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Unformatted text preview: 3.138 1. Since is open, there exists a ball ( x 1 ) ⊆ . Let = 1 + 2 . Then x + ( x 1 − x ) ∈ ( 1 ) ⊆ . 2. Let = − 1 . The open ball ( x 1 ) of radius centered on x 1 is contained in . Therefore is a neighborhood of x 1 . 3. Since is convex, for every y ∈ ( y ) ≤ (1 − ) ( x ) + ( z ) ≤ (1 − ) + ( z ) ≤ + ( z ) Therefore is bounded on . 3.139 The previous exercise showed that is locally bounded from above for every x ∈ . To show that it is also locally bounded from below, choose some x ∈ . There exists some ( x and such that ( x ) ≤ for every x ∈ ( x ) Choose some 1 ∈ ( x ) and let x 2 = 2 x − x 1 . Then x 2 = 2 x − x 1 = x − ( x 1 − x ) ∈ ( x ) and ( x 2 ) ≤ . Since is convex ( x ) ≤ 1 2 ( x 1...
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This note was uploaded on 01/16/2012 for the course ECO 2024 taught by Professor Dr.dumond during the Fall '10 term at FSU.
 Fall '10
 Dr.DuMond
 Macroeconomics

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