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midterm_2004Ans - Fall 2004 M IDTERM E XAM A NSWER KEY...

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Fall 2004 ARE211 M IDTERM E XAM - A NSWER KEY Problem 1 (25 points). Definition: Two metrics are equivalent if they define the same open sets, that is if a set is open with respect two the first metric whenever it is open with respect to the second. The next definition applies only to part c) of this question. Definition: Two metrics are uniformly equivalent given ε > 0, there exists δ > 0 such that for all x , y X , ρ ( x , y ) < δ = σ ( x , y ) < ε and σ ( x , y ) < δ = ρ ( x , y ) < ε a) Show that two metrics σ and ρ on a set X are equivalent if and only if given x X and ε > 0, there exists δ > 0 such that for all y X , ρ ( x , y ) < δ = σ ( x , y ) < ε (1) σ ( x , y ) < δ = ρ ( x , y ) < ε (2) We’ll first show that if either (1) or (2) fails, then we can construct an open set with respect to one metric that is not open with respect to the other. Assume that (1) fails, i.e., there exists x X , ε > 0 and a sequence ( y n ) such that for each n , ρ ( x , y n ) < 1 / n but σ ( x , y n ) ε . Let U = B σ ( x , ε ) . Necessarily U is open w.r.t. σ and contains x . However, the sequence ( y n ) converges to x w.r.t. ρ , but none of the y n belong to U . Hence x is a boundary point of U w.r.t. ρ and so cannot be open. A parallel argument can be constructed if (2) fails. Now suppose that both (1) or (2) are satisfied. We need to show that a set U is open w.r.t. ρ iff it is open w.r.t. σ . We will do so by picking an arbitrary set U that is open w.r.t. σ and showing that an point x U is an interior point of U w.r.t. ρ . This will show that every element of U is an interior point w.r.t. ρ , and thus that U is open w.r.t. ρ . Since U is open w.r.t. σ , there exists ε > 0 such that B σ ( x , ε ) U . From (1) , there exists δ > 0 , such that B ρ ( x , δ ) B σ ( x , ε ) U . That is x belongs to a ρ -open subset of U , and is hence a ρ -interior point of U . A parallel argument using (1) can be constructed to show that if U is open w.r.t. ρ , then it is also open w.r.t. σ . b) Show that the Pythagorian metric on R n is equivalent to the metric ρ , defined by ρ ( x , y ) = max {| x i - y i | : i = 1 ,..., n } We’ll refer to the Pythag metric as σ . Fix x R n and ε > 0 .and let δ = ε / n . It’s easier to prove that σ ( x , y ) ε = ρ ( x , y ) δ (3) ρ ( x , y ) ε = σ ( x , y ) δ (4) Clearly, if ρ ( x , y ) ε then σ ( x , y ) = ε 2 + K , for some nonnegative number K . Hence σ ( x , y ) ε . On the other hand, if σ ( x , y ) ε then necessarily | x i - y i | ≥ ε / n > ε / n , for at least one i . But this implies that ρ ( x , y ) > δ .
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2 c) Prove that the following metrics on R ++ are equivalent but not uniformly equivalent. ρ ( x , y ) = | x - y | σ ( x , y ) = | 1 / x - 1 / y | Hint: you can identify certain upper bounds and assume without loss of generality that ε > 0 is not greater than these bounds. To show that these are equivalent, pick x R and ε > 0 . Assume without loss of generality that ε < min ( 0 . 5 , 1 / x ) . We’ll set δ = ( min [ ε / 4 , ε x ( x - ε )] if x < 1 . 5 ε / x ( x + ε ) if x 1 . 5 .
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