# HW03 - C 6 f b a as b a 1 b a = b 1 b a a 1 b a 6 b 1 b a 1...

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Modern Analysis 1 Homework 03 1. Let ( M,d ) be a metric space. (i) Show that a new metric D is deﬁned on M by the rule x,y M D ( x,y ) = d ( x,y ) 1 + d ( x,y ) . (ii) Show that a subset of M is D -open if and only if it is d -open. Solution It is convenient to begin by noting that the function f : [0 , ) [0 , 1) : t 7→ t/ (1 + t ) is strictly increasing: in fact, if 0 6 r < s then s - r > 0 so that f ( s ) - f ( r ) = s - r (1 + s )(1 + r ) > 0 and therefore f ( s ) > f ( r ); note further that f is surjective, for if 0 6 T < 1 then T = f ( T/ (1 - T )). (i) Positivity and symmetry of D are plain. To establish the triangle inequality, let x,y,z M , write a = d ( y,z ) ,b = d ( z,x ) ,c = d ( x,y ) and write A = D ( y,z ) ,B = D ( z,x ) ,C = D ( x,y ) so that C = f ( c ) and so on. As f is strictly increasing, from c 6 b + a it follows that
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Unformatted text preview: C 6 f ( b + a ); as b + a 1 + b + a = b 1 + b + a + a 1 + b + a 6 b 1 + b + a 1 + a we deduce that C 6 B + A as required. (ii) Let p,x ∈ M and δ > 0. As f is strictly increasing, d ( x,p ) < δ ⇔ D ( x,p ) < f ( δ ) =: Δ so that B d δ ( p ) = B D Δ ( p ) . Thus d-balls and D-balls (of radius less than unity) are precisely the same sets, and so d-open and D-open have precisely the same meaning. Moral : In dealing with properties of metric spaces that may be formulated in terms of open sets, it may be assumed that the metric is bounded. 1...
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