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Unformatted text preview: 18.100 Practice Midterm Solutions 10/5/2008 Problem 1 a. Let x B . If x B , then d ( x ,x ) < r and so x C . Else, if x is a limit point of B , then there exists a sequence x n x with x n B . Then for every > 0, we can find N such that for all n N , d ( x n ,x ) < . By the triangle inequality, d ( x ,x ) d ( x ,x n ) + d ( x n ,x ) < r + . Since was arbitrary, we conclude that d ( x ,x ) r . This implies x C . b. The example can be found from Problem 4, Homework # 3. Namely, we defined a metric on X = R { + }{} such that induced metric on R was given by d ( x,y ) =  x y  1+  x y  . One way of defining the metric d on all of X is to declare that d ( ,x ) = d ( ,x ) = d ( , ) = 1 for all x R . Observe that for any y R X , we have B 1 ( y ) = R since B 1 ( y ) consists of points whose distance from y is strictly less than unity. However, B 1 ( y ) = B 1 ( y ) is not the ball consisting of points whose distance from y is less than or equal to unity, as one would get from the usual metric on...
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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.
 Fall '10
 Prof.KatrinWehrheim

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