AI Assistant
Transcript
00:04:590Paolo Guiotto: Okay, so last time we introduced these two important properties of SATSER.
00:16:10Paolo Guiotto: We defined the open set.
00:20:750Paolo Guiotto: Which are sets for which every point of the set
00:26:960Paolo Guiotto: Stays into the set itself with a ball, Centered at that point, huh?
00:34:410Paolo Guiotto: Equivalently, this means that all points are interior points of S.
00:39:980Paolo Guiotto: or the interiors of S coincide with S itself.
00:44:160Paolo Guiotto: We also defined a second type of set, which are closed sets.
00:49:420Paolo Guiotto: Which, by definition, is a set… Whose complementary is open.
00:56:60Paolo Guiotto: Now, since one can be confused by these definitions, there are certain errors that are frequently
01:06:670Paolo Guiotto: done with this definition. For example, one thing that SAT is either closed or open, or,
01:19:170Paolo Guiotto: At least one of the two, and it cannot be both. All these things are false.
01:25:80Paolo Guiotto: However, these definitions are relatively abstract and difficult to verify.
01:33:00Paolo Guiotto: In RN, usually, we define sets in a particular way, so it will be important to understand under which condition a set is open, closer by looking at the form, how the set is defined.
01:49:780Paolo Guiotto: So, normally.
01:57:720Paolo Guiotto: subsets… of… R. D… are defined… through…
02:13:260Paolo Guiotto: number… of… equations.
02:26:390Paolo Guiotto: Or… Well, let's say, and… Or… inequalities.
02:39:660Paolo Guiotto: So, for example, Let's see a few examples.
02:44:840Paolo Guiotto: Example.
02:50:230Paolo Guiotto: You want to describe a viscous entry at the origin with radius 1, no?
02:57:840Paolo Guiotto: So, these cut… So let's say, playing these cut…
03:08:150Paolo Guiotto: Sanford.
03:12:420Paolo Guiotto: 0, 0.
03:14:700Paolo Guiotto: And… That was… weed the…
03:23:320Paolo Guiotto: radius… are positive.
03:28:730Paolo Guiotto: So, graphically, we are talking about this.
03:32:920Paolo Guiotto: let's say the floor disc. So, the disc with the BMD, okay? How do we describe this? Well, formally, this is a bolar.
03:45:860Paolo Guiotto: This is the bowler, centered at 0, 0.
03:51:10Paolo Guiotto: with radius 1. So the definition is, formally, the set of vectors y in R2, such that the norm of Y minus the center, which is 0, is less or equal 1.
04:09:360Paolo Guiotto: This means that we are taking points Y with coordinates, say, UV, Then we will,
04:17:79Paolo Guiotto: use letter X and Y, but for this moment, since I'm using arrays, let's give different names to coordinates, such that… now, norm of u minus… norm of Y minus 0 is the norm of Y,
04:31:430Paolo Guiotto: So this means the root of u squared plus V squared less or equal than 1.
04:38:660Paolo Guiotto: And as you can see, this is an inequality. If you want, you can equivalently write this as the set of points UV in F2, where U squared plus V square is less or equal than 1.
04:53:150Paolo Guiotto: So you see, this is… the set is defined through this inequality.
05:00:30Paolo Guiotto: Okay? Let's see another example. The half plane…
05:12:100Paolo Guiotto: of points… with the… let's say… Abshesa…
05:23:30Paolo Guiotto: Greater than, 2.
05:25:920Paolo Guiotto: What is this? Well, this is… the figure is…
05:31:750Paolo Guiotto: Let's now use letters X and Y for the coordinates.
05:35:650Paolo Guiotto: You have to be a little bit flexible with the letters, so this is 2.
05:40:390Paolo Guiotto: This is where this, dot line…
05:45:720Paolo Guiotto: is the set of points where the abscessa is exactly equal to 2. We want those who the abscessa is greater than 2, so this is the half plane at right here, without the dotted line.
06:01:900Paolo Guiotto: So, formally, I could say that this is the set E made of points X, Y, in R2,
06:11:590Paolo Guiotto: Such that X is greater than 2.
06:15:660Paolo Guiotto: As you can see, this is, again, an inequality. This time it's a strict inequality, but let's say it's still an inequality.
06:24:140Paolo Guiotto: Okay?
06:25:290Paolo Guiotto: I want to, let's say, an equality example. The circle… centered, at, well, let's say, 1-1… Radius…
06:43:680Paolo Guiotto: 4.
06:45:470Paolo Guiotto: It is what?
06:46:980Paolo Guiotto: So I'm talking about this.
06:49:710Paolo Guiotto: In a plain ex-wife.
06:54:690Paolo Guiotto: The center is this one, 1-1.
06:58:210Paolo Guiotto: And the radius is 4, so let's say 1… 2, 3, 4.
07:03:820Paolo Guiotto: So, circa… Circle, so not disc. Circle means that it is this set.
07:14:460Paolo Guiotto: Only the red line that you see here.
07:18:340Paolo Guiotto: Okay, so this is the set of points, this one is the set of points, XY.
07:26:420Paolo Guiotto: In Aptu.
07:28:510Paolo Guiotto: Such that. Now, the distance to 11 is exactly equal to 4.
07:44:630Paolo Guiotto: Okay, so this is, if you want, this is the radius, it is 4. So this means that here we have points X, Y, and R2, such that the distance is the root of X minus 1 squared plus
08:01:940Paolo Guiotto: Y minus 1 square. This must be less… sorry, it must be equal, exactly equal to 4.
08:12:260Paolo Guiotto: If you want, equivalently, we can say that this is the set of points XY of 2,
08:20:70Paolo Guiotto: Where squaring everything becomes X minus 1 squared plus Y minus 1 squared equals 16.
08:30:110Paolo Guiotto: And this is… this time, this set is defined through an equality, an equation.
08:37:530Paolo Guiotto: Okay.
08:38:990Paolo Guiotto: We can do the same in three dimensions, no? For example, I do not write the names, because not… the sets are not necessarily so easy, no? But I could say, I don't know, I take the set E of points X, Y, Z in R3,
08:58:490Paolo Guiotto: That… such that they verify a number of equations, inequality, I don't know, like X plus YZ greater or equal to 4, maybe some of this, is, is, give… gives some condition without any points. It is clear that if I write X squared plus Y squared plus Z squared equal minus 1, there is nothing that verifies this.
09:21:650Paolo Guiotto: But this still makes sense, okay? The set will be empty.
09:25:740Paolo Guiotto: Okay?
09:26:950Paolo Guiotto: Normally, we write meaningful things, but let's say this is a possibility, no? As you can see here, this is an inequality. Sometimes we can have more than one condition, no? Because basically, these are conditions on the coordinates that must be verified.
09:45:380Paolo Guiotto: For example, let's say that I take the settee in the plane XY, so the set of points XY, such that X plus Y is greater or equal than zero, and Y is equal to…
10:05:490Paolo Guiotto: Explorer.
10:08:210Paolo Guiotto: So, this is a set of points where I have two conditions that must be both verified together. So, in any case, the first one is an inequality, the second one is an equation. This is just to say that it is a typical way
10:23:520Paolo Guiotto: how to define sets? In this case, we can even see what is this, because
10:28:920Paolo Guiotto: In the plane XY, the first condition, X plus Y greater or equal than 0 can be also written as Y greater or equal than minus X. Y equal to minus X is this, say, this dashed line.
10:47:500Paolo Guiotto: This is the liner.
10:49:260Paolo Guiotto: where Y is equal to minus X. Now, I have to take points XY, with the Y greater than minus X, so these are above that line, including the line, because there is the equal. So these points are included, and these points I'm…
11:06:900Audio shared by Paolo Guiotto: Coloring in red. These are…
11:10:60Paolo Guiotto: the points that verify this inequality, X plus Y greater than or equal to 0. You know why this is not nice? Because every time I have to stop the class to admit the person.
11:22:130Paolo Guiotto: It's impossible to make, to understand this thing.
11:25:960Paolo Guiotto: So, I repeat, this is for exceptional reasons. It cannot be that the classroom is half empty and people are online. And the second… there is a second condition, which is Y equals X squared.
11:42:290Paolo Guiotto: Now, what does it mean? These are points where the Y is X squared, so it's a parabola. The parabola Y equals X squared is this one.
11:52:990Paolo Guiotto: So it is something like this.
11:57:420Paolo Guiotto: More or less. Now, we have to take these points, so let's say that these points, the green set, is the set of points where Y is exactly equal to X square, so Y equal X square is the green set.
12:15:60Paolo Guiotto: Now, since we have, to take both conditions, so we have to take the points which are, at the same time in the green set and in the red set. So this means that we have to take these points.
12:30:80Paolo Guiotto: the points that are in this branch of the parabola, and the points that are on this other branch. So this blue is E.
12:40:580Paolo Guiotto: Okay? Of course, we can do figures when it is easy, and mostly for plain sets. For three-dimensional sets, it's hard, and for four-dimensional sets, it's basically impossible. So let's say that, however.
12:57:380Paolo Guiotto: So… quiet… Oh, man.
13:06:810Paolo Guiotto: Wait… to define sets.
13:12:560Paolo Guiotto: In… RD… Is that… is of this… General…
13:25:730Paolo Guiotto: form. Well, let's do by steps. For example, I could take the set of vectors X of RD, where a certain inequality… what is an inequality? Is a conditional
13:42:40Paolo Guiotto: on the coordinate greater or equal than zero, for example. You see, that's greater or equal than 4, you carry the 4 on the left-hand side, you can write something greater or equal than 0. So let's say that, in general, it will be a condition on the coordinates, X1, XD,
14:00:210Paolo Guiotto: That must be verified this greater or equal than zero, for example.
14:05:160Paolo Guiotto: Okay?
14:06:250Paolo Guiotto: Now, there can be sets where there are more conditions. So, for example, here we have two conditions. So, this means we have a first inequality, G1X1XD,
14:18:840Paolo Guiotto: greater or equal than zero. A second inequality, G2, X1, XD,
14:25:410Paolo Guiotto: greater or equal than zero, okay?
14:28:370Paolo Guiotto: then we can have an extension of this will be set of points of RD, where we have M conditions, so G1,
14:39:390Paolo Guiotto: Let's write just vector X greater or equal than zero, G2 vector X greater or equal than zero, and so on, DM of X greater or equal than 0.
14:53:870Paolo Guiotto: So this is a possibility, okay? We can have the case where these are strict inequalities, so something like,
15:02:670Paolo Guiotto: G1 of X strictly greater than 0. G2 of X strictly greater than 0. And so, GM of X
15:14:930Paolo Guiotto: strictly greater than zero. We can have the case when we have mixed inequalities. So, for example, some conditions are written as
15:25:200Paolo Guiotto: Do you wanna…
15:27:550Paolo Guiotto: of x strictly positive, then we have G2 of X greater or equal than 0, for example. Why not?
15:35:590Paolo Guiotto: Okay? There can be also equalities. They will be something like G1 of X equal.
15:45:290Paolo Guiotto: 0, G2 of X equal 0, and so on.
15:51:860Paolo Guiotto: Huh? Or any kind of combination of these things, okay?
15:56:640Paolo Guiotto: Now, this is a natural and common way to define sets in Ardino, through a certain number of… you have to look at these dates as conditions on the coordinates. So the point XYZ must refine this and this.
16:14:610Paolo Guiotto: Okay? Or only one, a state, inequality, etc. This is the natural way we define subsets of RD.
16:24:40Paolo Guiotto: So the question is, is there any condition on these functions, G, that can tell us
16:30:190Paolo Guiotto: That these sets are open or closed.
16:34:590Paolo Guiotto: So the problem is… He's there.
16:41:560Paolo Guiotto: And it's possibly simple.
16:46:860Paolo Guiotto: condition.
16:49:200Paolo Guiotto: on the functions G, Such that,
16:58:140Paolo Guiotto: Well, let's see.
17:01:80Paolo Guiotto: That… Belts.
17:07:190Paolo Guiotto: if… reset.
17:12:329Paolo Guiotto: He's open… Or closed, though.
17:18:869Paolo Guiotto: Now, we can, there is actually… of course, this is true, but let's first try to understand what should be the answer. Let's take this example.
17:32:850Paolo Guiotto: Now, we take a case, a simple case, so plain set.
17:38:880Paolo Guiotto: It's the disc. I first take the disc, center it at the origin with the radio sung, the close disc.
17:48:830Paolo Guiotto: reminder, we say closed this because at the end of the story, it will be a closed Saturn. So, this is the Saturn.
17:59:530Paolo Guiotto: This is the closed disk, and this one is similar, but with the strict inequality. It is the open disk that we already know, we have seen yesterday.
18:10:240Paolo Guiotto: It is open.
18:13:810Paolo Guiotto: So it is more or less the same. The only difference is that we do not have the points of the unitary circle.
18:22:00Paolo Guiotto: Which is the boundary for this set.
18:24:140Paolo Guiotto: Okay?
18:25:410Paolo Guiotto: Now, if you look at these two sets.
18:28:100Paolo Guiotto: Well, we know that the second one is open. We proved in general that an open ball is an open set.
18:34:900Paolo Guiotto: And what about the first one? What do you think? Is it open or closed?
18:41:330Paolo Guiotto: closed, because if you think to the complementary is whatever is outside except that disk, so it won't have the boundary, basically. So this one should look as a closed set.
18:53:860Paolo Guiotto: If we look at the analytical form, you see that the unique difference… mmm…
18:59:960Paolo Guiotto: The only difference is, at this stage, here we have a large sign, here we have a strict sign.
19:07:950Paolo Guiotto: Now…
19:09:280Paolo Guiotto: It turns out that basically this is the ruler, provided that the function that defines a constraint is a continuous function. So, it turns out that we have this proposition.
19:26:430Paolo Guiotto: Well, let's say we can state in words, and then write a bit formally.
19:33:320Paolo Guiotto: So, any set… defined.
19:41:320Paolo Guiotto: Bye.
19:43:150Paolo Guiotto: Yeah. Behind it, number.
19:50:80Paolo Guiotto: All.
19:53:90Paolo Guiotto: large… Large or weak… inequalities.
20:05:900Paolo Guiotto: So this means less or equal, greater or equal, okay?
20:10:580Paolo Guiotto: Less or equal, or… greater or equal Sorry.
20:17:770Paolo Guiotto: So sometimes you call them large inequalities, other times weak inequalities. In any case, that means the same.
20:25:730Paolo Guiotto: Or…
20:30:180Paolo Guiotto: by equalities.
20:35:130Paolo Guiotto: by a finite number of large, weak inequalities or equalities.
20:42:840Paolo Guiotto: So, where you have the equal sign.
20:45:980Paolo Guiotto: So, any set defined by a finite number of larger weak inequalities, or equalities, is closed.
20:58:540Paolo Guiotto: So this means, precisely.
21:04:630Paolo Guiotto: So Seth's like, no.
21:07:500Paolo Guiotto: I forgot the fundamental assumption, sorry. It's closed, let's add here, provided
21:20:90Paolo Guiotto: B.
21:22:110Paolo Guiotto: These are… these, inequalities, or equalities are called constraints.
21:34:30Paolo Guiotto: are continuous functions.
21:37:560Paolo Guiotto: So, are continuous.
21:41:860Paolo Guiotto: So… Yes, of course.
21:46:00Paolo Guiotto: formally.
21:49:360Paolo Guiotto: So, this means that if we have a set S,
21:53:700Paolo Guiotto: Such that S is made by vectors X of 5D,
21:58:820Paolo Guiotto: such that a certain function, G1 of X is greater or equal than 0. A certain other function, G2 of X is greater or equal than zero, and so on. A function GM of X is greater or equal than zero.
22:15:450Paolo Guiotto: So these functions, G1, G2, GM, are what we call the constraints.
22:29:480Paolo Guiotto: Because the idea is that, like, physically, these are conditions that must be verified to stay there, no?
22:37:690Paolo Guiotto: Like, physical constraints.
22:39:660Paolo Guiotto: So, a set S of this type, or S of type… X in RD. Where?
22:50:190Paolo Guiotto: G1 of X is equal to 0.
22:54:50Paolo Guiotto: G2 of X, is equal to zero, and so on. GM…
23:01:170Paolo Guiotto: of X is equal to 0.
23:06:760Paolo Guiotto: Well, actually, even a combination of them.
23:10:440Paolo Guiotto: Or, inside where you have.
23:13:850Paolo Guiotto: a number of large inequalities, say, G1X greater or equal than zero, etc, GKX greater or equal than zero, then you have a certain number of equalities.
23:30:300Paolo Guiotto: GK plus 1 of X equals 0 until GMX
23:36:520Paolo Guiotto: Equals zero, so any kind of combination of equalities and inequalities, provided the inequalities are large.
23:44:230Paolo Guiotto: And these functions, G1, GM, are continuous. So, formally, if S is. Are. Died.
23:56:470Paolo Guiotto: One of these.
23:58:850Paolo Guiotto: with.
24:01:230Paolo Guiotto: The functions G1, GM, continuous.
24:07:330Paolo Guiotto: Then, the set S is closed.
24:13:250Paolo Guiotto: Okay?
24:15:780Paolo Guiotto: So, it means that if you want to verify in concrete cases, like these ones that you have seen in the examples, if the set is open, closed, and look at this one.
24:26:610Paolo Guiotto: You can see this as a unique constraint, like G plus X plus YZ minus 4 greater or equal than 0. That's your function G for this example, G of XYZ.
24:40:250Paolo Guiotto: Now, you see that it is a large inequality, so you should bet on the fact that that's closed, Seth.
24:47:510Paolo Guiotto: To be closed, you are… it is sufficient that the function be continuous. Is that true? Yes, because it is a polynomial. So you see, you don't have to verify anything. You just look at the shape of the set, large inequalities.
25:01:440Paolo Guiotto: with continuous constraints gives immediately the settings closed, okay?
25:10:100Paolo Guiotto: The other case is when we have strict inequalities.
25:15:520Paolo Guiotto: Moreover…
25:23:600Paolo Guiotto: And he… set.
25:26:660Paolo Guiotto: defined.
25:28:800Paolo Guiotto: By a finite number.
25:36:430Paolo Guiotto: off.
25:38:180Paolo Guiotto: Maui, we have a spring.
25:42:30Paolo Guiotto: inequalities.
25:48:100Paolo Guiotto: So, it means that we have either is or IS sign, okay?
25:55:730Paolo Guiotto: Of course, there are not the equalities here.
25:59:30Paolo Guiotto: is, is… Open.
26:05:650Paolo Guiotto: provided… the… constraints.
26:16:610Paolo Guiotto: R… Continuous.
26:20:60Paolo Guiotto: So, formally, a set S of type.
26:26:970Paolo Guiotto: vectors X in R… D, such that G1 of X is strictly greater than 0, etc. GM
26:37:640Paolo Guiotto: of X is strictly greater than zero, this thing with G.
26:43:230Paolo Guiotto: with… the functions, G1, GM, continuous, then this set S is open.
26:56:30Paolo Guiotto: Okay?
26:57:230Paolo Guiotto: Now, I won't do the proof for such a generality, but we will see a simple proof in a… we will see a proof in a simplified case. So let's take an example of a set as defined as the set of points X of RD,
27:14:250Paolo Guiotto: where a unique constraint, G of X, is greater or equal than 0. Then you can extend the proof for the general case.
27:23:800Paolo Guiotto: So the goal is… Approver.
27:32:50Paolo Guiotto: that S is closed.
27:38:550Paolo Guiotto: If… the function G is continuous.
27:43:240Paolo Guiotto: And now here is where we use the contour characterization we have seen yesterday. So let's review this quickly. Yesterday, we have seen that
27:55:240Paolo Guiotto: characterization for closed… for closeness is that this property holds. Whenever you have a sequence of vector in your set.
28:05:420Paolo Guiotto: such that it has a limit, a finite limit, L, then also the limit must be intercept.
28:12:50Paolo Guiotto: Okay, that's what we verifying here.
28:14:530Paolo Guiotto: So, to check… Check that.
28:22:660Paolo Guiotto: S is closed.
28:27:480Paolo Guiotto: we apply… the… contour… optimization.
28:40:620Paolo Guiotto: of those.
28:44:430Paolo Guiotto: sense.
28:46:690Paolo Guiotto: So, I have to prove that.
28:50:90Paolo Guiotto: Let's write the property here. For every sequence Xn of vectors of S, Such that this sequence XN
29:00:480Paolo Guiotto: Goes to some limit.
29:02:520Paolo Guiotto: off.
29:04:30Paolo Guiotto: argue.
29:05:450Paolo Guiotto: Then, necessarily, also the limit belongs to the set S.
29:11:790Paolo Guiotto: Why?
29:13:120Paolo Guiotto: Well, let's see. Let's take a sequence of points of S. It means that the… each of them, XN, belongs to S.
29:23:360Paolo Guiotto: But S is the set of points where our function g is greater or equal than 0. So this means that G of Xn will be always greater or equal than 0 for every n.
29:38:720Paolo Guiotto: Now, let's pass to the limit.
29:41:180Paolo Guiotto: What happens when we send N to plus infinity? The point XN goes to point L, which is a finite point. It's not infinity.
29:51:830Paolo Guiotto: And the function G is continuous.
29:54:60Paolo Guiotto: So what happens to the continuous function? You know that when X goes… when X goes to X star, G of X goes to G of X star. That's the property of continuity. So by continuity, this G of XN will go to G of L.
30:11:90Paolo Guiotto: This is where continuity enters.
30:17:470Paolo Guiotto: But, since G of XN were positive, the limit must be positive greater or equal than zero. The limit must be greater or equal than zero, because of the
30:29:790Paolo Guiotto: So-called the payroll insert.
30:34:810Paolo Guiotto: All.
30:37:660Paolo Guiotto: the… sign.
30:40:940Paolo Guiotto: So, the limit of positive numbers is positive, cannot be negative.
30:45:100Paolo Guiotto: And therefore, what does it mean at the end? That G of L must be positive, so this means that also L belongs to the set S, because the set S is the set where G is positive.
30:56:10Paolo Guiotto: You see?
30:57:320Paolo Guiotto: And so we have exactly proved the deal, no? So we took a sequence in the set L that converges to some point L, and we proved that the point L is in the set
31:07:920Paolo Guiotto: S. So, this means that that property is verified.
31:11:800Paolo Guiotto: So, the conclusion is S is closed.
31:17:960Paolo Guiotto: Now, let's see, for example, the same proof works for the equality.
31:24:100Paolo Guiotto: So, similarly.
31:31:140Paolo Guiotto: The set S, which is defined by points X, where g of x is equal to 0, is closed.
31:41:50Paolo Guiotto: And you do the same argument.
31:43:910Paolo Guiotto: Same… argument.
31:50:760Paolo Guiotto: Now, let's see that the set where we have a strict inequality is open.
31:56:30Paolo Guiotto: Now, Let's… Let S be the set of points.
32:03:920Paolo Guiotto: in RD, where… Our G is strictly positive.
32:09:880Paolo Guiotto: And the claim here is, we claim…
32:15:750Paolo Guiotto: that this S is now open.
32:21:860Paolo Guiotto: Well, we could try to prove this by using the definition, or we could be smart.
32:28:220Paolo Guiotto: Now, you remind that the definition of closed set says that reminder.
32:37:940Paolo Guiotto: A set… let's give a different letter. A set, A… is closed.
32:47:910Paolo Guiotto: by definition.
32:52:270Paolo Guiotto: We say that the set is closed if and all if its complementary is open, right? A complementary is open.
32:59:670Paolo Guiotto: But you can use this definition also in the opposite direction. This says that A complementary is open if and only if A is closed, right? So if now change letters, call it this set S,
33:14:570Paolo Guiotto: So, S is open, if and only if, who is the closed set?
33:21:550Paolo Guiotto: the complementary of S, because A is the complementary of A complementary.
33:26:630Paolo Guiotto: You see the point?
33:29:890Paolo Guiotto: Set A is the complementary of A complementary. So I can say.
33:37:680Paolo Guiotto: By definition, set A is closed, even though if A complementary is closed. The same definition says also that a set is open, even though if the complementary of that set is closed.
33:51:420Paolo Guiotto: So, I could say that if you do the complementary here, so, if you do the complementary of S, so you are doing the complementary of the set of points X in RD, such that G of X is strictly positive.
34:07:650Paolo Guiotto: What is the complementary of this set?
34:17:00Paolo Guiotto: Sorry, I don't hear you.
34:20:40Paolo Guiotto: who is less or equal? G of X. So, it is the set of points, right, where G of X is less or equal than 0. Well, it is not greater or equal. If you want greater or equal change sign, this is the set of points X in RD, where minus G of X
34:39:730Paolo Guiotto: is greater than or equal than zero.
34:43:370Paolo Guiotto: And since G is continuous, of course, minus G is continuous as well. This is continuous, because…
34:52:130Paolo Guiotto: G is supposed to be continuous.
34:54:780Paolo Guiotto: And therefore, that set is now a set defined by a large inequality involving a continuous constraint, and this is closed by the first part of the proof.
35:07:50Paolo Guiotto: So we know that this is closed, and therefore S is open.
35:16:360Paolo Guiotto: And it's, NC.
35:19:230Paolo Guiotto: Now, here we have… so we have a simple, let's say, operational test to say if a set is,
35:28:550Paolo Guiotto: open. So, for example, no? Example…
35:33:880Paolo Guiotto: You have the set S made of points X, Y, Z in R3,
35:43:850Paolo Guiotto: Where we have X squared plus Y squared plus Z squared less or equal than 1. By the way, this is a…
35:51:730Paolo Guiotto: a ball centered at 000 radius 1, so a sphere, a unitary sphere. And then we have this… when we put the comma, we mean an intersection. Both conditions must be verified. So the second condition is X minus 1 half square
36:09:820Paolo Guiotto: plus Y squared less or equal 1 fourth.
36:15:270Paolo Guiotto: So, the question is, is S open Closed.
36:26:880Paolo Guiotto: Now…
36:30:400Paolo Guiotto: what can be said? We don't need to draw a figure. We could, it's a little bit complicated. If you want, we will do in a moment later. But let's try to see if we can respond to these questions.
36:46:690Paolo Guiotto: So…
36:47:990Paolo Guiotto: I would say that I noticed that there are large inequalities, right? So, it doesn't matter if you have less or equal than 1, because you can carry the 1 on this side, you have less or equal than 0. You understand that that's not the point, okay?
37:01:800Paolo Guiotto: So, if you want, and if you want in the standard form, I could write S equal 1 minus X squared plus Y squared plus Z squared greater or equal than 0. You see, this is the first constraint, G1, XYZ,
37:18:370Paolo Guiotto: greater or equal than zero. Then I have a second constraint, 1 fourth minus X minus 1 half square minus 1Y square, greater or equal than zero.
37:32:480Paolo Guiotto: Look here, you don't see Z, but since this is a condition on points, XYZ, if one of the coordinates, or more coordinates, are not
37:42:500Paolo Guiotto: In the function, it doesn't mean that this is a condition only in a coordinate. It's still a condition on the three points. One of the variables, in this case Z, is free.
37:53:260Paolo Guiotto: Can take whatever the value.
37:55:180Paolo Guiotto: So, and in fact, we will understand what is this kind of circulator that is confirmed.
38:00:730Paolo Guiotto: Sure.
38:03:80Paolo Guiotto: This is a set that can be written in this standard form. Of course, you see that the two functions are two polynomials, and therefore they are both continuous.
38:15:190Paolo Guiotto: And, the conclusion is that S is Closed.
38:21:820Paolo Guiotto: Then, can be also open.
38:25:390Paolo Guiotto: Well, here there is an answer, let's say, there is something that I told you, and it's not easy to be proved, but remind…
38:41:10Paolo Guiotto: That's it.
38:42:900Paolo Guiotto: It is possible to be both open and closed, but only in two cases. So either the set is empty, or the set is the full space. There is no other set that verifies this condition. That, E is… well, let's say S,
39:02:10Paolo Guiotto: is both open… Closed, huh?
39:10:390Paolo Guiotto: If, and only if… S is equal to empty, or S is the full space RD.
39:19:270Paolo Guiotto: This is a remarkable feature that We do not prove it.
39:24:260Paolo Guiotto: But it's useful. So, this set can be open if we prove that it is empty, or if we prove that it is the full space.
39:35:520Paolo Guiotto: So, is that set empty?
39:39:650Paolo Guiotto: Well, do not look at this, let's look at this one. Do you see any point that is in this set? You just need to find one point. One, not a lot of points. One point.
39:52:830Paolo Guiotto: How often is you…
39:57:510Paolo Guiotto: Okay, we are going to go complicated. Okay, I would say 000, because you put zeros and try to see if it makes easier. However, okay, for example, S is non-empty because…
40:15:510Paolo Guiotto: For example… .000… Verifies both, no?
40:23:580Paolo Guiotto: The first one becomes 0, less or equal 1. The second is minus 1 half square is 1 fourth, plus zero, less or equal, then, 1 fourth is true.
40:33:340Paolo Guiotto: But also, as you suggested, one half,
40:37:980Paolo Guiotto: then 0, 0. Also, this one, because if you plug this into the first, becomes 1 fourth less or equal than 1. In the second, it is 0 less or equal than 1 fourth, so it's correct, no? So, there are lots of points. Now, is S equal to RD?
40:58:130Paolo Guiotto: R, R3 in this case.
41:04:270Paolo Guiotto: Or is it different?
41:06:890Paolo Guiotto: Is it different? It means you just need to find a pointer which is sadly not in that set.
41:17:270Paolo Guiotto: You just choose X, yeah. One… 111.
41:22:770Paolo Guiotto: Because… Or… Example.
41:29:630Paolo Guiotto: One, one, one… well, I forgot to say that this belongs. This is not an answer.
41:37:10Paolo Guiotto: 111, the first condition is not verified, so we are sure that it is not in it.
41:42:660Paolo Guiotto: In essence.
41:43:910Paolo Guiotto: Okay, so this means that S cannot be empty, and neither the full space, so S cannot be…
41:54:980Paolo Guiotto: be opened.
41:57:270Paolo Guiotto: And that closes the question. So, what was asked is to say this. In this case, just to show you that it can be complicated, even if it is relatively easy.
42:13:800Paolo Guiotto: Because this is what? In this case, we have S is made of 2,
42:19:380Paolo Guiotto: conditions, X squared plus Y squared plus Z squared, less or equal 1, and then we have another X minus 1 half squared plus Y squared, less or equal 1…
42:34:520Paolo Guiotto: Now, the first one is a sphere, and even to draw a sphere in the space is not an easy task. So, it means that it's a unitary sphere, so if this is a point… this point here is 0.100,
42:53:880Paolo Guiotto: Here we have a 010, this is 001, so we have an object like that.
43:12:660Paolo Guiotto: Of course, it's the full sphere, no? Solid.
43:16:780Paolo Guiotto: It's… There's plenty of stuff inside.
43:21:390Paolo Guiotto: The second object is what?
43:24:940Paolo Guiotto: Well, if you look at the plane XY, this seems a condition in XY, so people would normally do this error here, because if you look at the plane XY for a strat on the…
43:36:830Paolo Guiotto: XY, and you forget of everything. You look at the second condition, this is a disc centered at 1 half 0, radius 1 half. So, the center is here, in plain XY, yeah?
43:52:00Paolo Guiotto: 1 half 0, this point, radius 1 half. So it is, these things.
44:01:190Paolo Guiotto: It is this, this.
44:03:160Paolo Guiotto: But this is if we are in two coordinates. Here, we are in three coordinates.
44:08:810Paolo Guiotto: So the set is actually in the play… in the space XYZ. So what do I see in the space?
44:14:600Paolo Guiotto: XY… in the space XYZ. So I still have this condition that says, in any case, X and Y are in that disk. What about Z? Z is arbitrary, so this means that this disk can be shifted up and down, and I am still in the Z. So this is… this kind of object is a cylinder.
44:34:570Paolo Guiotto: Okay? So, I should see something like, well, let's give this color.
44:46:790Paolo Guiotto: So this is the disk, now I shift up and down, and I get my cylinder.
44:54:850Paolo Guiotto: So, with the axis parallel to the Z axis, I'm doing a mess here, of course.
45:04:360Paolo Guiotto: Yes, also because it should be in the Z-axis. Okay, it's really awful.
45:13:110Paolo Guiotto: I don't know, it's correct.
45:14:830Paolo Guiotto: This is a big shift.
45:16:590Paolo Guiotto: So let's say that we have this thing.
45:20:770Paolo Guiotto: Now, we do the intersection.
45:23:70Paolo Guiotto: Yeah, you need to have some bit of fantasy, you know, because we're taking this ball, and you intersect with this solid cylinder, it's plenty, it's not just a pipe. So, we get something like this.
45:43:750Paolo Guiotto: I'm not, so particularly… Good in doing these kind of figures.
45:53:820Paolo Guiotto: This would be in the back.
45:58:280Paolo Guiotto: So it should be a solid, with,
46:03:40Paolo Guiotto: cylinder… it's a piece of sphere cut by this cylinder. So you understand that a figure is… is completely…
46:13:340Paolo Guiotto: beyond any possibility, so… Okay.
46:20:240Paolo Guiotto: Do you want to take a break?
46:24:350Paolo Guiotto: Okay, so we do… how much? 5 minutes?
46:28:920Paolo Guiotto: 5 minutes is okay for everyone.
46:31:750Paolo Guiotto: Okay.
46:35:150Paolo Guiotto: Is that…
46:39:940Paolo Guiotto: Okay, so… Now, before we, we launch in some of these exercises, I want also to introduce another
46:54:680Paolo Guiotto: important property of a satum.
47:02:880Paolo Guiotto: That this, this will, so… Let's introduce this definition.
47:11:670Paolo Guiotto: Effects… S.
47:15:360Paolo Guiotto: of RD.
47:17:500Paolo Guiotto: is bounded.
47:25:790Paolo Guiotto: Bounded, well, there are several equivalent definitions. Here, we give this one. If there exists a bound, M, Positive.
47:40:220Paolo Guiotto: Such that norm of X is…
47:44:130Paolo Guiotto: bounded above by this constant M for every point X of the set.
47:49:630Paolo Guiotto: S.
47:51:160Paolo Guiotto: So, points of the set cannot be at big distance to the origin, because the distance to the origin is the norm of X, which is bounded by this number M, no?
48:04:530Paolo Guiotto: A set which is both closed and bounded.
48:17:360Paolo Guiotto: Closed.
48:20:770Paolo Guiotto: Bounded.
48:26:240Paolo Guiotto: is… all the… Compact.
48:36:120Paolo Guiotto: Set.
48:37:900Paolo Guiotto: These sets are important because they are the base on, on which the…
48:44:670Paolo Guiotto: is grounded in one of the most important theorem of continuous function, which is the bias such theorem. So let's talk about this.
48:54:960Paolo Guiotto: So, compact.
48:59:750Paolo Guiotto: sets.
49:02:260Paolo Guiotto: Important.
49:10:00Paolo Guiotto: Eucos.
49:13:740Paolo Guiotto: off.
49:15:460Paolo Guiotto: Hmm.
49:16:430Paolo Guiotto: Extension.
49:23:340Paolo Guiotto: via stress.
49:28:530Paolo Guiotto: to your end.
49:30:290Paolo Guiotto: Now, so let me remind you what is Devices theorem. In one variable, So, a reminder… That, huh?
49:45:130Paolo Guiotto: we have this TRM.
49:49:790Paolo Guiotto: If we have a function F, which is continuous on a closed and bounded interval, AB, of the real line, then we are sure that this function has both minimum and maximum value on the interval AB.
50:07:550Paolo Guiotto: Then… F.
50:11:730Paolo Guiotto: us.
50:13:200Paolo Guiotto: both.
50:14:950Paolo Guiotto: I don't know what time you have used them, It's a, global, Or… absolute.
50:30:750Paolo Guiotto: Minimum and maximum.
50:34:470Paolo Guiotto: points.
50:36:460Paolo Guiotto: That is… There exists two points, at least two, maybe there are more than two.
50:44:450Paolo Guiotto: that we will call Xmin.
50:47:240Paolo Guiotto: and the X Max, in the interval AB,
50:53:290Paolo Guiotto: Such that, well, the function takes its minimum and maximum value, such that any other value, f of x, is between these two, fxmin
51:06:980Paolo Guiotto: and FX max.
51:12:80Paolo Guiotto: for every X in the interval AB.
51:17:00Paolo Guiotto: Well, this theorem is important, but actually, in dimension 1, you don't use so much, because to determine minimum and maximum points, you use differential calculus. This result is,
51:31:910Paolo Guiotto: Nice, but it's too weak, because it says there exists minimum and maximum, but it does not tell you any indication on how to determine these two points, okay?
51:44:300Paolo Guiotto: But if you add some more assumption, like the function is differentiable, then you can use the derivative, the signing of the derivative, you see where the function increases, decreases, and you figure out what are the minimum maximum values, okay? You also potentially have an equation, f prime equals 0, that
52:01:470Paolo Guiotto: Must be verified by minimum-maximum points, at least if they are in the interior.
52:07:120Paolo Guiotto: of the domain, etc. So, you use, in fact, differential calculus. But this theorem is important because from this theorem, for example, depends on the general properties of differential calculus on which that algorithm is based.
52:24:890Paolo Guiotto: Here, as you will see, this theorem is going to have a much more important role, so we have to… we will learn to use… not in this version, because this is the one variable case, so F here is a function of one variable.
52:43:260Paolo Guiotto: But what we want to do is to see the same kind of results. That is, if F is continuous.
52:50:920Paolo Guiotto: on some domain of RD,
52:55:540Paolo Guiotto: You see, for R1, intervals are natural domain, but for RD, there is not the prototype of a natural domain.
53:03:740Paolo Guiotto: And actually, you will see that the domain we have… the requirement we have to do on the domain is that the domain is closed and bounded.
53:12:280Paolo Guiotto: So it's that kind of set, compact set. So we will have a theorem that works like this. If F is continuous on a compact set, then F has both global or absolute minimum, maximum on that domain, okay? That's what we are going to see now.
53:29:500Paolo Guiotto: The goal is…
53:36:450Paolo Guiotto: to… Extend… these… Theorem.
53:44:990Paolo Guiotto: to decays.
53:49:490Paolo Guiotto: of a function f. Now, here we have to be a little bit more careful, because we are dealing with
53:54:930Paolo Guiotto: functions.
53:56:600Paolo Guiotto: of vector variable, vector valued. Now, this will apply to the case of functions f that are functions of vector variable, but numerical valued, because we have to say minimum value, maximum value. So, the values of F must be numbers, not vectors. Otherwise, we cannot say a vector is less than another vector, there is no ordering between vectors.
54:20:150Paolo Guiotto: So, of this type, where the function f is defined on some domain of RDL, and it is real valued.
54:30:430Paolo Guiotto: Now, it turns out that the extension of the previous theorem exists, and it works as follows. So, let F…
54:42:920Paolo Guiotto: B, continuous on the domain D,
54:47:30Paolo Guiotto: and assume that the domain D, you see that here I'm not saying anything on the domain, but in fact, I'm implicitly saying that the domain is of a specific type, a closed and bounded interval.
54:58:820Paolo Guiotto: If you remove that condition, the theorem becomes false. If the interval is open, not closed, it's not true that you have minimum-maximum. It might not be true, okay? If the interval is unbounded, it's not true. So, in fact.
55:14:480Paolo Guiotto: In that assumption written here.
55:18:600Paolo Guiotto: It's like if I'm doing two requirements. One is on the function, the function must be continuous, and the other one is on the domain. The domain must be a closed and bounded interval.
55:31:00Paolo Guiotto: So now here, you have a similar thing. I have a function which is continuous on the domain, and what is the key properties of the domain? And that's the property I said above. So, compact…
55:46:260Paolo Guiotto: That fees… closed.
55:53:190Paolo Guiotto: that… bounded.
55:59:370Paolo Guiotto: So, this is the, the, the, let's say, the feature that the domain must…
56:05:710Paolo Guiotto: So the function must be continuous on the domain compact. Then.
56:12:320Paolo Guiotto: Now, the conclusion is more or less similar with the domain that will be replaced by D. Then, F has
56:22:510Paolo Guiotto: both.
56:25:950Paolo Guiotto: absolute or global, I usually will use the…
56:34:30Paolo Guiotto: This one, but you can use absolute if you like.
56:37:670Paolo Guiotto: minimum, Maximum.
56:42:570Paolo Guiotto: points.
56:44:810Paolo Guiotto: That is.
56:48:270Paolo Guiotto: There exists two points that are now… they are in the domain, so they are vectors.
56:53:900Paolo Guiotto: X, mean… And that's Max, huh?
56:59:880Paolo Guiotto: in the domain D,
57:01:870Paolo Guiotto: such that the values of F, which are numerical, because our F is real-valued, so these are numbers, okay? So, it makes sense to say they are less or equal, greater or equal. They are between the minimum value
57:18:670Paolo Guiotto: which is the value of F at point X spin,
57:22:450Paolo Guiotto: and the maximum value, which is the value of F at point X mass.
57:29:770Paolo Guiotto: So, be careful, because these are called minimum Or… maximum… point.
57:42:490Paolo Guiotto: Okay?
57:43:890Paolo Guiotto: So the point where the function takes the minimum or the maximum value.
57:48:790Paolo Guiotto: This one, which are numbers, These are the minimum and maximum values, so mean, Max… value.
58:02:910Paolo Guiotto: So, the smallest possible value of F, the biggest… the biggest possible value of F.
58:09:230Paolo Guiotto: So, we also denote this, we also write, we also… Right.
58:22:470Paolo Guiotto: the value of F at point Xmin Will be called the minimum.
58:29:740Paolo Guiotto: of F on D.
58:32:380Paolo Guiotto: So if you see, this notation means the minimum value, not the minimum point, okay? The minimum value. Or, equivalently, this is… sometimes you see also this, the minimum of F of X
58:47:520Paolo Guiotto: when X belongs to D, these are equivalent notations, okay?
58:53:830Paolo Guiotto: Or if you want, you can also write the minimum of the set of values f of x.
59:01:870Paolo Guiotto: when X belongs to D. These are notations or equivalent notation, but the most frequently used is this one, because it, in a few symbols, it tells you what you are talking about. You are talking about the minimum value of the function on the domain D.
59:21:790Paolo Guiotto: And, of course, you have a similar notation for the maximum. The value of F at X maximum.
59:29:430Paolo Guiotto: will be what we call the maximum value of F only.
59:35:60Paolo Guiotto: Okay.
59:37:00Paolo Guiotto: So these are just notations.
59:40:250Paolo Guiotto: Now, this theorem we will not prove. The proof, is, is, is, really…
59:48:540Paolo Guiotto: A pure mathematical proof in the sense that it's not a constructive proof.
59:55:470Paolo Guiotto: This means that the theorem does not tell you a method to find the minimum and the maximum. It's a pure existence theorem. It says it works on a contradiction argument, so…
00:07:420Paolo Guiotto: it does not give any idea on how to determine the minimum-maximum points, okay? So, it has not immediately practical applications, because it says there is the minimum, there is the maximum, yes, but if I want to find what I have to do, I don't know.
00:24:100Paolo Guiotto: Okay, that will be done by differential calculus, which is the next topic. Exactly as it happens for one variable calculus. So you never use this theorem to determine minimum-maximum, you rather use differential calculus.
00:43:660Paolo Guiotto: results. Here, it's more or less the same, but since differential calculus tools are much weaker for function of several variables, so are not so strong as the tools you have in calculus in one variable, we will need to use a little bit more this result. We will see how to do.
01:03:530Paolo Guiotto: Back.
01:04:520Paolo Guiotto: The point is, however, that this result emphasizes the role of compact set. So, that's why it is important to know if a set is closed and bounded, because if you have to solve the optimization problem of determining minimum-maximum of the function on that domain.
01:24:130Paolo Guiotto: And you want to know if they exist, so if it is worth to spend time in searching for these points, this theorem is very important. And this theorem asks you, you should first verify the continuity, no? And that's something that we think it's quite easy.
01:40:940Paolo Guiotto: And we should check the nature of the domain, so if it is closed and bounded. Okay, so that's why it is important to have these qualitative properties. Now, so, I want to show you some exercise, which is at the end, and I suggest you to do.
01:59:650Paolo Guiotto: the relative exercise, it is… oh, by the way.
02:05:960Paolo Guiotto: I leave some exercise to do. We will meet on Tuesday.
02:11:910Paolo Guiotto: So, do… Exercises, let's say…
02:22:400Paolo Guiotto: one… I don't mind if I have already given this to you. 1814.
02:30:310Paolo Guiotto: Which is an abstract exercise on definitions of closed and open set, okay?
02:36:250Paolo Guiotto: So, this is a good exercise to practice. It's an abstract, it's not a concrete exercise. And then 1815, which is instead an exercise, now we see some of them.
02:49:880Paolo Guiotto: an exercise on determining the nature of a set. Is it closed, open, bounded, compact, etc. So…
03:00:330Paolo Guiotto: For, it says, for each… of the… Following… sets, huh?
03:13:780Paolo Guiotto: Say… if… E.
03:19:400Paolo Guiotto: Jeez.
03:20:890Paolo Guiotto: Open… closed the… Bounded.
03:28:650Paolo Guiotto: compact.
03:32:630Paolo Guiotto: Now, compact means that you have to verify close and bounded, so if you have together, you can say it is compact.
03:39:220Paolo Guiotto: I want to do this because there are standard things that are easy, of course, because it's a routine, and things which are less standard. In particular, checking this.
03:51:500Paolo Guiotto: If the set is bounded, it's not… it's almost never a standard task. So let's start with one of these, let's say, for example, number one.
04:04:100Paolo Guiotto: This is a domain… this is important, I repeat, because when we attack optimization problem, we must be able to understand what is the nature of the domain to use the appropriate strategy to solve the maximization-minimization problem, okay?
04:21:610Paolo Guiotto: So this is, let's say, a first task of a big problem we will discuss.
04:26:450Paolo Guiotto: in the next weeks. So this is the set of points, XY in R2, where there is one condition, X times Y, strictly positive.
04:39:280Paolo Guiotto: Now, in this case, if possible.
04:42:540Paolo Guiotto: It's not forbidden, it's not asked here. You can have an idea of what is the set by doing a feature. You can do this. So, if we can do, let's try to do, because maybe it will help with the intuition.
05:00:360Paolo Guiotto: So, what does it mean to stay in this setting? It means that… so the point XY belongs to D, it is written if and only if X times Y is positive.
05:13:180Paolo Guiotto: Now, this is an inequality, it's a one single inequality in two variables. How can I solve that?
05:19:680Paolo Guiotto: Well, a product is positive if and if both factors are positive, or both are negative. So this yields this alternative. Either X positive and Y positive, together.
05:37:600Paolo Guiotto: Or…
05:39:230Paolo Guiotto: X negative, Y negative. I cannot think anything simpler than this, because this gives a direct representation of the set.
05:50:150Paolo Guiotto: In the plane XY, I have either
05:54:130Paolo Guiotto: the first two conditions means the two coordinates are positive. This is the first quarter, here.
06:01:970Paolo Guiotto: I do not take the axis, because on the axis, one of the two, or both, are zero.
06:08:350Paolo Guiotto: And the other one, so this is this, and the other one says both negative. So in this case, this is
06:18:950Paolo Guiotto: We don't know why it is no more working.
06:27:840Paolo Guiotto: Red, red, red, red.
06:31:370Paolo Guiotto: Now we're gonna fall.
06:35:130Paolo Guiotto: Where capitalistic Bosnia.
06:37:370Paolo Guiotto: Okay, X negative, Y negative, so it means that we are here.
06:42:540Paolo Guiotto: So this is down here. Now, since I have either the first or the second, the set is made by the two. It's a union, okay? So, the main D is, all this, what I colored in red.
07:00:420Paolo Guiotto: Okay, so this was just to have an idea on how the set is made. Now, let's discuss this. What can be said about, is the set open?
07:10:320Paolo Guiotto: So what do I see? The set is defined by a strict inequality.
07:14:910Paolo Guiotto: Okay? And what is X times Y greater than 0? So you see the constraint is this thing.
07:21:510Paolo Guiotto: G of XY, there is one unique constraint greater than zero. It's a strict inequality, the function G is a polynomial, it is continuous. So, domain D is a set of type G greater than zero, where…
07:38:650Paolo Guiotto: GXY is X times Y. It's clearly a continuous function on R2, and therefore, the domain is definitely, open.
07:53:110Paolo Guiotto: This is the standard fact, no? The only thing is, you have to recognize the inequality, the equality, okay, the function, but it's a standard, you don't have to do anything here.
08:04:520Paolo Guiotto: And what about closed? We know that a set can be both open and closed, either if it is empty or if it is the full space, which is not the case here, because the set is not empty, as you can see, and it is not the full plane, because there are
08:20:370Paolo Guiotto: two half, two quarters which are not there. So, since,
08:26:330Paolo Guiotto: D is different from empty, and that 2 We cannot… B… also… Cannot be closed.
08:49:880Paolo Guiotto: So, e… You relieve.
08:55:319Paolo Guiotto: sets.
08:57:500Paolo Guiotto: Oh.
08:58:970Paolo Guiotto: D.
09:00:450Paolo Guiotto: both.
09:02:490Paolo Guiotto: open.
09:04:819Paolo Guiotto: Closed.
09:06:500Paolo Guiotto: R.
09:08:430Paolo Guiotto: empty.
09:10:790Paolo Guiotto: And…
09:14:800Paolo Guiotto: Okay, so it is not closed. Now.
09:18:439Paolo Guiotto: I can automatically now say that this set is not… is not.
09:32:69Paolo Guiotto: Compact, because to be compact, you must be closed.
09:35:720Paolo Guiotto: Okay?
09:37:130Paolo Guiotto: So, in particular, this D… is… not… Compact.
09:46:540Paolo Guiotto: because… It is… not.
09:52:600Paolo Guiotto: Closed. Remind, compact means both.
09:57:570Paolo Guiotto: Closed and bounded. If one of the two is false.
10:02:590Paolo Guiotto: It means that the set is not compact.
10:05:250Paolo Guiotto: Now, let's come to debounded.
10:09:230Paolo Guiotto: I told you that this is never a standard.
10:13:220Paolo Guiotto: There are certain insufficient conditions that tell something, but…
10:17:620Paolo Guiotto: That is not the test if and all if, okay?
10:20:900Paolo Guiotto: So, of course, if you look at this figure, what would you say?
10:25:550Paolo Guiotto: Is the set bounded or not?
10:28:660Paolo Guiotto: No, because bounded means vectors have a bound On… what is it?
10:37:970Paolo Guiotto: On the norm, okay?
10:40:270Paolo Guiotto: So if you want, geometrically, this means so that, remark. Let's do this remark.
10:51:260Paolo Guiotto: geometrically.
10:58:630Paolo Guiotto: I… let's say that this is useful when we see the set, because if we don't see the set, so if we don't have a graphical representation of the set.
11:09:960Paolo Guiotto: This is,
11:11:880Paolo Guiotto: basically useless, but it sometimes can be useful when we see the set, because we can draw it. So, geometically, a set S is bounded
11:25:30Paolo Guiotto: You see that if and only if the definition says there exists an M such that norm of X is less or equal than m for every X in S,
11:36:990Paolo Guiotto: But this means that every point of the set belongs to the ball centered in the origin and radius M.
11:46:350Paolo Guiotto: Right?
11:47:620Paolo Guiotto: Because… no, there is a… the arrow is on the wrong place here. Should be on X.
11:54:570Paolo Guiotto: Because, distance less frequent than M means a distance from X to 0.
12:02:690Paolo Guiotto: less or equal than M, so you can see that 0 as a center of a ball, where X must be because of this condition. So, in other words, it means that our set should be inside a bowl.
12:17:240Paolo Guiotto: with a certain radius. So there is a big ball that contains everything, okay? This means, you know? So, I can say that bounded can be equivalently said in this way, if there is a ball that contains S. And if you look at this set, you see that there is no ball, because whatever is the ball that you draw here.
12:41:910Paolo Guiotto: This ball, sooner or later, will finish, and there are lots of points outside, okay? So, let's say that,
12:52:910Paolo Guiotto: when you have to justify something, it would be always better to try to be, as much as possible, precise, okay? If you…
13:03:240Paolo Guiotto: if you… for example, in this case, you could say it is clear that it is unbounded, because it is evident, okay? So, I wouldn't penalize you if you…
13:16:520Paolo Guiotto: just, say something like this. Well, let's, but let's, let's give a more general strategy, no?
13:26:30Paolo Guiotto: Now, so, what to do in general? So, from… from…
13:38:640Paolo Guiotto: the… CEO.
13:43:170Paolo Guiotto: It is.
13:45:990Paolo Guiotto: Well, you know, in mathematics.
13:50:210Paolo Guiotto: a figure is never an evident fact, because a figure is a particular representation, okay, of something. It's never a general. But however, for you, it is evident that
14:03:460Paolo Guiotto: S is unbounded.
14:09:520Paolo Guiotto: So, if it is not bounded, we say unbounded.
14:13:590Paolo Guiotto: However, what is the general strategy? So let's do this. So, let's… this would finish the exercise, but let's do a remark.
14:24:860Paolo Guiotto: So how… Do we… verify…
14:36:140Paolo Guiotto: if, set S… is, bounded.
14:43:260Paolo Guiotto: Or… unbounded.
14:48:20Paolo Guiotto: Now, in the first case, if you want to, it's a sort of alternative. It's like, how do we prove that the limit exists, and it is this one, or it does not exist? It's, there are two ways, no? If you want to prove that the limit does not exist, what you do is
15:05:60Paolo Guiotto: You find two spatial sections along which the function behaves differently.
15:09:960Paolo Guiotto: or you have to prove that the function, no matter how you go to that point, does something. And here is similar, no? So, to prove that the set S is bounded, what you have to do is to prove that there is a bound for the normal vectors.
15:27:820Paolo Guiotto: So, to do this, you have… you have, 2.
15:36:990Paolo Guiotto: show.
15:39:120Paolo Guiotto: that there exists a constant M, such that norm of X is less or equal than
15:46:220Paolo Guiotto: M for every X in S.
15:50:650Paolo Guiotto: For example, this could be achieved if we prove, and it is equivalent, in fact, if we prove that there is a bound for each of the coordinates.
16:00:770Paolo Guiotto: Okay? So, equivalently.
16:09:950Paolo Guiotto: if I prove that, let's, if,
16:15:120Paolo Guiotto: Let's say the vector X is made of components X1, XD,
16:20:390Paolo Guiotto: Equivalently, if I access this, I… Avv.
16:27:360Paolo Guiotto: You have, not only. You have.
16:31:170Paolo Guiotto: to show… that you can find maybe different constants, M1,
16:38:470Paolo Guiotto: MD, such that you get bounds for each of the coordinates. So the jth coordinate is bounded by a constant MJ, for J equal 1 to D, and this happens for every point of the setting.
16:54:480Paolo Guiotto: S.
16:55:410Paolo Guiotto: So, in any case, the strategy is to bound the coordinates or the norm. Perhaps sometimes it could be useful to write the set in different coordinates, like polar coordinates, because for polar coordinates, raw is the norm.
17:12:340Paolo Guiotto: So if for some reason you arrive to bound the row, you are done.
17:16:440Paolo Guiotto: Do you want to ask something?
17:23:470Paolo Guiotto: Sorry? The end that you find, MFJ, that you find specifically the center of all of the…
17:29:680Paolo Guiotto: Clear in this place.
17:31:340Paolo Guiotto: No, no, no, it can be different, but then it doesn't matter, you can take the bigger, the biggest, you get the M, okay?
17:39:620Paolo Guiotto: What to do if you want to prove that it is unbounded? Now, here it's a different strategy, because, of course, you won't be able to bound the norm or to bound the coordinates. So what is sufficient to prove that the set is unbounded? Well, literally.
17:57:770Paolo Guiotto: The set is not bounded if you cannot find an M that bounds all the normal vectors.
18:06:170Paolo Guiotto: So it means that whatever is the value of M, there will be always a vector X in the sector with na greater than M.
18:18:250Paolo Guiotto: And if you apply this to… S unbounded.
18:25:650Paolo Guiotto: unbounded.
18:27:780Paolo Guiotto: If there is no M, such that the norm of X
18:34:10Paolo Guiotto: is less or equal than M?
18:36:890Paolo Guiotto: for every S in. So this means that no matter how you choose your M,
18:44:280Paolo Guiotto: you can find a vector X in the set S, such that its norm will be greater than
18:54:830Paolo Guiotto: Now, take M, the naturals. So this means that for every N natural, there exists a vector X,
19:05:610Paolo Guiotto: That now, I think, is going to depend on N, that belongs to S with norm of Xn larger than
19:16:850Paolo Guiotto: Do you know what does it mean?
19:19:110Paolo Guiotto: what I wrote means that
19:26:750Paolo Guiotto: So for every n, I can find the point of the set, with norm greater than N.
19:35:660Paolo Guiotto: But what happens to that sequence XN?
19:40:300Paolo Guiotto: 5 years.
19:41:820Paolo Guiotto: Yeah, it goes to infinity. So the equivalent is this, there exists a sequence of points of S such that this sequence goes to the infinity of the space.
19:57:280Paolo Guiotto: So if you are able to prove that, you have… you have… you obtained that the test is unbounded. So let's see, for example, in… in our exercise here, okay? So, for example.
20:11:50Paolo Guiotto: Ian, previous… exercise, you take the sequence XN made of, say, NN with N natural.
20:23:870Paolo Guiotto: Okay, let me redo the figure down here. So, the set is made of this quarter and this one, right? Now I'm taking the sequence NN. So, 1, 1 is here, 2, 2 is here, 3, 3 is here, 4, 4 is here, 5, 5 is here. So, in general, this is the point NN.
20:41:630Paolo Guiotto: What happens when n goes to infinity? Of course, this goes to infinity of the plane.
20:47:740Paolo Guiotto: And they are all points of S. So I have a sequence of points of S that goes to infinity. So this means that the set cannot be bound.
20:57:540Paolo Guiotto: Okay? So… The unique general thing we can say is.
21:04:320Paolo Guiotto: If we want to prove that S is bounded, we have to try to bound the coordinates.
21:09:990Paolo Guiotto: Or to bound the norm of the vector.
21:13:490Paolo Guiotto: If we want to prove that the set is unbounded, it is sufficient to find the sequence of points that goes to infinity, okay? Now, the point remains
21:24:990Paolo Guiotto: But you have to decide which one, so you have a skill to understand what is the case. So let's do another example, maybe a little bit more complicated than the previous one.
21:35:320Paolo Guiotto: But not particularly difficult. So, let's say to go a little bit faster. Number 2, we have domain D, we are still in plane XY in R2,
21:45:260Paolo Guiotto: Such that it looks like the previous one. Modulus of X times Y now is less or equal 1.
21:54:330Paolo Guiotto: Now, let's reverse the discussion. We first,
21:59:810Paolo Guiotto: do all the possible considerations on this D without looking at the set. Then we draw, okay?
22:06:950Paolo Guiotto: Now, we see that our D is what? D is defined by a large inequality, something like modulus XY minus 1 less or equal than 0. So this is my constraint, GXY. Now.
22:24:330Paolo Guiotto: when you compose X times Y, which is a polynomial, with the absolute value, which is a continuous function, we say that composition remains continuous. So, clearly, this G is a continuous function.
22:36:820Paolo Guiotto: on R2, so we see there is a large inequality. The set B is closed.
22:46:430Paolo Guiotto: You don't need to justify it, because we know this is a general fact, no? Now, can be open, we have to decide if this set is the empty or RD. That's the only possibility. But we see that the set D is non-empty, because, for example.
23:08:410Paolo Guiotto: Which point belongs to this?
23:11:40Paolo Guiotto: Wonderful.
23:12:10Paolo Guiotto: 11, for example, okay, and D is different from the plane.
23:18:50Paolo Guiotto: For example, a point that is not in… which is, not in the
23:27:110Paolo Guiotto: 2-1.
23:30:610Paolo Guiotto: Right. Okay, so this says T is not open.
23:38:780Paolo Guiotto: Now, bounded.
23:46:150Paolo Guiotto: Now, I know that maybe all of you are thinking to some figure, but you don't have to think to figures, because when it is complicated, you don't do any figure.
23:55:520Paolo Guiotto: So look at that relation. The idea is that that relation can be verified with, for example, X small and Y large. Even for X0, you can take Y whatever. For example, I see that points of type 0Y, whatever is Y real.
24:15:760Paolo Guiotto: They are, indeed. Do you agree?
24:19:100Paolo Guiotto: You see, because the product is 0, so whatever is Y, 0 less or equal than 1.
24:24:270Paolo Guiotto: And what is this? It's the y-axis.
24:28:540Paolo Guiotto: Okay, so if you want, so you want the sequence to show 0N belongs to D for every n natural, and clearly, when you stand n to plus infinity.
24:41:550Paolo Guiotto: this guy goes to the infinity of the plane. So this says that D is unbounded.
24:51:00Paolo Guiotto: And therefore, D is not compact.
24:58:270Paolo Guiotto: We could have said that
25:00:110Paolo Guiotto: No, D was closed, so if it was bounded, it would have been compound. Now, what is D? So that's all for the question, okay? The exercise is over. Now, just to do a picture of the surf, so what does it mean that XY belongs to D?
25:18:650Paolo Guiotto: It means that modulus XY is less or equal 1. Now, be careful, because it is possible to do something like 10 errors and have a wrong idea of what is the set. So, how should we proceed?
25:37:350Paolo Guiotto: I would say, what does it mean that modulus is less than 1? It means that the number is…
25:44:400Paolo Guiotto: Between minus 1 and 1. Okay, now let's focus on this, because these are two inequalities.
25:50:350Paolo Guiotto: Okay? That must be verified together.
25:54:340Paolo Guiotto: Okay?
25:55:820Paolo Guiotto: the temptation would be divide by X.
26:01:630Paolo Guiotto: But since it is an inequality, dividing or multiplying is dangerous, because, first of all, you should check if it is zero or not, you cannot divide by zero. And if it is positive, it preserves the inequality. If it is negative, it changes. So I would say.
26:19:830Paolo Guiotto: This is equivalent to… we distinguish three cases. X equals 0, X positive, X negative.
26:27:600Paolo Guiotto: For X equals 0 is minus 1, less or equal 0, less or equal 1. What is this? Always true.
26:35:340Paolo Guiotto: But that always refers to what?
26:41:690Paolo Guiotto: Because axis 0,
26:44:580Paolo Guiotto: But there is another coordinate, which is not… is silent there, you don't see why. This means for every Y, so this means that .0Y, we already knew this, they belong to D for every Y real.
26:59:840Paolo Guiotto: And this is… this is the Y axis, by the way. So, if you want to start to put these informations in the plane XY, what I'm saying is that the Y-axis, this thing, is in the domain.
27:16:10Paolo Guiotto: Then, for X positive. Now, for X positive, I can divide by X, I do not change the inequality, and it comes Y greater or equal than minus 1 over X, less or equal than plus 1 over X.
27:31:230Paolo Guiotto: Okay, 4X positive.
27:34:490Paolo Guiotto: So, for X posit, it means that we look at the right half plane.
27:39:230Paolo Guiotto: Okay, you see, that's X positive, and that's X negative.
27:44:180Paolo Guiotto: So we have to forget X negative, because what we are going to say is for X positive. For X positive, Y must be between these two lines, no? Y equal 1 over X, Y equal 1 over X is there in the set, because of the equality, and Y equal minus 1 over X in this one.
28:05:30Paolo Guiotto: So this is 1 over X and minus 1 over X. Both are in the set because of the equality that you have there.
28:13:420Paolo Guiotto: And then you have the Y between the two, so this means all this region.
28:21:570Paolo Guiotto: Okay?
28:22:850Paolo Guiotto: Now, for X negative, you have a similar discussion, but be careful, because when you divide by X,
28:31:320Paolo Guiotto: the inequality
28:33:00Paolo Guiotto: is inverted, because now we will have X, if you want, it's less or equal than minus 1 over X, greater or equal than 1 over X. And that's correct, because for X negative, 1 over X is this one, it is that one below.
28:52:190Paolo Guiotto: That's 1 over X. And minus 1 over X is above, it's this hyperbola.
28:57:910Paolo Guiotto: minus 1 over X. And you have to take the Y's on the two hyperbolas, because there is the equal, plus those who are between the two. So, again, this part here.
29:11:520Paolo Guiotto: So, what is colored in red is now our set P.
29:16:690Paolo Guiotto: From this figure, for example, you understand that it's amounted, clearly, no? You see, there are both the axes, which is actually what we already knew from this, okay?
29:28:410Paolo Guiotto: Okay, so…
29:30:280Paolo Guiotto: Let's, stop here, but, please do the exercise 1-8-15, apart for the 1-8-14, for, next class, which is on Tuesday.
29:53:310Paolo Guiotto: Okay, let's stop the recording.