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Unformatted text preview: MAT237Y Multivariable Calculus Summer 2009 Assignment # 3 Solutions 1. (a) Show that if A R n is convex, then the closure of A is convex. (b) Show that if A R n is convex, then the interior of A is convex. Hint: If r > 0 and x and y are such that B r ( x ) and B r ( y ) are subsets of A , show that B r ( z ) A for any z on the line between x and y . Solution. (a) Let A be convex. Two show that the closure A is convex, we must show that for any two points a, b A , and any t [0 , 1], the interpolated point c = ta + (1 t ) b is also in A . Since a, b are in the closure of A , we may find sequences ( a n ) n =1 , ( b n ) n =1 of elements in A which converge to a, b respectively. Let c n = ta n + (1 t ) b n . Since A is convex, we see that c n A . Moreover, we have lim n c n = lim n ta n + (1 t ) b n = t ( lim n a n ) + (1 t )( lim n b n ) = ta + (1 t ) b = c. Thus, c is the limit of a sequence of points in A , so that c A . (b) Let A be convex. To show that the interior int A of A is convex, we must show that for any two points a, b int A , and any t [0 , 1], the interpolated point c = ta + (1 t ) b is also in int A . Since a int A , we have for some r 1 > 0 that B r 1 ( a ) A . Simi larly, for some r 2 > 0, B r 2 ( b ) A . By taking r = min { r 1 , r 2 } , we thus have that B r ( a ) A and B r ( b ) A. Claim. B r ( c ) A . Proof of claim. For d B r ( c ), let h = d c , so that bardbl h bardbl < r . We thus have a + h B r ( a ) A and similarly, b + h B r ( b ) A . Also, d = c + h = ta + (1 t ) b + h = ta + (1 t ) b + th + (1 t ) h = t ( a + h ) + (1 t )( b + h ) , and since A is convex, this shows that d A , as required....
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This note was uploaded on 10/21/2010 for the course MATHEMATIC MAT237Y1 taught by Professor Romauldstanczak during the Fall '09 term at University of Toronto Toronto.
 Fall '09
 RomauldStanczak
 Calculus, Multivariable Calculus

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