prove-the-given-statement-about-subsets-a-and-b-of-mathbbrn-a-proof-for-an-exercise-may-use-result-2

Question

Prove the given statement about subsets $A$ and $B$ of $\mathbb{R}^{n} .$ A proof for an exercise may use results of earlier exercises.
If $A \subset B,$ then $\operatorname{conv} A \subset \operatorname{conv} B$

Gideon Idumah · Accepted Answer

No, let&#x27;s Suppose is a subset off B or A s Continue to be as we&#x27;re giving. No, we already know. We know that be is always a subset or he&#x27;s always contained in the convents all of being. So this implies that it is the substance of being on B is contained in the convicts all of being no, by definition, convicts All off being is a convict sets. Therefore we can use exercise. Seventh scene. This would imply dots the convicts because you are birds. It is a subset of the convicts. 11 just invited this way so as to be clear, so is the source of these. This is a convict sets. So that implies that the convicts all off the is this so sets off contacts or

Gideon Idumah · Answer

not to prove these. We know that into sexual being will always be a substance off a right because you&#x27;re intersections. Something would be it&#x27;s always you&#x27;re intercepting a woods be. It&#x27;s always contended day. So this implies that from exercise eight scene we have That&#x27;s the convex all off a intersection. B must be a subset of the comics all off, eh? Similarly, we know that intersection being will be a subset of Be so from exercise 18. Just doing the same thing over game we have that the convicts all off a intersection b will be, is so set off the convents all off B. Now look at these three questions. The left answer on something Sam is because this is Indy&#x27;s Andi. This is also indeed so. This implies that the convicts all off a intersection, being be a subset or contained in the convicts. All off a intersection. The convents all off. You can just put this now for the deep parts. Wanna showed us equality, Need notes. Oh, so let&#x27;s give an example. So take or consider four line segments warn out. Sue out 34 dots. Mick is square, so want to use an example of his square? No. Um, let&#x27;s A which is a cold and one comma l three on b which is out. So come on. L four be sets dogs, countries, a poor city lines. But that&#x27;s what means something like these. So this is everyone out soon off for So you have everyone l three outs for the opposite lines. This would imply that the convicts all off a intercision the record that this is a So this is a gun, then this is B. Now the contracts all off a intersection being would just be the convicts, all because it into sexual being. He&#x27;s just Did I go now on this side of the square off the square? I discussed this point this point this point This points So the convicts all off eight ounces of people be the convicts all off the boxes. That&#x27;s is this four points. This four points into Section B. Nazis. So Dobby, the convicts all off a into section will be that sees. Did the argot knows on sides off the square off the square. That is a dog going on the side of the square. No, we&#x27;re us. The convicts hold off into section convicts. All off being is the old square. So this is that all square where us D&#x27;s is decides? No, just the bodice. Is that say This side&#x27;s I&#x27;m Doug going out. So the dog owners on decides where us This one is the old square us old. So we see that equality doesn&#x27;t recorded, See doesn&#x27;t hold one is the old squared whereas one is the convicts all off the vortices

Gideon Idumah · Answer

least suppose it is contending. Be as we&#x27;re giving. No, we know that be is always contained in the are fine off B. This is world known, so this implies that&#x27;s a is contending, be on B. Is contending are finally being This would imply that a is contained in the fine of B. We also know that&#x27;s are fine of being is always fine rights by the definition so we can use exercise twins who warned. So conclude dots from these from star because this is our fine on DDE. Exceptions were not tells us that&#x27;s our fine of e would be contained in our fine off beat.

Michael Jacobsen · Answer

in this example, we&#x27;re going to be giving a direct proof of the statement that if BSF fine and if a is contained and be then the F fine hole of a must also be contained in B, let&#x27;s start out with the following at the following statement, which comes from the statement that B is f fine. What this means is that oh, fine combinations of vectors and be are also in B. So with this a statement here, let&#x27;s start with an element that comes from the a fine set of a We can denote it by why? And will be 1/2 in combination of vectors in a So let&#x27;s set, Why equal to see one v one plus c two v two plus all the way through C P V P. Where c one plus C for two all the way through C p equals one. So this is an and find combination. And let&#x27;s also is insist that V one through veep to through Visa P is in a so for this specific situation. This tells us that then why is in the f fine hole off the set? A now notice that we haven&#x27;t used yet the assumption that a is a subset of B because A is a subset of be It follows that the vectors the one V two through Visa P are also in be due to the set Containment A is a subset of B and we know also at the same time all at fine combinations of vectors and B are also in B and this is one such a fine combination. So this shows that the vector y, which has indicated above here, is in B. So what we&#x27;ve done is we&#x27;ve generated an arbitrary vector in the a fine hole of a and we&#x27;ve shown that that same vector must be in the set B. This shows then that the F find hole of a is entirely contained in a set B and this completes our proof.

Gideon Idumah · Answer

we need to give an example that shows that equality needs not old in the statement of exercise 25. So let&#x27;s a big call, too. X one comer x two Such That&#x27;s x two quarts zero zero is less than a goto x one less than a call to warn. So this is line segments from because extra zero So from zero comma zero sue one comma zero in our two. Similar Let&#x27;s be Let&#x27;s define P as x one comma x two Such That&#x27;s ex sued in court zero but less big X one from one lesson across the X one doesn&#x27;t cost a sou. So we said that this is a line segments from one comma zero sue to comma zero in our two. So we joined a dark. I was just something like this and eighties from zero said this is one on this is too. So this is a because extra zero on Then this is ah bee this line segments. So let&#x27;s not consider this implies that a intersection Be so I said this is a Andi. Let me use the going pain. This is a on then this is be my muse aerate pain from thoughts. This is B. Now this implies that&#x27;s it into Section B would just be this points. That&#x27;s where the off something in common now be one comma zero, which is just one points the are soon so because it&#x27;s just one point that implies that are fine off a intersection. B recorded our fine off a single. It is said that, as one point is just the same sets. So the sentence as a intersection being which is just sending us one. Come on, zoom. Now let&#x27;s look for um I find off, eh? He&#x27;s in line, right? That&#x27;s ghost True. The origin on points one comma zero on DDE Our find of being is in line. Dots goes true. One comma zero on Sue. Come on, zero. This lines are basically the same thing like I showed you. So these lines are the same. Why? Because the butcher percents the boots are percents the expanses This would imply Daddy are fine off, eh? It&#x27;s a section are fine off B. He&#x27;s in lye. Doug&#x27;s were percents. The Excellencies. We see that&#x27;s quality that is dense is not equal to Vince because so quite did see, doesn&#x27;t old because are fine off a intersection. B is a points on dhe are fine off a intersection are fine of being, is why.

Gideon Idumah · Answer

So for these elastic A to be called so zero comma zero. Because when our Tsu So this is is said&#x27;s duds quarantines only one points on the points is 00 So Circle Missouri is just one points because we are too now. Similarly took B to B whatsoever you want, say, for example, I can pick one comma, one that&#x27;s convenient for me. So, as it said&#x27;s dots contains only one points, which is one come award. Then this implies that a junior B would be zero comma, zero comma, one comma one It is said that contains two points. Contains two points, which is zeros comma zero on dhe, one comma one. Then this would imply that I find off a union be is in line because its contents two points is in line. True Zero comma zero on one comma one. I&#x27;ve, uh ah fine off, eh? Because toe a on dhe I find off b equals to be the reason why this is so is goes day, uh, said&#x27;s weeds. Only one element I recall that is a serum that sees that&#x27;s it sets with one element are fine off that sense. Sentinels are sets So this implies that the are fine off. A genuine are fine off be would be zero comma zero because there&#x27;s just a come on one. Come on. One which is It&#x27;s said&#x27;s dots contains sue points so we can see that&#x27;s our find off A You know, a friend of B is a set the contest to points where off Whereas I find off a union be line True zero comma zero So they&#x27;re not the same thing. So this is a set. This is in line. So this implies that&#x27;s I&#x27;m fine off a union be is not equal to our fine off eh? Union are fine off b.

Problem

Prove the given statement about subsets $A$ and $…

Problem 18 Easy Difficulty

Prove the given statement about subsets $A$ and $B$ of $\mathbb{R}^{n} .$ A proof for an exercise may use results of earlier exercises.
If $A \subset B,$ then $\operatorname{conv} A \subset \operatorname{conv} B$

Answer

Suppose that $A \subset B$ . We allready know that $B$ \subsetcomv(B), so $A \subset B$ Conv $( B )$ .
Also, by definition, corrv $( B )$ is compex set, so result follows from exercise 17.

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

Linear Algebra and Its Applications

Lily A.

Catherine R.

Anna Marie V.

Kristen K.

Vectors Intro

Vector Basics - Example 1

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Suppose that $A \subset B$ . We allready know that $B$ \subsetcomv(B), so $A \subset B$ Conv $( B )$ .Also, by definition, corrv $( B )$ is compex set, so result follows from exercise 17.

Linear Algebra and Its Applications

Lily A.

Catherine R.

Anna Marie V.

Kristen K.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Catherine R.

Anna Marie V.

Kristen K.

Suppose that $A \subset B$ . We allready know that $B$ \subsetcomv(B), so $A \subset B$ Conv $( B )$ .
Also, by definition, corrv $( B )$ is compex set, so result follows from exercise 17.

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

Linear Algebra and Its Applications