Problem_Set_6_Solutions - Problem Set 6 Solutions Economics...

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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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Problem_Set_6_Solutions - Problem Set 6 Solutions Economics...

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