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Macroeconomics Exam Review 49

# Macroeconomics Exam Review 49 - Solutions for Foundations...

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5. We have shown that conv 𝑆 conv 𝑆 , that is conv 𝑆 is closed. 6. conv 𝑆 is a closed and bounded subset of a finite dimensional space, and hence conv 𝑆 is compact (Proposition 1.4 and Exercise 1.215). 1.226 1. 𝑆 is bounded. Therefore, there exists some 𝑐 such that x = max 𝑖 𝑥 𝑖 < 𝑐 for every x 𝑆 . That is 𝑐 𝑥 𝑖 𝑐 so that x 𝐶 = { x = ( 𝑥 1 , 𝑥 2 , . . . , 𝑥 𝑛 ) ∈ ℜ 𝑛 : 𝑐 𝑥 𝑖 𝑐 for every 𝑖 } Therefore 𝑆 𝐾 . 2. Exercise 1.177. 3. 𝐶 is the convex hull of a finite set and hence is compact (Exercise 1.225) 4. 𝑆 is a closed subset of a compact set and hence is compact (Exercise 1.110). 1.227 A polytope is the convex hull of a finite set. Any finite set is compact. 1.228 The unit simplex Δ 𝑛 1 in 𝑛 is the convex hull of the unit vectors e 1 , e 2 , . . . , e 𝑛 , that is Δ 𝑛 1 = conv { e 1 , e 2 , . . . , e 𝑛 } = { ( 𝑥 1 , 𝑥 2 , . . . , 𝑥 𝑛 ) ∈ ℜ 𝑛 : 𝑥 𝑖 0 and 𝑥 𝑖 = 1 } This simplex has a nonempty relative interior, namely ri 𝑆 = { ( 𝑥 1 , 𝑥 2 , . . . , 𝑥 𝑛 ) ∈ ℜ 𝑛 : 𝑥 𝑖 > 0 and 𝑥 𝑖 < 1 } 1.229 Let 𝑛 = dim 𝑆 . By Exercise 1.182,
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