and therefore

𝜌(𝑥^𝑛, 𝑥) ≤ 𝛽^𝑛

1 − 𝛽𝜌(𝑥⁰, 𝑥¹) Similarly, for every 𝑛, 𝑚 ≥ 0

𝜌(𝑥^𝑛, 𝑥^𝑛+𝑚) ≤ 𝜌(𝑥^𝑛, 𝑥^𝑛+1) + 𝜌(𝑥^𝑛+1, 𝑥^𝑛+2) + ⋅ ⋅ ⋅ + 𝜌(𝑥^{𝑛+𝑚−1}, 𝑥^𝑛+𝑚)

≤ (𝛽 + 𝛽²+ ⋅ ⋅ ⋅ + 𝛽^𝑚)𝜌(𝑥^𝑛−1, 𝑥^𝑛)

≤ 𝛽(1 − 𝛽^𝑚)

1 − 𝛽 𝜌(𝑥^𝑛−1, 𝑥^𝑛) Letting 𝑚 → ∞, 𝑥^𝑛+𝑚→ 𝑥 and 𝛽^𝑚→ 0 so that

𝜌(𝑥^𝑛, 𝑥) ≤ 𝛽

1 − 𝛽𝜌(𝑥^𝑛−1, 𝑥^𝑛)

2.120 First observe that 𝑓(𝑥) ≥ 1 for every 𝑥 ≥ 1.Therefore 𝑓 : 𝑋 → 𝑋.For any 𝑥, 𝑧 ∈ 𝑋

𝑓(𝑥) − 𝑓(𝑦)

𝑥 − 𝑦 = 𝑥 − 𝑦 +²_𝑥−_𝑦² 2(𝑥 − 𝑦) =1

2 − 1 𝑥𝑦 Since _𝑥𝑦¹ ≤ 1 for all 𝑥, 𝑦 ∈ 𝑋

−1

2 ≤ 𝑓(𝑥) − 𝑓(𝑦) 𝑥 − 𝑦 ≤ 1

2 so that

𝑓(𝑥) − 𝑓(𝑦) 𝑥 − 𝑦

 = ∣𝑓(𝑥) − 𝑓(𝑦)∣

∣𝑥 − 𝑦∣ ≤ 1 2 or

∣𝑓(𝑥) − 𝑓(𝑦)∣ ≤1 2∣𝑥 − 𝑦∣

𝑓 is a contraction on 𝑋 with modulus ¹/2.

𝑋 is closed and hence complete (Exercise 1.107). Therefore, 𝑓 has a fixed point.That is, there exists 𝑥0∈ 𝑋 such that

𝑥₀= 𝑓(𝑥₀) = 1

2(𝑥₀+ 2 𝑥₀) Rearranging

2𝑥²₀= 𝑥²₀+ 2 =⇒ 𝑥²₀= 2 so that 𝑥₀=√

2.

Letting 𝑥⁰= 2

𝑥¹= 1

2(2 + 1) = 3 2 Using the error bounds in Corollary 2.5.1,

𝜌(𝑥^𝑛,√

2) ≤ 𝛽^𝑛

1 − 𝛽𝜌(𝑥⁰, 𝑥¹)

=(1/2)^𝑛 1/2 1/2

= 1 2^𝑛

= 1

1024 < 0.001

when 𝑛 = 10.Therefore, we conclude that 10 iterations are ample to reduce the error below 0.001.Actually, with experience, we can refine this a priori estimate.In Example 1.64, we calculated the first five terms of the sequence to be

(2, 1.5, 1.416666666666667, 1.41421568627451, 1.41421356237469) We observe that

𝜌(𝑥³, 𝑥⁴) = 1.41421568627451 − 1.41421356237469) = 0.0000212389982 so that using the second inequality of Corollary 2.5.1

𝜌(𝑥⁴,√

2) ≤ 1/2

1/20.0000212389982 < 0.001 𝑥⁴= 1.41421356237469 is the desired approximation.

2.121 Choose any 𝑥⁰∈ 𝑆.Define the sequence 𝑥^𝑛 = 𝑓(𝑥^𝑛) = 𝑓^𝑛(𝑥⁰).Then (𝑥^𝑛) is a Cauchy sequence in 𝑆 converging to 𝑥.Since 𝑆 is closed, 𝑥 ∈ 𝑆.

2.122 By the Banach fixed point theorem, 𝑓^𝑁 has a unique fixed point 𝑥.Let 𝛽 be the Lipschitz constant of 𝑓^𝑁.We have to show

𝑥 is a fixed point of 𝑓

𝜌(𝑓(𝑥), 𝑥) = 𝜌(𝑓(𝑓^𝑁(𝑥), 𝑓^𝑁(𝑥)) = 𝜌(𝑓^𝑁(𝑓(𝑥), 𝑓^𝑁(𝑥)) ≤ 𝛽𝜌(𝑓(𝑥), 𝑥) Since 𝛽 < 1, this implies that 𝜌(𝑓(𝑥), 𝑥) = 0 or 𝑓(𝑥) = 𝑥.

𝑥 is the only fixed point of 𝑓 Suppose 𝑧 = 𝑓(𝑧) is another fixed point of 𝑓.Then 𝑧 is a fixed point of 𝑓^𝑁 and

𝜌(𝑥, 𝑧) = 𝜌(𝑓^𝑁(𝑥), 𝑓^𝑁(𝑧)) ≤ 𝛽𝜌(𝑥, 𝑧) which implies that 𝑥 = 𝑧.

2.123 By the Banach fixed point theorem, for every 𝜃 ∈ Θ, there exists 𝑥𝜃 ∈ 𝑋 such that 𝑓𝜃(𝑥𝜃) = 𝑥𝜃.Choose any 𝜃0∈ Θ.

𝜌(𝑥𝜃, 𝑥𝜃0) = 𝜌(𝑓𝜃(𝑥𝜃), 𝑓𝜃0(𝑥𝜃0))

≤ 𝜌(𝑓𝜃(𝑥𝜃), 𝑓𝜃(𝑥𝜃0)) + 𝜌(𝑓𝜃(𝑥𝜃0), 𝑓𝜃0(𝑥𝜃0))

≤ 𝛽𝜌(𝑥_𝜃, 𝑥_𝜃0) + 𝜌(𝑓_𝜃(𝑥_𝜃0), 𝑓_𝜃0(𝑥_𝜃0)) (1 − 𝛽)𝜌(𝑥𝜃, 𝑥𝜃0) ≤ 𝜌(𝑓𝜃(𝑥𝜃0), 𝑓𝜃0(𝑥𝜃0))

𝜌(𝑥𝜃, 𝑥𝜃0) ≤ 𝜌(𝑓_𝜃(𝑥_𝜃0), 𝑓_𝜃0(𝑥_𝜃0))

(1 − 𝛽) → 0

as 𝜃 → 𝜃₀.Therefore 𝑥_𝜃→ 𝑥_𝜃0.

2.124 1.Let x be a fixed point of 𝑓.Then x satisfies x = (𝐼 − 𝐴)x + c = x − 𝐴x + 𝑐 which implies that 𝐴x = 𝑐.

2.For any x¹, x²∈ 𝑋

𝑓(x¹) − 𝑓(x²) = (𝐼 − 𝐴)(x¹− x²)

≤ ∥𝐼 − 𝐴∥x¹− x²

Since 𝑎𝑖𝑖 = 1, the norm of 𝐼 − 𝐴 is

∥𝐼 − 𝐴∥ = max

𝑖

∑

𝑗∕=𝑖

∣𝑎𝑖𝑗∣ = 𝑘

and

𝑓(x¹) − 𝑓(x²) ≤ 𝑘x¹− x²

By the assumption of strict diagonal dominance, 𝑘 < 1.Therefore 𝑓 is a contrac-tion and has a unique fixed point x.

2.125 1.

𝜑(𝑥) = { 𝑦^∗∈ 𝐺(𝑥) : 𝑓(𝑥, 𝑦^∗) + 𝛽𝑣(𝑦^∗) = 𝑣(𝑥) }

= {𝑦^∗∈ 𝐺(𝑥) : 𝑓(𝑥, 𝑦^∗) + 𝛽𝑣(𝑦^∗) = sup

𝑦∈𝐺(𝑥){𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦)}}

= {𝑦^∗∈ 𝐺(𝑥) : 𝑓(𝑥, 𝑦^∗) + 𝛽𝑣(𝑦^∗) ≥ 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) for every 𝑦 ∈ 𝐺(𝑥)}

= arg max

𝑦∈𝐺(𝑥){𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦)}

2. 𝜑(𝑥) is the solution correspondence of a standard constrained maximization prob-lem, with 𝑥 as parameter and 𝑦 the decision variable.By assumption the maxi-mand 𝑓(𝑥, 𝑦) = 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) is continuous and the constraint correspondence 𝐺(𝑥) is continuous and compact-valued.Applying the continuous maximum the-orem (Thethe-orem 2.3), 𝜑 is nonempty, compact-valued and uhc.

3.We have just shown that 𝜑(𝑥) is nonempty for every 𝑥 ∈ 𝑋.Starting at 𝑥₀, choose some 𝑥^∗₁ ∈ 𝜑(𝑥0).Then choose 𝑥^∗₂ ∈ 𝜑(𝑥^∗₁).Proceeding in this way, we can construct a plan x^∗ = 𝑥0, 𝑥^∗₁, 𝑥^∗₂, . . . such that 𝑥^∗_𝑡+1 ∈ 𝜑(𝑥^∗_𝑡) for every 𝑡 = 0, 1, 2, . . . .

4.Since 𝑥^∗_𝑡+1∈ 𝜑(𝑥^∗_𝑡) for every 𝑡, x satisfies Bellman’s equation, that is 𝑣(𝑥^∗_𝑡) = 𝑓(𝑥^∗_𝑡, 𝑥^∗_𝑡+1) + 𝛽𝑣(𝑥^∗_𝑡+1), 𝑡 = 0, 1, 2, . . . Therefore x is optimal (Exercise 2.17).

2.126 1.In the previous exercise (Exercise 2.125) we showed that the set 𝜑 of solu-tions to Bellman’s equation (Exercise 2.17) is the solution correspondence of the constrained maximization problem

𝜑(𝑥) = arg max

𝑦∈𝐺(𝑥){ 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦) }

This problem satisfies the requirements of the monotone maximum theorem (The-orem 2.1), since the objective function 𝑓(𝑥, 𝑦) + 𝛽𝑣(𝑦)

∙ supermodular in 𝑦

∙ displays strictly increasing differences in (𝑥, 𝑦) since for every 𝑥²≥ 𝑥¹ 𝑓(𝑥², 𝑦) + 𝛽𝑣(𝑦) − 𝑓(𝑥¹, 𝑦) + 𝛽𝑣(𝑦) = 𝑓(𝑥², 𝑦) − 𝑓(𝑥¹, 𝑦)

∙ 𝐺(𝑥) is increasing.

By Corollary 2.1.2, 𝜑(𝑥) is always increasing.

2.Let x^∗= (𝑥0, 𝑥^∗₁, 𝑥^∗₂, . . . ) be an optimal plan.Then (Exercise 2.17) 𝑥^∗_𝑡+1∈ 𝜑(𝑥^∗_𝑡), 𝑡 = 0, 1, 2, . . .

Since 𝜑 is always increasing

𝑥^∗_𝑡 ≥ 𝑥^∗_𝑡−1 =⇒ 𝑥^∗_𝑡+1≥ 𝑥^∗_𝑡 for every 𝑡 = 1, 2, . . . .Similarly

𝑥^∗_𝑡 ≤ 𝑥^∗_𝑡−1 =⇒ 𝑥^∗_𝑡+1≤ 𝑥^∗_𝑡 x^∗= (𝑥0, 𝑥^∗₁, 𝑥^∗₂, . . . ) is a monotone sequence.

2.127 Let 𝑔(𝑥) = 𝑓(𝑥) − 𝑥. 𝑔 is continuous (Exercise 2.78) with 𝑔(0) ≥ 0 and 𝑔(1) ≤ 0

By the intermediate value theorem (Exercise 2.83), there exists some point 𝑥 ∈ [0, 1]

with 𝑔(𝑥) = 0 which implies that 𝑓(𝑥) = 𝑥.

2.128 1.To show that a label min{ 𝑖 : 𝛽_𝑖 ≤ 𝛼_𝑖∕= 0 } exists for every x ∈ 𝑆, assume to the contrary that, for some x ∈ 𝑆, 𝛽_𝑖 > 𝛼_𝑖 for every 𝑖 = 0, 1, . . . , 𝑛.This implies

∑𝑛 𝑖=0

𝛽_𝑖>∑^𝑛

𝑖=0

𝛼_𝑖= 1

contradicting the requirement that

∑𝑛 𝑖=0

𝛽_𝑖= 1 for every 𝑓(x) ∈ 𝑆

2.The barycentric coordinates of vertex x𝑖 are 𝛼𝑖= 1 with 𝛼𝑗= 0 for every 𝑗 ∕= 𝑖.

Therefore the rule assigns vertex x𝑖 the label 𝑖.

3.Similarly, if x belongs to a proper face of 𝑆, it coordinates relative to the vertices not in that face are 0, and it cannot be assigned a label corresponding to a vertex not in the face.To be concrete, suppose that x ∈ conv {x₁, x₂, x₄}.Then

x = 𝛼1x1+ 𝛼2x2+ 𝛼4x4, 𝛼1+ 𝛼2+ 𝛼4= 1 and 𝛼_𝑖= 0 for 𝑖 /∈ {1, 2, 4}.Therefore

x +−→ min{ 𝑖 : 𝛽_𝑖≤ 𝛼_𝑖∕= 0 } ∈ {1, 2, 4}

2.129 1.Since 𝑆 is compact, it is bounded (Proposition 1.1) and therefore it is contained in a sufficiently large simplex 𝑇 .

2.By Exercise 3.74, there exists a continuous retraction 𝑟 : 𝑇 → 𝑆.The composition 𝑓 ∘ 𝑟 : 𝑇 → 𝑆 ⊆ 𝑇 .Furthermore as the composition of continuous functions, 𝑓 ∘ 𝑟 is continuous (Exercise 2.72). Therefore 𝑓 ∘ 𝑟 has a fixed point x^∗ ∈ 𝑇 , that is 𝑓 ∘ 𝑟(x^∗) = x^∗.

3.Since 𝑓 ∘ 𝑟(x) ∈ 𝑆 for every x ∈ 𝑇 , we must have 𝑓 ∘ 𝑟(x^∗) = x^∗∈ 𝑆.Therefore, 𝑟(x^∗) = x^∗ which implies that 𝑓(x^∗) = x^∗.That is, x^∗is a fixed point of 𝑓.

2.130 Convexity of 𝑆 is required to ensure that there is a continuous retraction of the simplex onto 𝑆.

2.131 1. 𝑓(𝑥) = 𝑥² on 𝑆 = (0, 1) or 𝑓(𝑥) = 𝑥 + 1 on 𝑆 = ℜ+. 2. 𝑓(𝑥) = 1 − 𝑥 on 𝑆 = [0, 1/3] ∪ [2/3, 1].

3.Let 𝑆 = [0, 1] and define

𝑓(𝑥) =

{1 0 ≤ 𝑥 < 1/2 0 otherwise

2.132 Suppose such a function exists.Define 𝑓(x) = −𝑟(x).Then 𝑓 : 𝐵 → 𝐵 conti-nously, and has no fixed point since for

∙ x ∈ 𝑆, 𝑓(x) = −𝑟(x) = −x ∕= x

∙ x ∈ 𝐵 ∖ 𝑆, 𝑓(x) /∈ 𝐵 ∖ 𝑆 and therefore𝑓(x) ∕= x

Therefore 𝑓 has no fixed point contradicting Brouwer’s theorem.

2.133 Suppose to the contrary that 𝑓 has no fixed point.For every x ∈ 𝐵, let 𝑟(z) denote the point where the line segment from 𝑓(x) through x intersects the boundary 𝑆 of 𝐵.Since 𝑓 is continuous and 𝑓(x) ∕= x, 𝑟 is a continuous function from 𝐵 to its boundary, that is a retraction, contradicting Exercise 2.132. We conclude that 𝑓 must have a fixed point.

2.134 No-retraction =⇒ Brouwer Note first that the no-retraction theorem (Ex-ercise 2.132) generalizes immediately to a closed ball about 0 of arbitrary radius.

Assume that 𝑓 is a continuous operator on a compact, convex set 𝑆 in a fi-nite dimensional normed linear space.There exists a closed ball 𝐵 containing 𝑆 (Proposition 1.1). Define 𝑔 : 𝐵 → 𝑆 by

𝑔(y) = { x ∈ 𝑆 : x is closest to y }

As in Exercise 2.129, 𝑔 is well-defined, continuous and 𝑔(x) = x for every x ∈ 𝑆.

𝑓 ∘ 𝑔 : 𝐵 → 𝑆 ⊆ 𝐵 and has a fixed point x^∗= 𝑓(𝑔(x^∗)) by Exercise 2.133. Since

𝑓 ∘ 𝑔(x) ∈ 𝑆 for every x ∈ 𝐵, we must have 𝑓 ∘ 𝑔(x^∗) = x^∗ ∈ 𝑆.Therefore, 𝑔(x^∗) = x^∗which implies that 𝑓(x^∗) = x^∗.That is, x^∗ is a fixed point of 𝑓.

Brouwer =⇒ no-retraction Exercise 2.132.

2.135 Let Λ𝑘, 𝑘 = 1, 2, . . . be a sequence of simplicial partitions of 𝑆 in which the maximum diameter of the subsimplices tend to zero as 𝑘 → ∞.By Sperner’s lemma (Proposition 1.3), every partition Λ_𝑘has a completely labeled subsimplex with vertices x^𝑘₀, x^𝑘₁, . . . , x^𝑘_𝑛.By construction of an admissible labeling, each x^𝑘_𝑖 belongs to a face containing x_𝑖, that is

x^𝑘_𝑖 ∈ conv {x_𝑖, . . . } and therefore

x^𝑘_𝑖 ∈ 𝐴_𝑖, 𝑖 = 0, 1, . . . , 𝑛

Since 𝑆 is compact, each sequence x^𝑘_𝑖 has a convergent subsequence x^𝑘_𝑖^′.Moreover, since the diameters of the subsimplices converge to zero, these subsequences must converge to the same point, say x^∗.That is,

𝑘lim^′→∞x^𝑘_𝑖^′ = x^∗, 𝑖 = 0, 1, . . . , 𝑛 Since the sets 𝐴_𝑖 are closed, x^∗∈ 𝐴_𝑖 for every 𝑖 and therefore

x^∗∈ ∩^𝑛

𝑖=0

𝐴_𝑖∕= ∅

2.136 =⇒ Let 𝑓 : 𝑆 → 𝑆 be a continuous operator on an 𝑛-dimensional simplex 𝑆 with vertices x0, x1, . . . , x𝑛.For 𝑖 = 0, 1, . . . , 𝑛, let

𝐴𝑖= { x ∈ 𝑆 : 𝛽𝑖≤ 𝛼𝑖}

where 𝛼0, 𝛼1, . . . , 𝛼𝑛 and 𝛽0, 𝛽1, . . . , 𝛽𝑛 are the barycentric coordinates of x and 𝑓(x) respectively.Then

∙ 𝑓 continuous =⇒ 𝐴_𝑖closed for every 𝑖 = 0, 1, . . . , 𝑛 (Exercise 1.106)

∙ Let x ∈ conv { x_𝑖: 𝑖 ∈ 𝐼 } for some 𝐼 ⊆ { 0, 1, . . . , 𝑛 }.Then

∑

𝑖∈𝐼

𝛼_𝑖= 1 =∑^𝑛

𝑖=0

𝛽_𝑖

which implies that 𝛽_𝑖 ≤ 𝛼_𝑖 for some 𝑖 ∈ 𝐼, so that x ∈ 𝐴_𝑖.Therefore conv { x_𝑖: 𝑖 ∈ 𝐼 } ⊆ ∪

𝑖∈𝐼

𝐴_𝑖

Therefore the collection 𝐴₀, 𝐴₁, . . . , 𝐴_𝑛 satisfies the hypotheses of the K-K-M theorem and their intersection is nonempty.That is, there exists

x^∗∈

∩𝑛 𝑖=0

𝐴𝑖 ∕= ∅ with 𝛽_𝑖^∗≤ 𝛼^∗_𝑖, 𝑖 = 0, 1, . . . , 𝑛

where 𝛼^∗ and 𝛽^∗ are the barycentric coordinates of x^∗ and 𝑓(x^∗) respectively.

Since∑

𝛽_𝑖^∗=∑

𝛼^∗_𝑖 = 1, this implies that

𝛽_𝑖^∗= 𝛼^∗_𝑖 𝑖 = 0, 1, . . . , 𝑛 In other words, 𝑓(x^∗) = x^∗.

⇐= Let 𝐴0, 𝐴1, . . . , 𝐴𝑛be closed subsets of an 𝑛 dimensional simplex 𝑆 with vertices x0, x1, . . . , x𝑛 such that

conv { x𝑖: 𝑖 ∈ 𝐼 } ⊆∪

𝑖∈𝐼

𝐴𝑖

for every 𝐼 ⊆ { 0, 1, . . . , 𝑛 }.For 𝑖 = 0, 1, . . . , 𝑛, let 𝑔_𝑖(x) = 𝜌(x, 𝐴_𝑖)

For any x ∈ 𝑆 with barycentric coordinates 𝛼₀, 𝛼₁, . . . , 𝛼_𝑛, define 𝑓(x) = 𝛽0x0+ 𝛽1x1+ ⋅ ⋅ ⋅ + 𝛽𝑛x𝑛

where

𝛽𝑖 = 𝛼𝑖+ 𝑔𝑖(x) 1 +∑_𝑛

𝑗=0𝑔𝑗(x) 𝑖 = 0, 1, . . . , 𝑛 (2.45) By construction 𝛽𝑖≥ 0 and∑_𝑛

𝑖=0𝛽𝑖= 1.Therefore 𝑓(x) ∈ 𝑆.That is, 𝑓 : 𝑆 → 𝑆.

Furthermore 𝑓 is continuous.By Brouwer’s theorem, there exists a fixed point 𝑥^∗ with 𝑓(x^∗) = x^∗.That is 𝛽_𝑖^∗= 𝛼^∗_𝑖 for 𝑖 = 0, 1, . . . , 𝑛.

Now, since the collection 𝐴0, 𝐴1, . . . , 𝐴𝑛 covers 𝑆, there exists some 𝑖 for which 𝜌(x^∗, 𝐴𝑖) = 0.Substituting 𝛽^∗_𝑖 = 𝛼^∗_𝑖in (2.45) we have

𝛼^∗_𝑖 = 𝛼^∗_𝑖 1 +∑_𝑛

𝑗=0𝑔𝑗(x^∗)

which implies that 𝑔𝑗(x^∗) = 0 for every 𝑗.Since the 𝐴𝑖 are closed, x^∗ ∈ 𝐴𝑖 for every 𝑖 and therefore

x^∗∈

∩𝑛 𝑖=0

𝐴𝑖∕= ∅

2.137 To simplify the notation, let 𝑧⁺_𝑘(p) = max(

0, z_𝑖(p))

.Assume p^∗ is a fixed point of 𝑔.Then for every 𝑘 = 1, 2, . . . , 𝑛

𝑝^∗_𝑘= 𝑝_𝑘+ 𝑧_𝑘⁺(p^∗) 1 +∑_𝑛

𝑗=1𝑧_𝑗⁺(p^∗) Cross-multiplying

𝑝^∗_𝑘+ 𝑝^∗_𝑘∑^𝑛

𝑗=1

𝑧⁺_𝑗(p) = 𝑝^∗_𝑘+ 𝑧_𝑘⁺(p^∗)

or

𝑝^∗_𝑘∑^𝑛

𝑗=1

𝑧_𝑗⁺(p) = 𝑧_𝑘⁺(p^∗) 𝑘 = 1, 2, . . . 𝑛

Multiplying each equation by 𝑧𝑘(p) we get

𝑝^∗_𝑘𝑧𝑘(p^∗)

∑𝑛 𝑗=1

𝑧_𝑖⁺(p) = 𝑧𝑘(p^∗)𝑧⁺_𝑘(p^∗) 𝑘 = 1, 2, . . . 𝑛

Summing over 𝑘

Each term of this sum is nonnegative, since it is either 0 or (

𝑧_𝑘(p^∗))₂

.Consequently, every term must be zero which implies that 𝑧𝑘(p∗) ≤ 0 for every 𝑘 = 1, 2, . . . , 𝑙.In other words, z(p^∗) ≤ 0.

2.138 Every individual demand function x𝑖(p, 𝑚) is continuous (Example 2.90) in p and 𝑚.For given endowment 𝝎𝑖

𝑚_𝑖=∑^𝑙

𝑗=1

𝑝_𝑗𝝎_𝑖𝑗

is continuous in p (Exercise 2.78). Therefore the excess demand function z𝑖(p) = x𝑖(p, 𝑚) − 𝝎𝑖

is continuous for every consumer 𝑖 and hence the aggregate excess demand function is continuous.

Similarly, the consumer’s demand function x_𝑖(p, 𝑚) is homogeneous of degree 0 in p and 𝑚.For given endowment 𝝎_𝑖, the consumer’s wealth is homogeneous of degree 1 in p and therefore the consumer’s excess demand function z_𝑖(p) is homogeneous of degree 0.So therefore is the aggregate excess demand function z(p).

2.139

Since preferences are nonsatiated and strictly convex, they are locally nonsatiated (Exercise 1.248) which implies (Exercise 1.235) that every consumer must satisfy his budget constraint

2.140 Assume there exists p^∗such that z(p^∗) ≤ 0.That is

z(p^∗) =

∑𝑛 𝑖=1

z𝑖(p) =

∑𝑛 𝑖=1

(x𝑖(p, 𝑚) − 𝝎𝑖)

=

∑𝑛 𝑖=1

x𝑖(p, 𝑚) −

∑𝑛 𝑖=1

𝝎𝑖≤ 0

or ∑

𝑖∈𝑁

x𝑖≤∑

𝑖∈𝑁

𝝎𝑖

Aggregate demand is less or equal to available supply.

Let 𝑚^∗_𝑖 =∑_𝑙

𝑗=1𝑝^∗_𝑗𝝎_𝑖𝑗 denote the wealth of consumer 𝑖 when the price system is p^∗ and let x^∗_𝑖 = x(p^∗, 𝑚^∗) be his chosen consumption bundle.Then

x^∗_𝑖 ≿ x_𝑖 for every x_𝑖∈ 𝑋(p^∗, 𝑚_𝑖)

Let x^∗ = (x^∗₁, x^∗₂, . . . , x^∗_𝑛) be the allocation comprising these optimal bundles.The pair (p^∗, x^∗) is a competitive equilibrium.

2.141 For each x^𝑘, let 𝑆^𝑘 denote the subsimplex of Λ^𝑘 which contains x^𝑘 and let x^𝑘₀, x^𝑘₁, . . . , x^𝑘_𝑛 denote the vertices of 𝑆^𝑘.Let 𝛼^𝑘₀, 𝛼^𝑘₁, . . . , 𝛼^𝑘_𝑛 denote the barycentric coordinates (Exercise 1.159) of x with respect to the vertices of 𝑆^𝑘and let y_𝑖^𝑘= 𝑓^𝑘(x^𝑘_𝑖), 𝑖 = 0, 1, . . . , 𝑛, denote the images of the vertices.Since 𝑆 is compact, there exists subsequences x^𝑘_𝑖^′, y^𝑘_𝑖^′ and 𝛼^𝑘′ such that

x^𝑘_𝑖 → x^∗_𝑖 y^𝑘_𝑖 → y^∗_𝑖 and 𝛼^𝑘_𝑖 → 𝛼^∗_𝑖 𝑖 = 0, 1, . . . , 𝑛

Furthermore, 𝛼^∗_𝑖 ≥ 0 and 𝛼^∗₀+𝛼^∗₁+⋅ ⋅ ⋅+𝛼^∗_𝑛= 1.Since the diameters of the subsimplices converge to zero, their vertices must converge to the same point.That is, we must have

x^∗₀= x^∗₁= ⋅ ⋅ ⋅ = x^∗_𝑛= x^∗ By definition of 𝑓^𝑘

𝑓^𝑘(x^𝑘) = 𝛼^𝑘₀𝑓(x^𝑘₀) + 𝛼^𝑘₁𝑓(x^𝑘₁) + ⋅ ⋅ ⋅ + 𝛼^𝑘_𝑛𝑓(x^𝑘_𝑛)

Substituting y^𝑘_𝑖 = 𝑓^𝑘(x^𝑘_𝑖), 𝑖 = 0, 1, . . . , 𝑛 and recognizing that x^𝑘 is a fixed point of 𝑓^𝑘, we have

𝑥^𝑘 = 𝑓^𝑘(x^𝑘) = 𝛼^𝑘₀y^𝑘₀+ 𝛼^𝑘₁y₁^𝑘+ ⋅ ⋅ ⋅ + 𝛼^𝑘_𝑛y^𝑘_𝑛 Taking limits

x^∗= 𝛼^∗₀y^∗₀+ 𝛼^∗₁y^∗₁+ ⋅ ⋅ ⋅ + 𝛼^∗_𝑛y^∗_𝑛 (2.46)

For each coordinate 𝑖, (x^𝑘_𝑖, y^𝑘_𝑖) ∈ graph(𝜑) for every 𝑘 = 0, 1, . . . .Since 𝜑 is closed, (x^∗_𝑖, y_𝑖^∗) ∈ graph(𝜑).That is, y^∗_𝑖 ∈ 𝜑(x^∗_𝑖) = 𝜑(x^∗) for every 𝑖 = 0, 1, . . . , 𝑛.Therefore, (2.46) implies

x^∗∈ conv 𝜑(x^∗) Since 𝜑 is convex valued,

x^∗∈ 𝜑(x^∗)

2.142 Analogous to Exercise 2.129, there exists a simplex 𝑇 containing 𝑆 and a retrac-tion of 𝑇 onto 𝑆, that is a continuous funcretrac-tion 𝑔 : 𝑇 → 𝑆 with 𝑔(x) = x for every x ∈ 𝑆.

Then 𝜑 ∘ 𝑔 : 𝑇 ⇉ 𝑆 ⊂ 𝑇 is closed-valued (Exercise 2.106) and uhc (Exercise 2.103).

By the argument in the proof, there exists a point x^∗ ∈ 𝑇 such that x^∗ ∈ 𝜑 ∘ 𝑔(x^∗).

However, since 𝜑 ∘ 𝑔(x^∗) ⊆ 𝑆, we must have x^∗ ∈ 𝑆 and therefore 𝑔(x^∗) = x^∗.This implies x^∗∈ 𝜑(x^∗).That is, x^∗ is a fixed point of 𝜑.

2.143 𝐵 = 𝐵₁× 𝐵₂× . . . × 𝐵_𝑛 is the Cartesian product of uhc, compact- and convex-valued correspondences.Therefore 𝐵 is also compact-convex-valued and uhc (Exercise 2.112 and also convex-valued (Exercise 1.165). By Exercise 2.106, 𝐵 is closed.

2.144 Strict quasiconcavity ensures that the best response correspondence is in fact a function 𝐵 : 𝑆 → 𝑆.Since the hypotheses of Example 2.96 apply, there exists at least one equilibrium.Suppose that there are two Nash equilibria s and s^′.Since 𝐵 is a contraction,

𝜌(𝐵(s), 𝐵(s^′) ≤ 𝛽𝜌(s, s^′) for some 𝛽 < 1.However

𝐵(s) = s and 𝐵(s^′) = s^′ and (2.46) implies that

𝜌(s, s^′) ≤ 𝛽𝜌(s, s^′)

which is possible if and only if s = s^′.This implies that the equilibrium must be unique.

2.145 Since 𝐾 is compact, it is totally bounded (Exercise 1.112). There exists a finite set of points x₁, x₂, . . . , x_𝑛 such that

𝐾 ⊆

∩𝑛 𝑖=1

𝐵𝜖(x𝑖)

Let 𝑆 = conv {x1, x2, . . . , x𝑛}.For 𝑖 = 1, 2, . . . , 𝑛 and x ∈ 𝑋, define 𝛼𝑖(x) = max{0, 𝜖 − ∥x − x𝑖∥}

Then for every x ∈ 𝐾,

0 ≤ 𝛼𝑖(x) ≤ 𝜖, 𝑖 = 1, 2, . . . , 𝑛 and

𝛼𝑖(x) > 0 ⇐⇒ x ∈ 𝐵𝜖(x𝑖) Note that 𝛼_𝑖(x) > 0 for some 𝑖.Define

ℎ(x) =

∑∑𝛼𝑖(x)x𝑖

𝛼_𝑖(x)

Then ℎ(x) ∈ 𝑆 and therefore ℎ: 𝐾 → 𝑆.Furthermore, ℎ is continuous and

∥ℎ(x) − x∥ =

∑ 𝛼𝑖(x)x𝑖

∑𝛼_𝑖(x) − x



=

∑

𝛼∑𝑖(x)(x𝑖− x) 𝛼𝑖(x)



=

∑𝛼_𝑖∑(x) ∥x_𝑖− x∥

𝛼𝑖(x)

≤

∑∑𝛼_𝑖(x)𝜖 𝛼𝑖(x) = 𝜖 since 𝛼_𝑖(x) > 0 ⇐⇒ ∥x_𝑖− x∥ ≤ 𝜖.

2.146 1.For every x ∈ 𝑆^𝑘, 𝑓(x) ∈ 𝑆 and therefore 𝑔^𝑘(x) = ℎ^𝑘( 𝑓(x))

∈ 𝑆^𝑘. 2.For any x ∈ 𝑆^𝑘, let y = 𝑓(x) ∈ 𝑓(𝑆) and therefore

ℎ^𝑘(y) − y < 1𝑘 which implies

𝑔^𝑘(x) − 𝑓(x) ≤ 1𝑘 for every x ∈ 𝑆^𝑘 2.147 By the Triangle inequality

x^𝑘− 𝑓(x) ≤ 𝑔^𝑘(x^𝑘) − 𝑓(x^𝑘) + 𝑓(x^𝑘) − 𝑓(x)

As shown in the previous exercise

𝑔^𝑘(x^𝑘) − 𝑓(x^𝑘) ≤ 1𝑘 → 0 as 𝑘 → ∞.Also since 𝑓 is continuous

𝑓(x^𝑘) − 𝑓(x) → 0 Therefore

x^𝑘− 𝑓(x) → 0 =⇒ x = 𝑓(x) x is a fixed point of 𝑓.

2.148 𝑇 (𝐹 ) is bounded and equicontinuous and so therefore is 𝑇 (𝐹 ) (Exercise 2.96). By Ascoli’s theorem (Exercise 2.95), 𝑇 (𝐹 ) is compact.Therefore 𝑇 is a compact operator.

Applying Corollary 2.8.1, 𝑇 has a fixed point.

In document Foundation of Mathematical Economics Solutions (Page 90-107)