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# ps5_solution - TA Yankun Wang Economics 617 Problem Set 5...

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TA: Yankun Wang Economics 617 Problem Set 5 Solution Key 1. [Distance] Let d denote the Euclidean distance function on R n × R n . De fi ne: g ( x, y ) = d ( x, y ) / [1 + d ( x, y )] for all x, y in R n Show that: (a) g ( x, y ) 0 for all x, y in R n (b) g ( x, y ) = 0 if and only if x = y (c) g ( x, y ) = g ( y, x ) (d) g ( x, z ) g ( x, y ) + g ( y, z ) for all x, y, z in R n Solution: (a) d ( x, y ) 0 for all x, y in R n , and thus 1 + d ( x, y ) 1 for all x, y in R n . As a result, g ( x, y ) 0 for all x, y in R n . (b) By the de fi nition of g ( x, y ) , g ( x, y ) = 0 if and only if d ( x, y ) = 0 . But d ( x, y ) = 0 if and only if x = y. (c) Since d ( x, y ) = d ( y, x ) , it is straightforward to see that g ( x, y ) = g ( y, x ) . (d) g ( x, z ) g ( x, y ) + g ( y, z ) d ( x, z ) 1 + d ( x, z ) d ( x, y ) 1 + d ( x, y ) + d ( y, z ) 1 + d ( y, z ) d ( x, z )[1 + d ( x, y )][1 + d ( y, z )] d ( x, y )[1 + d ( x, z )][1 + d ( y, z )] + d ( y, z )[1 + d ( x, y )][1 + d ( x, z )] d ( x, z ) d ( x, y ) + d ( y, z ) + 2 d ( x, y ) d ( y, z ) + d ( x, y ) d ( y, z ) d ( x, z ) Since d ( x, y ) + d ( y, z ) d ( x, z ) , d ( a, b ) 0 for any ( a, b ) R n × R n , the last inequality is always true. Thus g ( x, z ) g ( x, y ) + g ( y, z ) for all x, y, z in R n . 2. [Interior, Exterior and Boundary] 1

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Let A be a set in R 2 , de fi ned by: A = { ( x 1 , x 2 ) R 2 : ( x 1 ) 2 ( x 2 ) 2 > 1 } (a) Describe the interior, exterior and boundary of A (algebraically). (b) Graph the sets in (a) in an appropriate diagram. Solution: (a) Algebraically, Int ( A ) = { ( x 1 , x 2 ) R 2 : ( x 1 ) 2 ( x 2 ) 2 > 1 } ; Ext ( A ) = { ( x 1 , x 2 ) R 2 : ( x 1 ) 2 ( x 2 ) 2 < 1 } ; Bond ( A ) = { ( x 1 , x 2 ) R 2 : ( x 1 ) 2 ( x 2 ) 2 = 1 } . (b) The graph is as follows: 3. [Limits of Sequences and Continuity] Let { a n } be a sequence, with a n S R m for each n { 1 , 2 , 3 , .... } , and lim n →∞ a n = p. Suppose f : S R is continuous on S, and p S, show that: lim n →∞ f ( a n ) = f ( p ) Intuition: The intuition behind this problem is as follows: We know that as n becomes very large, a n becomes very close to p. We also know that as a n gets very close to p, f ( a n ) gets very close to f ( p ) . Now we are asked to prove that as n gets large, f ( a n ) gets very close to f ( p ) . Proof: Given any ε > 0 , by the de fi nition of continuous functions, we know that there exists δ > 0 , which might depend on ε, such that k a n p k < δ implies k f ( a n ) f ( p ) k < ε. (We have used the fact that if a function is continuous on its domain, it is continuous at each point on its domain by the de fi nition of continuous functions. ) Now, for this speci fi c δ > 0 , by the fact that lim n →∞ a n = p, we know that there exists N N , where N might depend on δ, such that for every n > N , k a n p k < δ. Essentially we have proved that for any ε > 0 , there exists N N , where N might depend on ε, such that for every n > N , k f ( a n ) f ( p ) k < ε. Therefore, lim n →∞ f ( a n ) = f ( p ) .
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ps5_solution - TA Yankun Wang Economics 617 Problem Set 5...

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