Metric Spaces - Lectures 21, 22 & 23 Oxford Mathematics 2nd Year Student Lecture

This is Lecture 21 of A2, Metric Spaces, and this is the last lecture corresponding to the chapter of the course on sequential compactness. That's Chapter 8 in the notes. And in this lecture, we're going to talk about spaces of functions and when they may or may not be sequentially compact as metric spaces. So we're only going to look at quite nice examples. We'll only be looking at functions on a space X, which is itself a sequentially compact metric space. And actually, you lose little in this lecture by imagining that X is actually just the closed interval zero one. One nice thing about sequentially compact metric spaces is that the space C of X of continuous real valued functions coincides with the space C sub B of X of bounded continuous real valued functions, because as we've shown, any continuous function on a sequentially compact metric space is bounded. And what that means is that we have the sub norm or infinity norm available to us on C of X. Furthermore, C of X is a complete metric space with this norm, as we've already shown. Now, the first thing to observe is that the space C of X is never actually a sequentially compact metric space by itself. And the reason for that is actually kind of obvious. You can just consider the following sequence f n of functions in which f n is the constant function n. That's always continuous, of course, and it quite obviously has no conversion subsequence. So the problem is that C of X is not sufficiently bounded. There are constant functions which are arbitrarily large and that prevents it being sequentially compact. But there's another rather more interesting reason why C of X is never sequentially compact. In fact, the unit born in C of X is not sequentially compact because of the following example. Let's specialise to the case of the closed interval zero one and let's look at the sequence f n of functions, continuous functions on that interval defined in the following way. So f n will be supported on the interval from one over n plus one up to one over n. So that means that it's zero outside of that interval. On that interval, well, it's going to just be a spike. It'll take the value one at the midpoint of that interval and I'll make it piecewise linear from the edges of the interval to the midpoint. So it takes the value zero at one over n plus one, the value zero at one over n, and then you draw a line from the point with coordinates one over n plus one comma zero to the point T n comma one and you draw a line from T n comma one to one over n comma zero. So that's a sort of spike or tent function supported on the interval from one over n plus one to one over n. It's continuous. Now, I claim that the distance between two different functions f m and f n in that sequence in the subnorm is actually one. And the reason for that is that whilst f m takes the value one at the midpoint T m of this interval from one over n plus one to one over n, at that same point, f n is zero because T m, it lies in the interval from one over n plus one to one over n. And therefore, it's outside the interval from one over n plus one to one over n. So the subnorm of f m minus f n is actually one. And therefore, the sequence of f n's doesn't have a convergent subsequence. So we've got two examples here showing that the space C of X of bounded continuous functions, even on the closed interval zero one, is not sequentially compact. Now, what we want is a positive theorem stating that under certain conditions, certain collections of functions are sequentially compact. And to get such a positive result, we're going to have to respond to these two examples. So we'll make a couple of definitions that are specifically designed to exclude the two examples that I've just described. And that's because, as I said, we want to produce interesting subspaces of the space C of X, which are sequentially compact. So here are the two definitions that we'll use. First of all, the definition of uniform boundedness. So let X be a sequentially compact metric space and let curly f be a collection of continuous functions on X. Then we say that curly f is uniformly bounded if it's bounded as a subset of C of X in the subnorm. So if it's contained in some fixed ball in that space. Or to spell it out, if there is a number m, such that the absolute value of f of X is less than or equal to m, uniformly for all little x and for all functions little f in the family curly f. So that's uniform boundedness. Second definition is designed to exclude examples like that sequence of spike functions. So somehow, although each of those spike functions was continuous, as m gets larger, they somehow become less and less continuous, if you like, because the gradient of the spikes gets greater and greater. And this definition is designed to exclude that. So let X be a sequentially compact metric space and let's take a family, curly f, of continuous functions on X. Now, remember that in a sequentially compact metric space, any continuous function is automatically uniformly continuous, which means that in the epsilon delta definition of continuity, you can choose delta to depend only on epsilon and not on the points of the underlying space. Equicontinuity means that more than that, you can choose delta to be uniform in the choice of function f as well, as f ranges over this family curly f. So if you like, you can think of it as a kind of uniform uniform continuity or uniform continuity in a family curly f. So to spell it out, for every epsilon greater than zero, there is a positive delta such that whenever the distance from x to y is less than delta, the distance from f of x to f of y is less than epsilon. And that's got to be true for all points x and y at distance less than delta and for all functions little f and curly f. So that's the definition of what it means for a family f, curly f, to be equicontinuous. Now we come to the major theorem on this topic, which is an important theorem in analysis, called the Arzela-Ascoli theorem. And what this theorem states is that essentially the only way that a family of functions can fail to be sequentially compact is if it's either not uniformly bounded or not equicontinuous. Perhaps simpler to state it in the positive direction, if you're equicontinuous and uniformly bounded, then your closure is sequentially compact. So let's state it. Let X be a sequentially compact metric space. As I said, you lose little really by imagining that X is the closed interval 0, 1, but the proof will work for any sequentially compact. And let's take a family, curly f, of continuous real value functions on X and suppose that it's equicontinuous and uniformly bounded. Then any sequence of elements of this space, curly f, has a convergent subsequence. And in particular, if this family, curly f, is closed, well, then the limit of any such convergent subsequence lies in curly f and therefore it's sequentially compact. So to summarise, if you've got a family f that's equicontinuous, uniformly bounded and closed, then it's sequentially compact. Now, this theorem is actually an if and only if. I'm only going to deal with this direction, which is by far the more substantial direction. And the other thing I should say about it is that the proof of this theorem is non-examinable. So actually, this is now the end of the examinable portion of the lecture. But I just can't state theorems without wondering what their proofs go like. So I am going to now talk about how the proof of this theorem goes. It's not mega difficult, but it is a little bit tricky, but it's quite interesting. So let's have a look at the proof of the Arzela-Ascoli theorem. Let's look at the closure of this family f. Now, that's a closed subset of a complete metric space, namely the space C of X of bounded continuous functions on X. And so it's itself complete. That's something I showed earlier in the course. Now, if I want to show that a metric space is sequentially compact, well, it's enough to show that it's complete and totally bounded. In fact, complete and totally bounded is the same thing as sequentially compact, as I showed in the last lecture. So all I need to do is show that this closure f bar is totally bounded. And the first claim is that actually it's enough to show that f is totally bounded. Now, here, I'm not really using any particular properties of the fact that the curly f consists of functions. This is really just a general lemma about metric spaces, that if a set is totally bounded, then so is its closure. Well, let's prove the claim. Suppose epsilon is greater than zero. And suppose that f is totally bounded. So then there is some finite collection of balls of radii epsilon over two and centred on little f i's, which cover this family curly f. And the important thing about this collection of balls is that it's a finite collection of balls. So now take an arbitrary g in the closure of f. Well, if it's in the closure of f, then there's a function f in curly f, such that the distance from f to g is less than epsilon over two. Well, this little f is in curly f and therefore it's in one of these balls of radius epsilon over two about these f i's. But then the distance from g to f i by the triangle inequality is bounded above by the distance from g to f plus the distance from f to f i, which is less than epsilon. So the sum of epsilon over two and epsilon over two, epsilon. And what this means is that because g was arbitrary, if I take the balls with the same centres, but now with radii epsilon rather than epsilon over two, then actually those balls cover the closure f bar. And that completes the proof of the claim. So I've reduced the proof of Arzela-Ascoli to showing that this family f is totally bounded. And that's what I now need to show. So let's let epsilon be greater than zero to show that f is totally bounded. I'm going to show how I can cover f by finitely many balls in the subnorm of radius epsilon. Well, let's write down everything I know. So I know that curly f is a uniformly bounded family, and that means that there's an m such that the absolute value of f of x is less than or equal to m uniformly for all little x and for all little f. So that's just the definition of uniformly. I also know that curly f is equicontinuous. So let's write down the definition of that with epsilon over four replacing epsilon. So there's a positive delta such that if little x and little y are in big X and if the distance between them is less than delta, then the distance from f of x to f of y is less than epsilon over four. And that's for all little f in the family. Well, finally, I'm going to use the sequential compactness of the space X itself. So X is assumed to be sequentially compact and therefore it's totally bounded. Any sequentially compact metric space is totally bounded. And so it's covered by a finite collection of balls of this, of radius delta, where delta is the quantity that I've just, has just come up in applying the definition of equicontinuity. So there's some finite collection of balls with centres little x i and radius delta. So i ranges one, two up to some finite number k, which cover the space X. OK, so that's, that's the sort of set up. Now let's get right into the heart of the argument. I'm going to divide the interval from minus m up to m. So this is basically the interval on which the values of all of the f's lie. I would divide it into some finite number, capital K, of intervals, all of length less than epsilon over four. OK, so I can clearly do that with capital K being something like ten over epsilon, ten times m over epsilon or something like that. But anyway, you can divide it into some finite number of intervals, all of length less than epsilon over four, and I'll label them i one up to i sub capital K. Now, for each function that assigns a value between one and capital K, to each number between one and little k, so for each function alpha that does that, there might or there may not be a function little f in our family curly F such that little f of x i lies in the interval i, big I, sub alpha little i. OK, so this is maybe, this is for all little i between one and little k. This is maybe a slightly complicated concept here. What I'm doing is I'm basically saying, is there a function little f whose values at these centres of the balls, little x i, are roughly some values or not? So there may or there may not be. So if there is such a function, we'll just pick one and we'll give it a name f sub alpha. If there isn't one, we'll just choose f sub alpha to be an arbitrary function. OK, so I've got a finite collection of functions f sub alpha. The reason it's a finite collection is that there are only finitely many functions alpha at most big K to the power little k, such functions. So it's a finite collection of functions alpha. And I claim that the balls with centres those functions f sub alpha and with radii epsilon cover curly F. And if I can show that, then I'm done, because that's a finite collection of balls of radius epsilon in the subnorm that cover my family F. And therefore, since epsilon was arbitrary, curly F is totally bounded. So that's what I need to prove. OK, so it remains to prove that claim that those balls cover curly F. Well, to do that, let's take an arbitrary function little f in curly F. And I want to show that it's in one of the balls. Well, let's look at the values of that little f at the points x i. So there will be a function alpha that encodes where those values f of x i lie in terms of these intervals capital I. So choose alpha such that f of x sub little i lies in the interval big I sub alpha of little i for all i equals one, two, three, and so on up to k. Now consider the function f sub alpha as defined on the previous slide. By definition, that function has the same property. So that function, it's also the case that when you evaluate it at x sub little i, its value lies in capital I sub alpha of little i for all values of i. And in particular, the distance between, because those intervals, that the big I intervals, they all have width less than epsilon over four. So in particular, the distance between f of x i and f alpha of x i is less than epsilon over two for all values of i, one, two, three, up to little k. So I'll call that statement one. We'll be using that in a moment. So now, I mean, I should say, so this is this is quite nice because now I know that at least at the points little x i, f is close to f alpha. So it's beginning to look like I'm putting my f close to one of these f alphas. But now I need a statement that's true at every point x in the space, not just at these special points x i. So now let little x be an arbitrary point of big X. Now, by the choice of the little x i, well, there will be an i such that the distance from x to x i is less than delta. So remember that we chose the x i using the total boundedness of the metric space X. By definition, the balls of radius delta centred on those x i's cover big X. But also from the definition of delta, because this family is equicontinuous, I know that the distance from f of x to f of x i is less than epsilon over four. And the distance from f alpha of x to f alpha of x i is less than epsilon over four. So this is applying the equicontinuity of the family and using the fact that both f and f alpha lie in this family curly F. So we'll call this statement two. Now, if you put one and two together and apply the triangle inequality, well, it's quite easy to see that the distance from f of x to f alpha of x is less than epsilon. So the sum of epsilon over two, epsilon over four and epsilon over four. And if you think about what this is now, so little x is an arbitrary point of the space big X. So I've shown that an arbitrary point of big X, the distance from f of x to f alpha of x is less than epsilon. In other words, the subnorm f minus f alpha is less than epsilon. Or put another way, f lies in the ball of radius epsilon about f sub alpha. So I've proven the claim that those balls cover curly F. And that completes the proof of the Arzela-Ascoli theorem. So it's undeniably a little bit tricky, but it's not that terrifying. However, it is also non-examinal. That's the end of chapter eight. We've got one more chapter of the metric spaces part of the course, chapter nine, divided into two lectures. And there I'm going to talk about the notion of compactness. This is the penultimate lecture of the metric spaces part of A2, lecture 22. And this is the first of two lectures corresponding to chapter nine of the notes. In this lecture, we're going to introduce the notion of compactness. Now, I've already said that compactness turns out to be the same thing as sequential compactness, and that sometimes people call sequential compactness compactness right from the beginning, but that I've chosen not to do that. So I'm going to introduce this notion of compactness. And then in the final lecture, we'll show that it really is the same thing as sequential compactness. Now, this is a definition that's really hard to motivate. I think it's a perfect example of one of these fundamental things in maths that you just have to somehow get used to, rather than really trying to understand from first principles. So this really is one of the most completely fundamental notions in all of mathematics, the notion of compactness. Let X be a metric space and suppose I have a collection of open subsets of X. So we'll call that curly U and we'll let the sets in it be U sub i, i ranging over some index set capital I. And this is, I'm not by any means assuming that this is a finite collection. This capital I could be any indexing set whatsoever, could be uncountable. So we say that this curly U is an open cover of X if the sets U i cover X. So, in other words, if X is the union over little i in the indexing set big I of the U sub i's. So that's an open cover. Now, a sub cover is a covering of X by a subset of those U i's. So if J is a subset of the indexing set capital I and if the sets U sub little i for i in J also cover X, so if their union is also X, then we say that this is a sub cover of the cover curly U. So in other words, it may be that some of the sets U i in the cover curly U are redundant and I can remove them and what I'm left with still covers X. And then finally, if this new set J is a finite set, then we say that this is a finite sub cover. So I'm just using finitely many of these sets U i to cover X. And here's the definition of compactness. A metric space is said to be compact if every open cover has a finite sub cover. So no matter how I try and cover X with a collection of open sets, it might be an uncountably infinite collection of open sets, I can refine that collection to just a finite sub collection, which still covers X. So that's the definition. Now, we're talking about this for metric spaces, but for those of you who tuned into the non-examinable comments I made about topological spaces, you'll see that all I've used here is the notion of open set. There's no notion of distance explicitly. And so this definition makes perfect sense in the general context of a topological space. That's why it's so powerful. That's why it's such a very fundamental. Well, let's give an example of a space that's not compact, just to make sure that we're on board with what the definition is. The real line is not compact. And here is an open cover of it with no finite sub cover. I'll just take cover it by the open intervals from minus n up to n. So it's clear that if you take any finite sub collection of those, that won't cover the whole real line. We'll see in the next lecture a positive example of a subset of the reals that is compact. Now, this notion of compactness, it can be a little bit confusing how this interfaces with subspaces. And that's because there's a very common abuse of nomenclature in play in this situation. So I just want to describe that. Sometimes we'll have a metric space X and we'll have a subspace Y of X. And we want to talk about whether Y is compact. And by convention, when you talk about an open cover of Y, what you usually mean is a collection of open subsets of X. So a collection U sub i of open subsets of X, whose union contains Y. Now, that's different to what really should be meant by an open cover of Y, where the open set should be somehow internal to Y. But this is a very common abuse of notation, and it leads to the same sort of concepts for reasons that I'll explain in a moment. So we say that a sub collection of these U i's is a sub cover if Y is contained in their union. And then we say that Y is compact if and only if every open cover has a finite sub cover. But as I said, this is not a different definition, as it turns out. And the reason is that we proved earlier that the open sets in Y are exactly the same thing as the open sets in X intersected with Y. So I guess this is an exercise to check this carefully. But the definition I've given here of Y being compact is equivalent to the definition on the previous slide if Y is considered as a metric space in its own right with the subspace. So this is, this I find somewhat confusing, but it's a totally standard abuse of notation to use the word open cover in a slightly different way in the setting where we're talking about a subspace Y of a space X. OK, let's have a look at the link between compactness and sequential compactness. As I've said, we're going to show that the two notions are the same. One of those directions is slightly easier than the other, quite a bit easier, in fact, which is that compact implies sequentially compact. So that's the proposition. A compact space is sequentially compact. We'll need a lemma in the proof, which I will isolate, and that's the following. Suppose that X is compact and that I have some non-empty closed sets S i and that they're nested. So S 1 contains S 2, contains S 3. Then the conclusion is that their intersection is non-empty. So you might want to compare this just out of interest with a lemma that I proved in chapter six, I think it was, about complete metric spaces, where the hypothesis and the conclusion were somewhat similar. OK, let's prove it. Suppose that the intersection is in fact empty. Well, then the complements of the S i's, because they're all open sets, because the S i's are all closed sets, and they are a cover of X. Because if there's a point that's not in the union of the complements, well, then that point must be in the intersection of all of the S n's. So we've got an open cover of X. By compactness, there's a finite sub cover of it. And in particular, that means that for some value of n, the sets S 1 complement up to S n complement cover X. But those sets are nested. So S 1 contains S 2, contains S 3 and so on. And therefore the nesting of the complements goes the other way round. S 1 complement is contained in S 2 complement and so on is contained in S n complement. So if the sets S 1 complement up to S n complement cover X, well, that just means that the set S n complement covers X. Or in other words, that X is S n complement. But that's impossible because if X is the complement of S n, well then S n has to be empty. But I've assumed that S n is non-empty. So this is a nice little lemma that uses basically just the definition of compactness and the fact that the complement of a closed set is an open set. OK, so now we can turn to the proof of the proposition that compactness implies sequential compactness. So let X be a compact metric space and suppose that I have a sequence X n of elements of X. So n equals one, two, three and so on. Now I'm going to define some sets A n. I'll define A n to consist of what I'd call the tail of the sequence X n. So this is the set X n, X n plus one, X n plus two and so on. So A n is the set of all elements of the sequence from X n onwards. And obviously those are nested. So A 1 is the whole sequence that contains A 2, which is the sequence with X 1 removed, that contains X 3 and so on. And therefore, if I look at their closures, those are also nested. So the closure of A 1 contains the closure of A 2 contains the closure of A 3 and so on. And those closures, well, they're all non-empty closed sets. And therefore, from the lemma on the last slide, we see that their intersection is non-empty. So the intersection of the A n bars is non-empty. Let's take a point A in that intersection. And then I'm going to inductively construct a subsequence of my original sequence X n. So I will inductively construct a sequence X sub n sub k, k equals one, two, three and so on, such that the distance from X sub n sub k to A is less than one over k. So assuming I've done that, well, it's then clear that that subsequence converges to A. And then I'll have finished the proof because I'll have taken an arbitrary sequence X n and constructed from it a subsequence which converges. And that's precisely what it means to be sequentially compact. So let's turn to the details of that inductive construction. Suppose I've already constructed the integers n 1 up to n k. So I've constructed the first k elements of that sequence. Now, A, well, it lies in the intersection of all of the A n bars. And in particular, it lies in A sub n sub k plus one bar. Or in other words, in the closure of the set X sub n sub k plus one, X sub n sub k plus two and so on. So in particular, there's some element of that set or that sequence, which is at distance less than one over k plus one from A. So there'll be one of those X i's, i being ranging from n k plus one, n k plus two and so on, which is at distance less than one over k plus one from A. And we can just take that to be our X sub n sub k plus one. So it's a simple sort of inductive construction of this subsequence. And that's the end of the proof. Well, this is lecture 23, the last lecture of the metric spaces part of A2. We're going to do two things in this lecture. First of all, we'll prove what's called the Heine-Borel theorem, which is the statement that any closed bounded subinterval of R is compact. And then we'll show that compactness and sequential compactness are one and the same thing in metric spaces. So let's look at the Heine-Borel theorem. And this is the statement that the closed interval a b is compact. Well, let's have a look at the proof. Suppose I have an open cover of it. And here I'm thinking of this closed interval a b as being a subspace of the real line. And so that abusive nomenclature that I talked about in the last lecture is in play here. So I'm thinking of the UIs as open in R, not as open in a b itself. So those are open subsets of the real line which cover a b. And what I want to do is to show that some finite subcollection of those also covers a b. That's what it means for a b to be compact. So we'll define the set S as follows. We'll define it to be the set of all x in the interval a b, for which the closed interval from a up to x does have a finite subcover by these UIs. So what we ultimately want to do is to show that x equals b. That S contains everything between a and b. Well, whatever S is, it's certainly non-empty. And that's because a lies in S. Because this one of the open sets Ui must contain a. And also it's bounded above by just by definition, because it consists only of points between a and b. And therefore, it has a supremum, which I'll call c. And that supremum lies in the closed interval from a up to b. Now, in fact, that supremum is not equal to a. It's strictly bigger than a. The reason for that is that a is contained in one of these open sets. Let's say it's contained in Uj. And Uj being open, therefore, contains a little interval of positive length about a. So I'm, for technical reasons, going to work with a closed interval. So it contains a closed interval from a minus eta up to a plus eta for some positive eta. And in particular, that means that a plus eta lies in S. And so there is at least one number strictly larger than a that lies in S. Therefore, c, the supremum of S, is strictly bigger than a. We'll assume that c is strictly less than b. We'll get a contradiction. Well, curly U is an open cover of closed a, b. And c, whatever it is, does lie in one of the sets Uj, because all of the sets Ui cover closed a, b. Therefore, c lies in at least one of them. So suppose c lies in Uj. Well, Uj is open. And so there's a little interval about c. Let's call it closed c minus epsilon up to c plus epsilon, where epsilon is positive. That's also contained in Uj. And let's assume that I choose epsilon so small that c minus epsilon is bigger than a. I can do that because c is strictly bigger than a. And also that c plus epsilon is strictly less than b, which I can do because I'm assuming that c is strictly less than b. Now, c minus epsilon must be contained in S, because if not, it would be an upper bound for S and it's obviously smaller than c. And therefore, by the definition of S, the closed interval from a up to c minus epsilon is covered by finitely many of the sets from this collection curly U. But then if I take that finite subcollection of sets, together with a single further set Uj, I've remarked that Uj contains the interval from c minus epsilon up to c plus epsilon. So if you take that finite subcollection of sets from U that covers a up to c minus epsilon and throw in Uj, well, then you get a covering of a up to c plus epsilon, and it's still a finite collection of sets, a finite subcollection of curly U. Well, that is a problem because then c plus epsilon lies in S. But I've been assuming that c is an upper bound for S, so that c is the supremum of S. So what's the contradiction? Well, we must have been wrong to assume that c is strictly less than b. And therefore, the only possibility is that c equals b. Well, now we need to finish off the argument from here. Well, these sets in curly U cover closed AB and therefore b must be in one of those sets. So let's say that it's in U sub i. U sub i is open and therefore it contains an interval of positive length about b. Let's call that interval b minus kappa up to b plus kappa, where kappa is strictly positive. Now, the supremum of S is b and therefore b minus kappa must lie in S. Otherwise, b minus kappa would be an upper bound for S, strictly less than b. So b minus kappa lies in S, which means that the interval from a up to b minus kappa has a finite sub cover from this collection curly U. But then finally, if I take that finite sub collection and throw in the set U i, what I've got is still a finite collection of sets from curly U, but now it covers the whole of the interval AB, closed AB. So that's the proof of the Heine-Borel theorem, that every open cover of closed AB has a finite sub cover, or in other words, closed AB is compact. Well, now we come to the very final part of this part of the course. And this is a statement that a sequentially compact metric space is compact. And we showed in the last lecture that compact spaces are sequentially compact. And this, of course, is the converse to that statement. Now, only the statement is examinable. But so there is a statement, a sequentially compact metric space is compact. But let me just observe what this implies if you put it together with the main result of chapter eight. So it implies the following big theorem, which is sort of the biggest theorem in the basic theory of metric spaces. Let X be a metric space, then the following are equivalent. So one, X is compact. Two, X is sequentially compact. And three, X is complete and totally bounded. So let me just remind you that in chapter eight, we showed that sequentially compact is equivalent to complete and totally bounded. In the last lecture, we showed that a compact space is sequentially compact. And finally, now I'm going to show that a sequentially compact space is compact. So we'll turn to that now. And this proof is non-examinable. So let X be a sequentially compact metric space. Well, I know that X is complete and totally bounded. And this will be somewhat reminiscent of the proof that a complete and totally bounded metric space is sequentially compact. I'm going to use the totally bounded hypothesis for radii being one over powers of two. So for each m, natural number one, two, three, and so on, I will take a collection of balls, finite collection of balls, b one to the m up to b sub k m to the m, the radius of those balls being two to the minus m, and they cover the space X. So that's just applying the totally bounded hypothesis at these different scales, two to the minus m. OK, so suppose I have an open cover of X by some set U i, where i ranges over some indexing set big I. And suppose this has no finite sub cover. I want to get a contradiction. Well, by the pigeonhole principle, one of these balls at level one, so one of the balls b sub j to the one, is not covered by finitely many of the U i. Because if each of those balls was covered by finitely many of the U i, then because there are only finitely many of the balls, the whole of X would be covered by finitely many of the U i, which we're assuming is not the case. So one of those balls is not covered by finitely many of the U i, and let's call that ball b to the one. Well, now let's consider the balls b sub j to the two, which intersect that ball b to the one. Now, one of those is not covered by finitely many of the U i, because if it was, well, then b to the one would be. And I will call that ball b to the two. And now carry on in the same fashion. So consider the balls b j to the three, which intersect this ball b to the two and so on. So I will summarise on the next slide what properties this construction has. So what we get is a sequence b to the one, b to the two and so on of open balls. b to the m has radius two to the minus m. It intersects the previous ball. So b to the m intersect b to the m plus one is non-empty for all m. And none of the balls in this list are covered by finitely many of the U i. So that's what this construction will give us. Well, those balls, I haven't really given them names. They have radius two to the minus m. Let's give their centre a name. So let x m be the centre of the ball b to the m. Now, I claim that those x m's are quite close together. In fact, I'm going to prove that they're a Cauchy sequence. So the balls b to the m and b to the m plus one intersect. Let's say that they intersect in some point t. And so the distance from x m to x m plus one, well, it's at most the distance from x m to t, which is at most two to the minus m, plus the distance from x m plus one to t, which is at most two to the minus m plus one. And that is rather crudely less than twice two to the minus m. But now to compute the distance between x m and x n, when n is greater than or equal to m, I can use the triangle inequality. So the distance from x m to x n is less than or equal to the distance from x m to x m plus one, plus dot dot dot, plus up to the distance from x n minus one to x n. That's the triangle inequality applied several times. But I've just shown that those distances, so the first one is bounded by twice two to the minus n and the last one is bounded by twice two to the minus n minus one. And what I have there is a geometric series whose sum is certainly bounded by four times two to the minus m. So the distance from x m to x n when n is greater than or equal to m is bounded by four times two to the minus m. And it's clear that this makes the sequence of centres of these balls, the sequence x m, a Cauchy sequence. But x is being assumed to be complete. We've used total boundedness. Now let's use completeness. I have a Cauchy sequence. Therefore, it has a limit, which I will call x. Let's begin the next slide by just summarising where we've got to. So we've got a ball b to the m, which is the ball centre x m, radius two to the minus m. And I've now shown, first of all, that that sequence of centres is a Cauchy sequence and therefore that it has a limit, which is x. And by construction, these balls b to the m are not finitely covered by the set u i. So there's no finite sub cover from this family curly u. But those sets, they do cover x by assumption. And so one of them must contain the point x. Let's say without loss of generality, it's u one that contains x. Well, u one is open and therefore there's a ball of positive radius about x. Let's call it the ball of radius epsilon centred on x that's also contained in u one. Now, if I choose n really big, n is large enough, then the distance from x n to x is less than epsilon over two, just because the limit of the x n's is x. Also, if I choose n really big, two to the minus n will be less than epsilon over two. So choose n big enough that both of those things hold. But then what we see is that the ball b to the n is contained in the ball centred on x and radius epsilon. That follows from the triangle inequality. So the ball b to the n is the ball of radius two to the minus n, which is less than epsilon over two about x n. But x n is at distance less than epsilon over two from x. So every point in that ball by the triangle inequality is contained in the ball centred on x with radius epsilon. But that ball is contained in u one and therefore b to the n is contained. But that's utter nonsense because then b to the n, it's actually covered by just one of the sets u i. But I was assuming in my construction that it wasn't covered by even finitely many of the sets u i. So I've got a contradiction. I was wrong to assume that there was an open cover of x with no finite sub cover. And therefore, x is indeed a compact metric space. And that completes the proof that a sequentially compact metric space is compact.