prove-that-the-convex-hull-of-a-bounded-set-is-bounded

Name: prove that the convex hull of a bounded set is bounded
Uploaded: 2019-08-16
Duration: 1 min 13 s
Channel: Numerade
Description: Prove that the convex hull of a bounded set is bounded.

Question

Prove that the convex hull of a bounded set is bounded.

James Chok · Accepted Answer

Okay. So he will prove that the convicts hole all a bounce. It is bounded. So that s b abounded. Founded set. Then since our just rights properly founded sense. So since s is a bounce set that exists a delta well that there exists a radius Great. Enjoy such that this set s is contained in a bull centered at zero with various are now This ball is convex was continents Also we have that the convex home off s is a subset off the bull sent it at zero. Don&#x27;t s there. Aw so therefore you&#x27;re convicts. Come on Convicts Haul of s is also founded since it&#x27;s contained in their bounds

Chris Trentman · Answer

were asked to prove that the sum of conduct sets is again a convex set. So let A and B he convict&#x27;s now let the vectors x and y B in the set some a plus B. This means that we confined in a in C in A and A B and a D in the set B Such that X is equal to a plus B and why is equal to C plus D. Now let t lie between zero and one Them we have that the point on the line segment x y It&#x27;s going to have the form W, which is tee times X plus one minus t times why you could do the other way around. This is the same as tee times a plus B plus one minus t times C plus d. Of course, this is equal to grouping together vectors from the same set. So we have t a plus one minus t see plus t B plus one minus t the now because a and beer convex it follows that ta plus one minus t See, this is an element of the line segment between A and C. This lies in a and we have that T B plus one minus TD. Since B is convicts and this is an element of the line segment between B and D. This also lies in B, so it follows that w lies in a plus. B. Therefore, it follows that a plus B is convicts, since it contains any points online segment between and between X and Y.

Chris Trentman · Answer

rest to prove that a point in a context set is an extreme point of the set if and only if the set minus that point is still convex. So we&#x27;re going to let the be an element of s where s is Convex set. And now let&#x27;s prove the yeah, it direction first. So suppose that V is an extreme point of this And let t b the set of all vectors X in s such that X is not equal to our extreme vector. The now if x and why are in t them it follows that the line segment by X y is an s. Since Esa&#x27;s convicts now the is an extreme point of s so follows the V does not lie on the line segment y z. If it did, it would have to be one of the endpoints x or y, which it&#x27;s not not lying line saying the next why and therefore falls X Y is a proper subset of tea. So it follows that t is convex is what we wanted to prove now to prove the other direction Who&#x27;s that he is not in extreme points were actually going to prove the contra positive, then this means that we confined two vectors X and Wyness such that the lies in line segment X, Y and V is not equal to X and V is not equal to why has to find the interior of the line segment. So it follows that from the previous paragraph x and Y R N t. But the line segment X Y is not a subset of tea because V allies on the line segment India&#x27;s not in T so follows. T is not convicts. We have proven what we wanted to prove.

Chris Trentman · Answer

were asked to find an example of a bounded conduct set s in our to such that the profile of set is non empty. But the convex hole of the profile is not the same as the set. So this is different from a previous problem in the previous problem. Exercise 10. We&#x27;re looking for an example of a compact contact set with this property, but now we&#x27;re looking for a bounded convex set. So an easy example is to consider a square. We&#x27;re only two of the sides are included in the set, So we&#x27;re going to let s be a square with just two adjacent edges. In other words, it&#x27;s going to look something like this. The dotted lines indicate that this boundary is not included. Now we see that the context hole Well, Katie profile is non empty because mm profile P includes the three. Yeah. Mm hmm. Corners shall guess I&#x27;ll put in quotes because technically, two of them aren&#x27;t really corners. But the three corners of this square, So the profile looks something like this is the only extreme points. So you have these three points for the profile, and we have that the convex hole of P is a triangle, so forming the convex hole which are drawing green, you simply draw a triangle between these three points and it&#x27;s clear at the convicts. Whole of P is not the same as the set s.

Gideon Idumah · Answer

so two shoulders the sides s which is a gold toe x In our end such that f of X is in C. To show that this set is a convict sense convicts subsets off our aim would peak sue points See, let&#x27;s call them X one x two in this under. We sure that&#x27;s the line segments the line segments between Damn, Isn&#x27;t this no leads X one be in this By definition, it&#x27;s implies deaths f off X one isn&#x27;t seen Now this implies dunce f x one de quarto. Why one for why one in c all rights? Because we say f excellent isn&#x27;t seen. So I miss fo x one is altering Why one for which why one has got to see this would imply that X one accord to f involve us off what I want. You just take the inverse of what side? Similarly, let x to be element of Bess. This implies that f of x two by definition, beaten See, which implies that f of x two. Because why too for a white soon in C. This implies that X Toole accord to f involve us off. Why? I want so to show that the line segments between them it&#x27;s between s we must show We must show That&#x27;s one minus C x one lost The ex too must be in s for every zero lesson opposed to see less than a course one. So let&#x27;s show these now off one minus c x one lost Seen X to disappear one Miners seem our x one is f ive us of y one loss seem f universe off white too. This is nothing Bad&#x27;s when you multiply it. Woman listed by these are FBI of us off Why Juan minus C f f s of y one lost seen f inverse off. Why, too, which of course toe f any of us or why one miners? Because f is Nina f impossible genius So I can pull this inside. We&#x27;ll see why one loss as one of us off. See why, Sue? So this is just equal to f ive us because this sort of putting things f is linear. Ft of us is also Minya. So that&#x27;s why I can do all these operations. I can pull out every bus I have what I want minus tse. Why one lost see Why soon? So this is recalled to f Reverse off one minute. Just see why Want plus c? Why soon Now, just pull these down. What obvious? That things is a board to bees? No, Because of these, you call that See his converts. So let&#x27;s go back. That&#x27;s so since C is convicts the points one miners see why one lost See white soon Most be unseen all rights Because this is the convicts combination. Because why one on White? You isn&#x27;t seen now from this three questions, Tom and Nicole is a question star. No, from star. What would our duds one minus c? Why? Want lost? See White? So I&#x27;m moving this f in verse to this side. So this is equal to f off this. So that&#x27;s is this will be equal soon. F off one minus c. That&#x27;s one lost X zoom. Now we said this isn&#x27;t seen, so this most b c so we&#x27;ll show that f off one minus c x one Must be in C by definition. Oh s If this isn&#x27;t see by definition of s, this would imply that one minus X one lost. See X two Must be s on. This implies that this is a conference

Chris Trentman · Answer

so given a conduct set and to positive numbers, C and D were asked to show that CS plus D s going to be equal to C plus D s. So if we in other words, stretch the elements of a convex set by sea and stretch the elements of a complex set ID and then add thes two new sets together we get a set, which is three elements of s stretched by C plus D. So we&#x27;ll let s be convex set. Let&#x27;s see, be greater than zero and d be greater than zero. Now let X be an element of CS plus ts and we confined S one and s two in s such that X is equal to see X one are sorry CS one plus D s to then it follows. The X is also equal to C plus d times. See over C plus d. We can divide by C plus Decency plus D is greater than zero since both c and D or greater than zero s one plus de over C plus D s to now because S is a convex set. It follows that see, oversee plus D s one plus D over C plus D. As to linear combination of elements from s also lies an s. So it follows that X is an element of C plus D s. So we&#x27;ve proven one direction. That is, that C s plus D s is a subset of C plus D s. Now to prove the other direction Similar exercise. So we&#x27;re going toe Let X be in the set C plus D s. This means we confined. A vector s in the set s such that X is equal to C plus D. Yes, and of course, this is equal to CS plus D s. After distributing the scaler, we could say, even distributing the vector across the scale Er&#x27;s And this is clearly an element of C S plus D s. So X is an element of CS plus ts and therefore it follows that C plus D S is a subset of CS plus D s. Combining this statement and the previous statement we get that CS plus D s equals C plus D s, which is what we wanted to prove

Problem 30 Medium Difficulty

Prove that the convex hull of a bounded set is bounded.

Answer

Theorem 9 in section 8.3 implies that conv $S \subset B(p, \delta)$ and conv $S$ is
bounded.

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Lily A.

Kayleah T.

Kristen K.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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Theorem 9 in section 8.3 implies that conv $S \subset B(p, \delta)$ and conv $S$ isbounded.

Linear Algebra and Its Applications

Lily A.

Kayleah T.

Kristen K.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Lily A.

Kayleah T.

Kristen K.

Joseph L.

Theorem 9 in section 8.3 implies that conv $S \subset B(p, \delta)$ and conv $S$ is
bounded.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications