18.100mid1_sol

18.100mid1_sol - 18.100 Midterm Solutions 10/9/2008 Problem...

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Unformatted text preview: 18.100 Midterm Solutions 10/9/2008 Problem 1 (a) Given a space X , a metric on X is a function d : X × X → R + such that (1) d ( x,y ) = d ( y,x ) ≥ 0 for all x,y ∈ X ; (2) d ( x,y ) = 0 iff x = y ; (3) d ( x,y ) ≤ d ( x,z ) + d ( z,y ) for all x,y,z ∈ X . (b) Property (1): Clear since absolute value is nonnegative. Property (2): | x 1 − y 1 | = ... = n | x n − y n | = 0 iff x i = y i for all i . Property (3) follows from the fact that absolute value satisfies the triangle inequality: d ( x,y ) = | x 1 − y 1 | + ... + n | x n − y n | ≤ | x 1 − z 1 | + | z 1 − y 1 | + ... + n | x n − z n | + n | z n − y n | = d ( x,z )+ d ( z,y ) . (c) d ( x, 0) < 1 implies | x 1 | + 2 | x 2 | < 1, so our region is bounded by the lines x 1 + 2 x 2 = 1, x 1 − 2 x 2 = 1, − x 1 + 2 x 2 = 1 and − x 1 − 2 x 2 = 1. 1/2-1/2-1 1 (d) No. Take x = (1 , 1) and y = (0 , 1). Then g ( x,y ) = | 1 − 1 | = 0 but x negationslash = y . Problem 2 (a) For ( x n ) the possible limit points are 1 (take n = 3 m and m even) − 1 (take n = 3 m and m odd) 1 / 2 (take n odd 3 negationslash | n ) − 1 / 2 (take n even and 3 negationslash | n ) For ( y n ) the possible limit points are 1 if n is even and − 1 if n is odd. For ( z n ) the only possible limit point is 6 / 7....
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This note was uploaded on 12/07/2011 for the course MATH 18.100B taught by Professor Prof.katrinwehrheim during the Fall '10 term at MIT.

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18.100mid1_sol - 18.100 Midterm Solutions 10/9/2008 Problem...

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