Let S be any set and ?> 0. Define T = {t ? ? : |t s|< ? for some s ? S} . Prove

Question

Let S be any set and ?> 0. Define T = {t ? ? : |t s|< ? for some s ? S} . Prove that T is open.

Explanation / Answer

(a) S is a polyhedron. It is the parallelogram with corners a1 + a2 , a1 - a2 , -a1 + a2 , -a1 - a2 , as shown below for an example in R2 . c2 a2 a1 c1 For simplicity we assume that a1 and a2 are independent. We can express S as the intersection of three sets: • S1 : the plane de?ned by a1 and a2 • S2 = {z + y1 a1 + y2 a2 | aT z = aT z = 0, -1 = y1 = 1}. This is a slab parallel 1 2 to a2 and orthogonal to S1 • S3 = {z + y1 a1 + y2 a2 | aT z = aT z = 0, -1 = y2 = 1}. This is a slab parallel 2 1 to a1 and orthogonal to S1 Each of these sets can be described with linear inequalities. • S1 can be described as T vk x = 0, k = 1, . . . , n - 2 where vk are n - 2 independent vectors that are orthogonal to a1 and a2 (which form a basis for the nullspace of the matrix [a1 a2 ]T ). • Let c1 be a vector in the plane de?ned by a1 and a2 , and orthogonal to a2 . For example, we can take aT a2 c 1 = a 1 - 1 2 a2 . a2 2 Then x ? S2 if and only if -|cT a1 | = cT x = |cT a1 |. 1 1 1 3 • Similarly, let c2 be a vector in the plane de?ned by a1 and a2 , and orthogonal to a1 , e.g., aT a1 c 2 = a 2 - 2 2 a1 . a1 2 Then x ? S3 if and only if -|cT a2 | = cT x = |cT a2 |. 2 2 2 Putting it all together, we can describe S as the solution set of 2n linear inequal- ities T vk x = 0, k = 1, . . . , n - 2 T -vk x = 0, k = 1, . . . , n - 2 cT x = |cT a1 | 1 1 T -c1 x = |cT a1 | 1 cT x = |cT a2 | 2 2 -cT x = |cT a2 |. 2 2 (b) S is a polyhedron, de?ned by linear inequalities xk = 0 and three equality con- straints. (c) S is not a polyhedron. It is the intersection of the unit ball {x | x 2 = 1} and the nonnegative orthant Rn . This follows from the following fact, which follows + from the Cauchy-Schwarz inequality: xT y = 1 for all y with y = 1 ?? 2 x Although in this example we de?ne S as an intersection of halfspaces, it is not a polyhedron, because the de?nition requires in?nitely many halfspaces. (d) S is a polyhedron. S is the intersection of the set {x | |xk | = 1, k = 1, . . . , n} and the nonnegative orthant Rn . This follows from the following fact: + T x y = 1 for all y with n ? |yi | = 1 ?? |xi | = 1, i=1 We can prove this as follows. First suppose that |xi | = 1 for all i. Then xT y = ? i ? xi yi = i ? |xi ||yi | = |yi | = 1 i if i |yi | = 1. Conversely, suppose that x is a nonzero vector that satis?es xT y = 1 for all y ? with i |yi | = 1. In particular we can make the following choice for y: let k be an index for which |xk | = maxi |xi |, and take yk = 1 if xk > 0, yk = -1 if xk < 0, and yi = 0 for i = k. With this choice of y we have ? xT y = ? i xi yi = yk xk = |xk | = max |xi |. 4 i Therefore we must have maxi |xi | = 1. All this implies that we can describe S by a ?nite number of linear inequalities: it is the intersection of the nonnegative orthant with the set {x | - 1 ? x ? 1}, i.e., the solution of 2n linear inequalities -xi = 0, i = 1, . . . , n xi = 1, i = 1, . . . , n. Note that as in part (c) the set S was given as an intersection of an in?nite number of halfspaces. The di?erence is that here most of the linear inequalities are redundant, and only a ?nite number are needed to characterize S. None of these sets are a?ne sets or subspaces, except in some trivial cases. For example, the set de?ned in part (a) is a subspace (hence an a?ne set), if a1 = a2 = 0; the set de?ned in part (b) is an a?ne set if n = 1 and S = {1}; etc.

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