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511-Midterm-2008F-ANS

# 511-Midterm-2008F-ANS - Economics 511 Professor John...

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Economics 511 Professor: John Nachbar Fall 2008 Midterm Answers 5. Theorem. Let ( X, d x ) and ( Y, d y ) be metric spaces and suppose f : X Y and g : X y are both continuous. Let E X be dense in X . (a) If { y t } is a sequence in Y and y t converges both to y * and to ˆ y then y * = ˆ y . (b) If f ( x ) = g ( x ) for x E then f = g . (c) f ( E ) is dense in f ( X ) . Proof. (a) Suppose y t converges to both y * and ˆ y . Take any ε > 0. Then there is a T * such that for all t > T * , y t N ε/ 2 ( y * ) and a ˆ T such that for all t > ˆ T y t N ε/ 2 y ). By the triangle inequality, for any t > max { T * , ˆ T } , d y ( y * , ˆ y ) d y ( y * , y t ) + d y ( y t , ˆ y ) < ε . Since ε was arbitrary, d y ( y * , ˆ y ) = 0, hence y * = ˆ y . Here’s an alternative proof. By contraposition. Suppose y * 6 = ˆ y , hence d ( y * , ˆ y ) > 0. Suppose also that that y t y * . Define 2 ε = d ( y * , ˆ y ). Since y t y * , there is a T such that for all t > T , d ( y t , y * ) < ε . By the triangle inequality, for all t > T , 2 ε = d ( y * , ˆ y ) d ( y t , y * ) + d ( y t , ˆ y ) < ε + d ( y t , ˆ y ), hence d ( y t , ˆ y ) > ε , which shows that y t 6→ ˆ y .

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