Metric Spaces - Lectures 7 & 8 Oxford Mathematics 2nd Year Student Lecture

OK, this is Lecture 7 of A2, Metric Spaces, and this lecture covers the whole of Chapter 3 of the course. Now, when you introduce a new mathematical structure, as we have done with metric spaces, it's natural to ask about maps between those structures. So what kind of maps preserve the essential structure of a metric space? And that's what we're going to talk about in this lecture. So we'll first of all talk about a very rigid notion of structure preserving map between metric spaces, and that's the notion of an isometry. So if you've got two metric spaces, x and y, with respective distance functions d x and d y, then a function f from x to y is an isometry if it preserves the distance function. So if the distance between f of x and f of y, and of course, that's the y distance, is equal to the distance between x and y. So that's the capital X distance. So it really just does preserve the distance. Now, it's a very easy exercise to show that isometries are necessarily continuous. In fact, you can take delta to equal epsilon in the definition of continuity. They're also automatically injective. That's also a straightforward exercise, but they're not automatically surjective. So an example is if you take the sequence space little l 2 or one of the other sequence spaces, you can consider this right shift map. So that takes a sequence x 1, x 2, x 3 and so on, then shifts everything one place to the right and then puts a zero at the start. So that is an isometry. It preserves the distance, the l 2 distance, but it's not a surjective map because everything in the image starts with a zero. The reason I mention this is that some authors, when they say isometry, they mean an isometry that's also surjective. We'll call that a bijective isometry when we use this term. So they're slightly different concepts, but there doesn't seem to be any complete agreement about which is which. Some authors use the word isometry to mean what we're going to call a bijective isometry. Now, in the prelims course on geometry, there's quite a lot of discussion of the specific case when x equals y equals r to the n. And in that course, it was shown that all isometries have this form f of x is a x plus b, where a is an orthogonal matrix. And one particular consequence of that is that isometries from R n to R n are automatically bijective isometries. So there's no need to distinguish between the two terms in that context. So that's the most structured notion of a map that preserves metric spaces structure. Now, here's a much looser notion of structure preserving map, the notion of a homeomorphism. So suppose again that you've got two spaces, x and y, and that you have a continuous function between those spaces. And we'll say that it's a homeomorphism if it's a bijection. And moreover, if it's inverse, f inverse, which is a map from y to x is a continuous map. Now, something that's important to note here is that just being continuous and a bijection is not enough. So you really do need to additionally ask that the inverse be continuous. And here's an example to show that. So take x to be this slightly curious subset of the real line. So it consists of the half open interval 0, 1 with 0 in but 1 not in, together with the closed interval 2, 3. And then I'll take y to be the closed interval 0, 2. And then you can consider this map f from x to y. So what it does is it takes the closed interval 2, 3 and shifts it one to the left so that it sits nicely up against that half open interval 0, 1. So this is a continuous map from x to y and it's a bijection, but it's not, its inverse is not a continuous map. So points just to the left of 1 under the inverse will be mapped to themselves, but points just to the right of 1 will be mapped over to things just bigger than 2. So this is the inverse, f inverse is not continuous at the point 1. So that example shows that in the definition of homeomorphism, we really do need to separately specify that f and its inverse are continuous maps. So let's give a few examples of homeomorphisms. Here's one, the closed disc of radius 1. So the set of all points whose Euclidean distance from the origin in R2 is less than or equal to 1 is homeomorphic to the closed square minus 1, 1 cross minus 1, 1. And this is a proof by picture is the best thing to do here. What you can do to define a continuous map from the disc to the square, just work radially and along each arc, just stretch into the appropriate edge of the square. So this picture, I think, shows quite clearly what's going on. You could, if you wanted, write down an actual formula that does this. It would have to be a little bit of a sort of case by case formula according to which angle you were at around the origin. And you could very carefully check if you wanted that it's continuous. But I think it's clear just by looking at it that this is going to be a continuous map. With a continuous inverse. Let's look at a few other examples. It turns out that the open interval minus 1, 1 is homeomorphic to the whole real line. Now, that's perhaps a bit surprising at first sight because minus 1, 1 is a bounded metric space and the real line is not a bounded metric space. So the notion of boundedness is not invariant under homeomorphism. So why is this true? Well, you can give an explicit homeomorphism between the two. So this map f, which sends x to x divided by 1 minus the absolute value of x, is a continuous map from minus 1, 1 to R. And it has an inverse that you can also write down explicitly. The function g of x is x divided by 1 plus the absolute value of x. In a somewhat similar vein, the reciprocal function h of x equals 1 over x gives you a homeomorphism between open 0, 1 and open 1, infinity. So another homeomorphism between a bounded metric space and an unbounded metric space. A very famous example from topology is that a coffee cup is homeomorphic to a donut. I've written the American spelling of donut there as well, because if you search using that spelling, you'll find some nice YouTube videos of this homeomorphism. So coffee cup being turned into a donut. I recommend you do that if you haven't seen this, if you've never heard this fact before. The homeomorphism between metric spaces is a much looser notion of spaces being the same than an isometry. So certainly every isometry is a homeomorphism, but the converse is absolutely not true. In particular, an isometry would certainly take a bounded metric space into another bounded metric. Now, the final part of this lecture is not really examinable. It's not on the schedules, but I think it's interesting to talk about. So I'm going to talk about it. So this is the notion of equivalence between two different metrics on the same underlying set. So suppose I have a set X and that I've got two different metrics on X, D and D'. Then I'm going to say that they're equivalent if the identity map, which is denoted by the Greek letter iota here, from X with the metric D to X with the metric D' is a homeomorphism. So that's automatically a bijection. The key point here, then, is that this map is continuous and that its inverse is continuous. So what does that mean? Well, if you think about it for a little while and think about the definition of continuous, what it means is that every open ball in the D metric, so every open ball of radius epsilon, contains an open ball of radius some other epsilon primed in the D' metric. So that epsilon primed there would be the delta in the definition of continuity. So an open ball in one metric must contain an open ball in the other metric centred on the same point. So that is equivalent to this property. And it turns out that if you've got two equivalent metrics, then various things about those metric spaces coincide. For example, the notion of limit is the same in the two metric spaces. I'll leave the proof of that as an exercise. This is being a non-examinable part of the course. Now, one of the reasons I wanted to talk about this is that we've already seen a very basic example of some equivalent metrics. So we've seen three metrics on R to the n, Euclidean space R to the n. D1, D2 and D infinity. So recall that these are all induced by norms. D1 is induced by the L1 norm, which is the sum of the absolute values of the coordinates. D2 comes from the Euclidean norm. Sum of the squares of the coordinates, all square rooted. And D infinity is the sub norm, so it comes from the L infinity norm on R to the n. I claim that those are all equivalent. In fact, they're all what is known as strongly equivalent. Now, two metrics on a space X are strongly equivalent if this holds. So if you call the metrics D and D primed, essentially they're the same up to multiplicative constants. So if there's an upper bound for D, it's upper bounded by a constant times D primed. And similarly, D primed is upper bounded by a constant times D, where that constant does not depend on X and Y. So it's a very straightforward exercise to show that strongly equivalent implies equivalent. I mean, the ball of radius epsilon is going in the D metric will contain the ball of radius epsilon over C in the D primed. Now, the three metrics that we're talking about, as I've said, they all come from norms, the L1 norm, the L2 norm and the L infinity norm. And to say that the three metrics are equivalent is equivalent to a statement about the norms. It's equivalent to saying that the norms are bounded above and below by constants times each other. So in other words, there is a constant C such that the norm of X is less than C times the norm primed of X for any pair norm and norm primed. So there are going to be six possible pairs here that we need to look at. So, for instance, we need to show that the L1 norm of X is bounded by a constant times the L2 norm of X. And we need to show that the L infinity norm of X is bounded by a constant times the L1 norm of X. There's six things to show. So I've just rewritten here, written down again what I just said. So we need to prove these inequalities for each pair of these norms that we've got an upper bound for one in terms of the other. Now, there are actually four of those inequalities are pretty easy to write down. So here are two. So, first of all, the L infinity norm is less than or equal to the L1 norm. If I've got a sequence of numbers, the sum of them is at least the biggest one. That's the first inequality. And also the sum, if there are n numbers, the sum of them is at most n times the largest one. So I regard those as obvious inequalities. That's dispatched two of the pairs of norms. And for almost identical reasons, you have a similar statement with the L2 norm and the L infinity norm. So again, the sum of the squares of some numbers is at least the largest one squared. So when you square root, you get that the L infinity norm is less than or equal to the L2 norm. And similarly, the sum of the squares is at most n times the largest one squared. And if you square root that, you get that the L2 norm is less than or equal to square root n times the L infinity norm. So we still need two more inequalities relating the L1 norm and the L2 norm. But actually, we can get them for free by putting these two together. So, for instance, from the first inequality and the second one, we get that the L1 norm of X is less than or equal to n times the L infinity norm of X, which is less than or equal to n times the L2 norm of X. So that gives us a relation between L1 and L2. And we can similarly get one in the other direction by combining these two inequalities as well. Or you could use the Cauchy-Schwarz inequality to get the optimal constants. So the three different metrics on R to the n turn out to be equivalent. This is lecture eight of the A2 Metric Spaces course, and this is the first lecture that corresponds to chapter four of the printed notes. Now, in this lecture, we're going to introduce one of the most important concepts in the theory of metric spaces, and that's the notion of an open set. So let's start by giving the definition. Let X be a metric space. And suppose that I've got a subset U of X, which could be empty, it could be the whole space X or anything in between. Then I say that it's open, more precisely, I say that it's open in X if every point y in that set U, there's a positive delta such that the open ball of radius delta about y is contained in U. So for every point in the set U, there's a whole ball around that point that's also contained in U. That's the definition of open set. So let's give some examples. So what we normally term the open interval, open zero one, is an open set in R. And that's because if you take any point x that's strictly between zero and one, there will be some little interval of width delta about x, which is also contained in zero and one. You can always fit such an interval in and therefore open zero one is an open set. But the closed interval zero one is not. And that's because, for instance, there's no open ball centred on the point zero that lies entirely within that set. Every such open ball contains a rational number. The rational numbers are not an open subset of the reals. That's because any open interval on the real line contains irrational numbers. In fact, uncountably many irrational numbers. And therefore, you certainly cannot have any open ball centred on a rational number consisting entirely of rational. That's just not true. But finally, if x is a set with the discrete metric, then absolutely every set is open. So at first sight, that may seem slightly surprising. But the reason is that if little x is a point in big X, then the ball of radius one about little x, that consists just of the point x, if you think about it, and therefore it's contained in big X as well. Now, it's very important to stress that this notion of open set is what we call a relative notion. So when you say that U is open, you really have to be thinking of U as a subset of x. And you really should be saying U is open in x. So let me clarify a bit what I mean by that. If I take the set, the closed interval zero one, well, as I've just remarked, that's not an open set in the reals. But if you think of it as a metric space by itself, which you can with the subspace metric, then it is an open set considered as a subset of itself. In fact, if x is a metric space, then x itself is always open. Whether something is open depends on the underlying metric space x in which I'm working. Therefore, it's a relative notion. Now, we've talked about open balls. In fact, the notion of open ball was important even in the definition of open set. But to justify calling them open balls, we need to prove a little lemma. We need to prove that every open ball in a metric space is an open set. And just to emphasise, that's not a completely trivial statement. Just because we call them open balls doesn't mean that they are open. So that's something that has to be proven. So let's prove that statement. We'll give the ball a name. We'll call it the ball centred on a with radius epsilon. And let's take a point x in that ball. Then by definition of the ball, the distance from x to a is less than epsilon. And if you're less than epsilon, then you're at least, you're also less than epsilon minus some positive quantity. So there's an epsilon dash greater than zero, such that the distance from x to a is in fact less than epsilon minus epsilon. And I claim that the open ball centred on x with this smaller radius epsilon dashed is contained in the original open ball centred on a. So let's prove that. Suppose I take an arbitrary point z in that smaller ball. By definition, the distance from z to x is less than epsilon dashed. And therefore, by the triangle inequality, the distance from z to a is less than or equal to the distance from z to x plus the distance from x to a. And that's less than epsilon primed plus epsilon minus epsilon primed, which is equal to epsilon. Another key concept in the theory of metric spaces is that of a closed set. So closed sets are the complements of open sets. So a set F in a metric space X is closed if and only if the complement X backslash F, also denoted F to the C, is an open subset of X. So it's very important to stress that the property of closed is not the same as being not open. In fact, even just on the real line, you can see that any permutation and combination of open and closed is possible. The real line itself is both open and closed. So I think it's clear that it's open. To see that it's closed, its complement is the empty set. And if you think about the definition of being open, well, in the case of the empty set, there's nothing to check. So the empty set is open. Now, the open interval zero one is open. We remarked on that already, but it's not closed. So its complement contains, for instance, the interval starting at one and going off to infinity. And that is not open. Well, it's that together with the negative real numbers. And that's not an open set because, for instance, there's no ball about the point one that lies entirely in that set. The closed interval zero one is closed, but it's not open. And the half open interval zero up to one is neither open nor closed. So you can be any of the four possibilities, open or not, closed or not. Now, we've talked about closed balls. We'd better prove an analogue of our earlier lemma saying that closed balls are in fact closed. And that will avoid a lot of confusion once we prove that. So once again, we define the notion of a closed ball. We called it a closed ball, but we haven't shown that it's a closed set. And a consequence of this is that singleton sets are closed. The proof of this will go along rather similar lines to the proof that open balls are open. We'll again give the ball a name. So we'll call it the closed ball centred on a with radius epsilon. And just to remind you, this is the set of all x whose distance to a is less than or equal to epsilon. That's the definition. So we'll show, which is what we're going to do, that the complement of this is open. So let x be a point in that complement. Well, then by definition, the distance from x to a is greater than epsilon. So there's an epsilon dashed positive such that the distance from x to a is also greater than epsilon plus epsilon dashed. And I claim that the open ball centred on x with radius epsilon dashed is contained in the complement of the closed ball that we're interested in. So then I'll have shown that for an arbitrary point x in the complement of the set I'm interested in, there's an open ball about that x that's also contained in the complement of the set I'm interested in. And that means that that complement's open and therefore the original set is closed. So there's quite a bit of sort of complementation to think about here. So let's prove this claim. Well, let's take an arbitrary point z in that smaller ball centred on x with radius epsilon. By definition, the distance from z to x is less than epsilon primed. And so by the triangle inequality, slightly rearranged triangle inequality, the distance from z to a is at least the distance from x to a minus the distance from z to x. And that is greater than epsilon plus epsilon dashed minus epsilon dashed, which is epsilon. And then finally, the statement about singleton sets. Well, that's just the observation that the singleton set A, so the set consisting of just the element A, is the closed ball of radius zero centred on A. Let's turn back now to open sets and we'll establish some of the key properties of open sets. So let X be a metric space. Then something we've already mentioned is that the sets X and the empty set are open. Second point is that an arbitrary union of open sets is open. So more formally, for any indexing set capital I, and for any collection U sub little i, with i being in capital I, of open sets, the union of those U sub little i's is an open set. And then the notion of being open is also closed under finite intersections. So if the U sub little i are open sets, where the indexing set big I is now just a finite set, then that intersection is open. So let's prove statement two. Let's suppose we have a point that's in the union of a whole load of open sets. Well, then in particular, it must be in one of those sets. Let's say X lies in U sub i. Now U sub i is open. And so there must be some open ball, radius epsilon, about X that's contained in U sub i. But if it's contained in U sub i, then it's certainly contained in the union of all of the U i's. And so we found a little ball about X that's contained in the union of the U i's. And that's precisely what it means for that union set to be open. The statement about intersections is a little bit more interesting. Suppose X is in the intersection of some U sub little i's, where the little i is now ranging over a finite indexing set, capital I. So for each of those little i's, X must lie in U sub little i. That's what it means for X to lie in the intersection. And so for each of those little i's, there must be a ball centred on X with some positive radius, epsilon i, that's contained in U i. Now, the epsilon i's might be different for different U i's. They're all positive. And because there's only finitely many of them, their minimum is also positive. So we'll define epsilon to be that minimum. It's a positive number. And then all of the ball of radius epsilon about X is then contained in the ball of radius epsilon i about X. And that's contained in U i. And that's true for every i. So that ball of radius epsilon about X is contained in the intersection of the U i's. So you'll see that we made really critical use of the fact that the indexing set I is finite here, because whilst it's true that the minimum of a collection of finitely many positive quantities is finite, that's not true for infinite epsilon i's. Let me make a few further comments on open sets. OK, so if you wanted to, in fact, the fact that the set X and the empty set are both open follows from the other statements two and three, but I'm not personally particularly interested in producing minimal lists of statements. Point three, which is the finite intersection property, the proof doesn't work if the indexing set A is not finite. And moreover, the statement fails. So here is an example. Let's consider these open sets on the real line. So U i is the open interval of length of width one over i about zero. So that's the interval from minus one over i up to one over i. Those are all open. Their intersection contains just the point zero, the singleton zero, and that's not an open set. So it may fail that the intersection of some arbitrary collection of open sets is open. And we'll be coming back to this point in a couple of lectures. And in much more detail, you'll see this in the topology course. But the collection of all of the open sets in a metric space X is called the topology on X. So the last thing I want to do in this chapter is just to go over quickly the analogue of that previous proposition for closed sets. And this really follows just by taking complements of that whole lemma. So let me just tell you the result of doing that. So let X be a metric space. And suppose I have a collection F i of closed sets. Then once again, the whole space X and the empty set are closed. And now an arbitrary intersection of closed sets is closed. And a finite union of closed sets is closed. So this just follows from the previous proposition by taking complements and using De Morgan's laws. I'll leave the detailed check of that as an exercise, but that's the corresponding properties of closed sets.