ans_midterm08

ans_midterm08 - ECON 831 Solution Midterm 1. (a) Consider...

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ECON 831 Solution Midterm 1. (a) Consider the following sequence of subsets S k in the metric space ( R ,d 1 ) : S k = (1 - 1 k +1 , 1 + 1 k +1 ) . The following conditions are satis ed: (i) For any k , consider the real number 1 . It is such that, for any k < , 1 S k because 1 - 1 k +1 < 1 < 1 + 1 k +1 . Hence, S k 6 = . (ii) For any k : S k +1 = (1 - 1 k +2 , 1 + 1 k +2 ) with 1 k +2 < 1 k +1 . Hence, we have, for any k , 1 - 1 k +1 < 1 - 1 k +2 < 1 + 1 k +2 < 1 + 1 k +1 . Hence, S k +1 S k . (iii) We have diam ( S k ) = 1 + 1 k +1 - ( 1 - 1 k +1 ) = 1 2( k +1) 0 . (iv) T k =1 S k = T k =1 (1 - 1 k +1 , 1 + 1 k +1 ) = (1 - lim k ( 1 k +1 ) , 1 + lim k 1 k +1 ) = (1 , 1) = . (b) The above sequence is not closed in the metric space ((0 , 1) ,d 1 ) . Consider now the sequence of subsets S k of (0 , 1) : S k = (0 , 1 k +1 ] . It is easy to see that conditions (i) to (iv) are still satis ed. (v) For any k , the complement of S k in (0,1) is S c k = ( 1 k , 1) . We can easily show that this set is open in (0 , 1) because for any point in S c k , I can nd an ² -neighborhood in (0,1) s.t. every point in that neighborhood remains in S c k . Hence, the complement of S c k in (0,1) is closed in (0,1) ie S k closed in (0 , 1) . 2.
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This note was uploaded on 02/19/2010 for the course ECON 831 taught by Professor Antoine during the Fall '09 term at Simon Fraser.

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ans_midterm08 - ECON 831 Solution Midterm 1. (a) Consider...

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