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Prove the given statement about subsets $A$ and $B$ of $\mathbb{R}^{n} .$ A proof for an exercise may use results of earlier exercises.a. $[(\operatorname{conv} A) \cup(\operatorname{conv} B)] \subset \operatorname{conv}(A \cup B)$b. Find an example in $\mathbb{R}^{2}$ to show that equality need not hold in part (a).

On the other hand, $\operatorname { conv } ( A \cup B )$ is segment from $( 1,1 )$ to $( 2,2 ) ,$ so equalitydoes not hold.

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 3

Convex Combinations

Vectors

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

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for part a of this exercise. It is enough to prove that the comics whole off a is a subset of the comics whole off a union. Be onboard that the comics all of B is also a subset of the comics. Haul off a union, Be okay. So let's start by taking eggs in the complex. All of a that is X can be written at the comics combination of elements of A on the coefficients of this combination are all non negative. They add up to one Andi. Every ai is an element of a A. But since it is a subset off the union off A and B well, then every AI is an element off a union. Be on one is a complex combination of elements of a union. Be therefore, eggs is in the convicts whole off a NMB. And that means that a complex all off A is a subset of the comics hole off a Airbnb. Yeah, on we can do the same for the to prove that the convicts, all of BCE's upset off the convicts. All of a union me for part B. Let's consider this example in our to it's steak. These four points they wanted to B one B two on Let's Set eight with a set that contains a one and a two. Be the set that contains be Wanna be, too. Then the comeback trail of A is the line statement that join a one with a two on the complex. All of B is another language segment, the one that joins 31 and B two Andi Zehr shown in green on in Orange, respectively. However, a Union me is a set with four points. They want it to be one B two on its comics. Hole is I mean, the complex off a union be Issa quadrilateral on. In this case, it's going to be shown in blue. So quite a lateral whose burgesses are the elements of a union beef? Well, in part a, we proved that the union off the convicts Hall of a on the convicts, all of B is a subset off the comeback soul off a union be. But this example shows that this inclusion is not necessarily inequality

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