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Unformatted text preview: ISyE 7682 FALL 2010 – 2nd Homework Problem 0.1 Let C i ⊆ ℜ n i be a convex set for i = 1 , 2. Show that aff C 1 × aff C 2 = aff ( C 1 × C 2 ) , (1) ri C 1 × ri C 2 = ri ( C 1 × C 2 ) . (2) Solution . We first show (1). It is easy to see that if A 1 ⊆ ℜ n 1 and A 2 ⊆ ℜ n 2 are two affine sets then A 1 × A 2 ⊆ ℜ n 1 × ℜ n 2 is an affine set. Using this observation, we conclude that aff C 1 × aff C 2 is an affine set which contains the set C 1 × C 2 . Since aff ( C 1 × C 2 ) is the smallest affine set containing C 1 × C 2 , we conclude that aff ( C 1 × C 2 ) ⊆ aff C 1 × aff C 2 . To show the reverse inclusion, let ( x, y ) ∈ aff C 1 × aff C 2 be given. Then, x = α 1 x 1 + . . . + α k x k and y = β 1 y 1 + . . . + β l y l , where x 1 , . . . , x k ∈ C 1 , y 1 , . . . , y l ∈ C 2 , α 1 + . . . + α k = 1 and β 1 + . . . + β l = 1. It is easy to see that ( x, y ) = k summationdisplay i =1 l summationdisplay j =1 α i β j ( x i , y j ) , 1 = k summationdisplay i =1 l summationdisplay j =1 α i β j , from which it follows that ( x, y ) ∈ aff ( C 1 × C 2 ). We next show (2). First observe that in the definition of relative interior, we may use any ball as long as the metric that defines the ball is equivalent to the Euclidean distance. In the product space ℜ n 1 ×ℜ n 2 , it is useful to consider the balls which are cartesian product of Euclidean balls of ℜ n 1 and ℜ n 2 . Hence, we have the following equivalences: ( x, y ) ∈ ri ( C 1 × C 2 ) ⇔ ∃ δ > 0 such that [ B ( x, δ ) × B ( y, δ )] ∩ aff ( C 1 × C 2 ) ⊆ C 1 × C 2 ⇔ ∃ δ > 0 such that [ B ( x, δ ) × B ( y, δ )] ∩ [aff C 1 × aff C 2 ] ⊆ C 1 × C 2 ⇔ ∃ δ 1 > 0 and δ 2 > 0 such that...
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 Fall '08
 Staff
 Convex set, C∞

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