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Unformatted text preview: Problem Set 6 Solutions Economics 6170 TA: Christopher Handy October 27, 2008 Problem 1 (a) Consider an arbitrary ¯ x ∈ S ∩ T . Now, since ¯ x ∈ S and S is open in R n , there exists r S > 0 such that B (¯ x,r S ) ⊂ S . And since ¯ x ∈ T and T is open in R n , there exists r T > 0 such that B (¯ x,r T ) ⊂ T . Take r = min { r S ,r T } > 0. Then B (¯ x,r ) ⊂ B (¯ x,r S ) ⊂ S and B (¯ x,r ) ⊂ B (¯ x,r T ) ⊂ T . Therefore B (¯ x,r ) ⊂ S ∩ T , so S ∩ T is open in R n . (b) We can write A = A 1 ∩ A 2 where A 1 = { ( x 1 ,x 2 ) ∈ R 2  x 1 > ,x 2 > } A 2 = { ( x 1 ,x 2 ) ∈ R 2  x 1 x 2 > 1 } Note that A 2 contains some points where x 1 < 0 and x 2 < 0. We want to show that both A 1 and A 2 are open in R 2 , which will show by part (a) that A is open in R 2 . Claim: A 1 is open in R 2 . Proof: Let ¯ x ∈ A 1 and take r = min { ¯ x 1 , ¯ x 2 } > 0. We want to show B (¯ x,r ) ⊂ A 1 . This is the same as showing that for any x / ∈ A 1 , we have x / ∈ B (¯ x,r ). So, consider some x / ∈ A 1 and assume without loss of generality that x 1 ≤ 0. Then we have that x 1 ≤ < r ≤ ¯ x 1 , so ¯ x 1 x 1 ≥ r . That means that d ( x, ¯ x ) = p ( x 1 ¯ x 1 ) 2 + ( x 2 ¯ x 2 ) 2 ≥ r , so we have x / ∈ B (¯ x,r ), which is what we wanted to show. Therefore A 1 is open. Claim: A 2 is open in R 2 . To show that A 2 is open in R 2 , it is easiest to show that ∼ A 2 is closed in R 2 . First, define the function f ( x ) = 1 x 1 x 2 for all x ∈ R 2 and note that f is continuous on R 2 . Then we can write ∼ A 2 as ∼ A 2 = { ( x 1 ,x 2 ) ∈ R...
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This note was uploaded on 10/03/2009 for the course ECON 6170 taught by Professor Mitra during the Fall '08 term at Cornell.
 Fall '08
 MITRA
 Economics

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