# ans2-1 - ECON 831 Solution Homework #1 1. (i) (ii) - Study...

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ECON 831 Solution Homework #1 1. (i) (ii) - Study at 0: Γ(0) = (0 , 1] . Preliminary remarks: NB1: there are only 2 open sets O in [0 , 1] s.t. (0 , 1] O , (1) [0 , 1] (which is open and closed); (2) (0 , 1] . NB2: for any ² > 0 , N ², [0 , 1] (0) = { y [0 , 1] / d (0 ,y ) < ² } = [0 ) (when ² 1 ) and [0,1] (when ² > 1 ). Proof: for any y (0 , 1] , Γ( y ) = (0 ,y ) Γ(0) . Hence, Γ is upper-hemicontinuous at 0. - Study at y (0 , 1) : Γ( y ) = (0 ,y ) and for ² "small enough" N ², [0 , 1] ( y ) = ( y - ²,y + ² ) . We prove that Γ is NOT upper hemicontinuous. Consider (0 ,y ) : it is an open set in [0 , 1] and it contains the image of y by Γ . However, for any ² > 0 , there exist a point y * (to the right of y ) s.t. 1 y * > y and y * N ², [0 , 1] ( y ) and Γ( y * ) = (0 ,y * ) * (0 ,y ) . Hence, Γ is not upper hemicontinuous at y . - Study at 1: Γ(1) = (0 , 1) . Preliminary remark: an ² -neighborhood around 1 in [0 , 1] takes the form N ²,X (1) = (1 - ², 1] for ² "small enough". Proof: for any y * s.t. 1 - ² < y * < y , Γ( y * ) = (0 ,y * ) (0 ,y ) . Hence, Γ is upper- hemicontinous at 1. 1

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2. (i) Consider an arbitrary x X , an arbitrary open set O Y s.t. Φ( x ) O . We need to show that there exists ² > 0 s.t. Φ( N ²,X ( x )) O . Φ( x ) O 1 ( x ) Γ 2 ( x )) O Γ 1 ( x ) O and Γ 2 ( x ) O Hence, there exist ² 1 > 0 s.t. Γ 1 ( N ² 1 ,X ( x )) O (b/c Γ 1 is upper-hemicontinuous and O is an open set in Y that contains the image of x by Γ 1 ). Similarly, there exist ² 2 > 0 s.t. Γ 2 ( N ² 2 ,X ( x )) O . De ne ² = min { ² 1 2 } then N ²,X ( x ) N ² 1 ,X ( x ) and N ²,X ( x ) N ² 2 ,X ( x ) Γ 1 ( N ²,X ( x )) Γ 1 ( N ² 1 ,X ( x )) O and Γ 2 ( N ²,X ( x )) Γ 2 ( N ² 2 ,X ( x )) O 1 ( N ²,X ( x )) Γ 2 ( N ²,X ( x ))) O Φ( N ²,X ( x )) O We conclude that Φ is upper hemicontinuous at x X because we have worked with an arbitrary open set O . (ii) - proof of compact-valued: By assumption, Γ 1 and Γ 2 are compact-valued. Hence, for any x X , Γ 1 ( x ) and Γ 2 ( x ) are 2 compact sets. We have Γ 1 ( x ) Γ 2 ( x ) Γ 1 ( x ) . Hence, to prove that Ψ( x ) is compact, we only need to show that it is closed. Since, Γ 1 ( x ) and Γ 2 ( x ) are compact sets, there are also closed. And a nite intersection of closed sets is also closed, hence
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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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ans2-1 - ECON 831 Solution Homework #1 1. (i) (ii) - Study...

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