In this video we will study some of the properties of compact sets that also will help us to understand how they look like in any metric space. The main theorem we are going to state and prove is the following one. So we have again a metric space that is going to be fixed and K a subset of that metric space. Turns out that if this set is compact then the set has to be closed and has to be bounded.
The second property is if K is closed and X is compact. Then, K is also compact. So, let's go through the proof of these two statements. And the only thing we can use here, and that's the only thing we have, is the definition of compactness through open sets. So, let's try first of all to prove the first one.
Let's prove that K is closed by showing that the complementary of this set is open. To do so, let's take a point in the complementary. And what we want to show is that this point is interior, so we want to find radius such that the ball of radius r and center in P is still included in the complementary of our set. Okay, so the idea is we have our set here, K, and we have a point P there. and how we can ensure, I mean, it's clear by this picture that I made that that thing is going to be, let's say, far away from the set but the thing is, how can we ensure that?
well, the idea is the following one the first thing that we observe is that this set has to be the union of all its points. And since p is not in k, then the distance between p and q is let's say our P is positive for all Q okay yeah so let's say that this distance divided by 4, let's say a small number, is still true. So if this is positive, if I divide by 4, that quantity would be still positive.
In other words, so if I have a pointer Q, then the ball, so why am I dividing this by 4? the thing is, if that is the distance divided by 4 there it is and I take a ball with that distance divided by 4 these two balls would be completely separated from each other and that is what we have The ball centered at P and radius rP is disjoint from the ball with the same radius but centered at Q And now, since Q is the center of this ball, this also trivially holds Agree? So I have K is covered by a union of open balls.
And here is where we can use the compactness assumption, because by compactness, there exists R, sorry, That should be Q all the time, sorry for the mess. Q1 are Q in a finite family such that K is still included in this finite number. of balls.
And, ok, this is literally an open set as is a union of balls. Remember these numbers are positive. and they also satisfy this property that each of these balls would be disjoint to a particular ball but then so for each so i'm going to have first of all a finite number of these balls Cover my whole set.
And for these centers, each of them, they are going to have a different radius, r, q associated. Since they are finite, let r just be the minimum for j equal to 1 till n of this radius. And so it satisfies since B of R, P is going to be a subset of B RQJ, P, it turns out that what is going to happen is that this ball intersection this open set is also empty. But if this ball intersection W is empty, then BRP intersection K has to be empty.
which is the same as saying that this ball is included in the complementary of K. So what we have shown is that for any point in the complementary of K, there exists a positive number, and this is positive because this is a finite number of points for which the ball, center at that point and that radius, is included still in the complementary of the set. In other words, that the complementary... is open and if it is open that's equivalent to say that K is closed.
To finish the proof of the first statement we need to show also that our set is bounded. Well, we can use a similar idea to this thing that we have done Let's say that how to prove that our set is K is bounded Well, the idea is going to be just cover our set Let's say cover our set by poles of a fixed radius. And by compactness, it's going to be a final number of elements in the compact set such that I've said this can be covered by a finite number of these balls of radius one.
Okay, so So how we can, so essentially the same picture as before we have our set and we just cover by a final number of balls of the same radius. How can we say that it's bounded? For all x in K, there exists j between 1 and n, for which the distance between x and q, j, is smaller than 1. So that's what we have here.
In particular... The distance between x and q1 is going to be smaller or equal than the distance between x and qj plus the distance between qj and q1. But this first distance is always more than one, and that's going to be smaller than the maximum.
between all these distances. And let's define this as m. In other words, for all x in k, There exists an x, there exists a constant m, such that distance is smaller than m.
That is, k is included in the bool of center q and radius, let's say, 2m. In other words, k is bounded. Okay, now we need to show that any closed subset of a compact set is also compact.
And again, the only thing we have here to prove this is the definition of compactness. So, let's continue with the proof, but this time of two. We know that X is compact, we know that K is closed. Well, if it's closed and compact, if it's closed, then its complementary is open.
Agree? Now let's take the alpha. an arbitrary open cover.
And, okay. And since this is an open cover, and its complementary is open, so it turns out that x It's going to be covered by the union of all those B-alpha, the union, the complementary of K. Why?
Because K and its complementary is the total. But since E is compact and these are open sets, X is compact. X can be covered by a finite number of these things, and probably, maybe, by its complementary of k.
But if the whole space can be covered by this finite union of sets, then K, that is K intersection X, can then be covered by this finite family. That's it, because we have shown that for an arbitrary open cover of the set, there exists a finite subcover of the set K.