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0:01
in this lecture we are going to define the notion of a metric space. In real analysis, a metric space is kind of like the broadest, most inclusive type of domain that we work on because it doesn't take much to have a metric space, as you will see over the course of this lesson. So what does it take to have a metric space? You need a set M of elements. It could be real numbers, it could be complex numbers, but it could also be like markers in the room,
0:26
as we will actually use in an example. So you need a set M, and you need a function D which measures distance between elements. So that's why I'm saying here that D is a real-valued function of two inputs because you plug in the two inputs, and you get out the real-valued notion of distance. Okay, so you have a metric space if you have these two things: a set and a distance, so that the distance satisfies four very natural properties. The first one is positivity.
1:06
It's very strange to tell distance with negative numbers, right? So the distance from me to the camera is not -3 meters, it's like about 3 meters. So D of X comma Y is greater than or equal to zero for all X and Y in the metric space. I'll leave off writing that distance could be zero, but when is it zero? If and only if you're measuring the distance from an element to itself. We call that non-degeneracy. So that means that D of X comma Y is zero
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if and only if X and Y are the same. We call that non-degeneracy because this word indicates that D distinguishes between distinct objects. So if it kind of like couldn't tell the difference between two different objects X and Y, kind of collapses them onto each other, and said their distance was zero, that would be a degenerate function. So here we're saying that it separates different elements, you can tell the difference between them.
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The next one is symmetry, right? So the distance from the green marker to the orange marker is the same as the distance from the orange marker to the green marker. Straightforward, right? So it's a natural property. Symmetry. And then the last property is that detours make your journey longer. So if I want to go from the green marker to the orange marker, it would be a detour if I went and visited
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a different marker along the way. I'm trying to say the triangle inequality. I'll give you a more mathematical illustration of it in just a second. So the triangle inequality is that the distance from X to Y is less than or equal to going through a different element Z. So it's less than or equal to the distance from X to Z plus the distance from Z to Y. If I imagine that X, Y, and Z are points in R^2, so imagine the board here is just Euclidean two-dimensional space,
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say here's X and here's Y. This straight line in Euclidean space is an illustration of their distance, so the length of this line is the distance from X to Y. If I go through any other point—supposed to be a straight line segment, so is this one—okay, you get the idea. It's the triangle inequality because the distance from X to Y in Euclidean space—that's the length of this line segment—is less than or equal to the length of this
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segment plus that one. So that's the triangle inequality. Okay, it looks like a triangle. Alright, this is an illustration of one of these properties in two-dimensional space, but what I would like to do now is look at a totally different metric space which you may have never worked in before. I said it's not too hard to have a metric space, the requirements are not that difficult to satisfy. So here's an example. Let M be the set of markers in the room. That's
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actually these five markers. And let D be what's called the discrete metric. So this is a metric that I will refer to when we work in metric spaces pretty often. So if you're following my lessons, please keep this in mind. This is our first introduction to it. The discrete metric is defined as D of any X and Y from the set M is zero if they are the same element, otherwise it's one. So it's either zero or one. D of orange marker, comma orange marker is zero. D of orange marker,
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comma green marker is one. The distance between these elements is still one, it's still one, it's still one, it's always one. So this is our notion of distance between distinct elements. Let's verify that this is a metric space. Well, to be a metric space, you have to have a set of elements, we have that, so M is ready to go. We need to check that the function D here satisfies the four required properties. I think the first three I can do out loud, you
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may want to write them down, but we can probably go through those out loud, and then I will write down a quick sketch—or actually, I'll probably do the full proof for the triangle inequality. Okay, the first property we need to check is positivity. This function only has two possible outputs, zero and one, neither is negative, so by construction we have positivity for this function. Also by construction we have non-degeneracy, so if two elements are the same, their distance is
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zero. On the other hand, if their distance is zero, they must be the same. Don't forget non-degeneracy is an if and only if statement, but again, this function is basically set up to match non-degeneracy. Same thing with symmetry, so D of X comma Y is going to be the same as D of Y comma X. If they're the same, both of those values will be zero. If they're different, both of those values will be one. Okay, so the first three are pretty much automatic, we can do them out loud.
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Almost always, when the triangle inequality is one of the properties of a function like this one, it's worth taking a little bit more time to treat it carefully. So let me pause in case you are still thinking about this metric, and go ahead and write down the triangle inequality. Think about the kinds of values that you would encounter on the left and right-hand sides of the triangle inequality, and then I will come back and actually write out a little proof.
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Alright, if you write down the triangle inequality, this is what you should have. I'll go ahead and start the formal proof. So I'll say we want to show that for any X, Y, and Z in the metric space or in the set M, because we don't yet know it's a metric space, the distance from X to Y is less than or equal to the distance from X to Z plus the distance from Z to Y. If I was writing this on paper, I might put this centered on its own line, but I don't have too much space to
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work with here. For this particular metric, let's consider the values that are possible on both the left and right-hand sides of this inequality. On the left-hand side, either zero or one, that comes straight from the definition. On the right-hand side, we could be looking at 0+0, we could be looking at 0+1, or similarly 1+0, or we could be looking at 1+1. Okay, so 0, 1, 0, 1, or 2. I typically prefer direct proofs, and I think if I were actually writing this up and not hoping to
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do the fastest proof possible because I'm running out of board space, I would probably write this up directly, but I'm going to do a proof by contradiction because it might be the fastest route. Suppose the—maybe I should say what the values are for the reader, and then I'll go into the contradiction. So notice the left-hand side—so abbreviation left-hand side—is either zero or one, while the right-hand side is either zero, one, or two. The only way that the triangle
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inequality could fail is if the left-hand side is strictly bigger than the right-hand side. So if the left-hand side is zero, we're done. If it's one and the right-hand side is zero, that's the only way the triangle inequality could fail here for the sake of contradiction. So for the sake of contradiction, suppose Dxy is one, but dxz plus dzy is zero. You probably see what's going on now. Now, if the right-hand side is zero, we must have added 0+0. That's the
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only way we could get 0 out of this sum. But that means x equals z and z equals y, which means x equals y by transitivity. But then that means the left-hand side is zero. This latter implies x equals z and z equals y, but then x equals y. This would produce—the left-hand side is zero. So let me say this implies—alright,
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so this implies the distance between x and y is in fact zero, because they really should be the same element. But then that contradicts this statement here. So this is my symbol for contradiction, and I just kind of quickly wrapped up this proof. You could write this maybe avoiding the contradiction language. I just kind of did this right on the spot. You also could write it without
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contradiction, just kind of show that the triangle inequality must be satisfied directly. But I think this is a nice little argument to recognize for this particular metric that the triangle inequality can only fail in one particular situation which literally cannot happen. Alright, I'll give you a minute to finish writing this down or if you'd like to tackle it a little bit differently, you can do so. When I come back,
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we will talk a bit about what geometry might look like in this metric space. The exercise I would like for us to complete now is to describe circles in this discrete metric space. So first, what is a circle? If we just think back to the classic circle that you would draw on paper in the XY plane, a circle is a set of points which are the same distance from a fixed center point. So we can characterize circles in terms of a distance from like an anchor point, the
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center point. Okay, so what would a distance look like in this discrete metric space? Well, we only really have two distances, zero and one, which are realized between objects in this setting. So here's my metric space. It's these markers together with the discrete metric. I need a center point. Let's make it this pink marker, P. Okay, can you think for a moment what is the unit circle centered around P? So if I want to draw the unit circle around P, what would that look like? Are
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you ready? It would be all of the other markers. So every other marker in this discrete metric space is distance one from P. So these markers form the unit circle around P. These markers—I just switched out pink for orange—these markers form the unit circle around this orange marker. Okay, what about the circle around the orange marker of radius two? What does that look like? That is the empty set. No markers at all are distance two from this orange marker. So there
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is nothing in the circle. The circle consists of no elements from this discrete metric space. Same thing for the circle of radius two. If I wanted like the solid disc of radius one, so all of the markers in the metric space whose distance from this central orange marker is less than or equal to one, what would that look like? It would be all of them, right? So all of the markers in the metric space have a distance less than or equal to one for this orange marker. Okay,
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so I just wanted to kind of lift that up for you as an illustration of how geometry can change when you move from one metric space to another. So our notion of circles in Euclidean space, it's like we have circle radius one, circle radius two, solid disc of radius one in the plane does not contain all of the plane, but when we look at this particular metric space, it looks very different. Okay, let's also look at another example, and then we will wrap up with an important result.
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For another example of a metric space you may not have seen before, let M be the XY plane. So in other words, elements in M are points X comma Y. And let our notion of distance be what we call the taxi cab metric or the Manhattan distance between two points. So it's defined this way. What we will do now is just a quick check that we understand how to read this function. We will verify that it has the four properties of a metric so that we can call this a metric
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space. Then I just want to take a quick moment to sketch a circle in this new metric space. Okay, so how does this work? Suppose I take the point -3,1 down here in the third quadrant—that would be like U1 comma U2 in my definition—and I want to measure its taxi cab distance to the point 2,3 right here—that's the point V1 V2.
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What is the taxi cab distance between these two points? So if I do the distance between -3,1 and 2,3, that's going to be the absolute value of the difference in their x coordinates -3 minus 2. Take the absolute value of that, add to that the absolute value of the difference in their y coordinates. So -1 minus 3. Alright, so we get 5 + 4 is 9. Alright, so the distance—we want to call it a distance, we still have to verify the four properties—but let me use that already. The
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distance between these two points with the taxi cab metric is 9. That is not the same as the length of the straight line connecting them. That would be the Euclidean distance, that's not what we're looking at here. Instead, what we're saying is that the distance between these two points is the horizontal distance that's being measured by this absolute value of one x coordinate minus the
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other. That horizontal distance plus the vertical distance, so that's being measured as the distance between their y components. This is called taxi cab distance because if you think about it, imagine this was a city grid like Manhattan and you had kind of north-south running streets, east-west running streets, and maybe these blocks here are like apartment buildings and hospitals and schools and things like that. If you were a taxi driver navigating the city and
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you wanted to figure out some idea of how far this point is from this one, it doesn't really make sense to imagine going from one to the other in a straight line because that's not how you can get there. So you can't cut through buildings, you have to go along city streets. So this taxi cab distance is measuring how many blocks the taxi would drive down in order to get from one point to the other. So 5 here is like 5 blocks to the east plus 4 blocks to the
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north. That's a more natural way to think about a trajectory in the city compared with how, say, a bird might fly from one roof to another. Okay, so that's why it's called the taxi cab distance. Take a moment and if you don't do anything else, just say the names of the four properties out loud and kind of assess which ones are immediate, which ones do you think I'm going to write down. I will come back and we will talk through them and we'll do a little bit more with this metric.
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The four properties that we need to check are positivity, non-degeneracy, symmetry, and the triangle inequality. Let's think about each one visually first, and then we will write probably every single one down just to have a complete argument. Positivity seems very immediate, right? So no way that if I'm telling you how many blocks I'm going to drive to get from one location in the city to another, there's no way I'm going to say -4 blocks, and that's going to agree with the way
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this function is defined. Non-degeneracy, if I want to go from one intersection back to itself, like what's the distance between one point in the city and itself is zero. On the other hand, if I say that the distance from another point to this one is zero, is it clear that I'm looking at comparing the same point to itself? Is this metric non-degenerate is what I'm trying to say? Yes,
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that's also going to be true. Sorry, it's kind of an awkward if and only if thing to phrase in this context. Symmetry: is the distance from here to here the same as the distance from here to here? That is actually true for this function, but if you think about it, it's not necessarily true in a real city because you might have something like a one-way street, right? So in the place where I
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live, if I want to drive from the north side of town to the south side, I actually have kind of a different trajectory based on traffic and sort of unusual traffic patterns through the center of town. But for this function and this idealized city, symmetry will hold. Triangle inequality: you know, if I want to get from this point to this one, the most direct thing I can do is travel only east and north the right amount of blocks. If I take any kind of detour,
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like I go south or I go west, I'm going to add on to the length of my journey. That would be the triangle inequality for the city. Let's work through all of those statements now algebraically.
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Positivity doesn't quite feel like it's worth writing down, but let me say something. Since the absolute value of A minus B is greater than or equal to zero for all real numbers A and B, in particular pairs of x coordinates and pairs of y coordinates,
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the distance between the point U1 U2 and the point V1 V2 is made up of the sum of those two. So it's greater than or equal to zero, and I'm just going to write by construction.
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So that means it follows right out of the definition with very little analysis necessary. Non-degeneracy, which I did not describe very well a minute ago, is an if and only if statement. Let's see if we can write it more clearly than I said it out loud. Notice the distance from any point—I'm going to describe it as this vector U—back to itself
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is the absolute value of U1 minus U1 plus U2 minus U2, the absolute value of that—sorry, that should be a minus—which is zero, zero plus zero, which is zero. That's one direction of non-degeneracy. For the other direction, conversely,
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suppose that the distance from U to V was zero. How could we get zero from adding together two absolute value expressions? Both would have to be zero. Oh, this marker is worse. Let's see. A little bit better. Then the absolute value of U1 minus V1 equals zero, and also the absolute value of U2 minus V2 is zero, which implies that U1 U2, that point, equals V1 V2. Okay, what I really want to highlight here by writing out non-degeneracy is
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that it's if and only if, and you need to address both directions. Okay, let me step aside just so I can erase this. Then we will continue with symmetry and the triangle inequality. Alright, for symmetry, we can write that the distance from V to U is the absolute value of V1 minus U1 plus the absolute value of V2 minus U2, but we have symmetry inside the absolute values. So that's the same thing as the absolute value of U1 minus V1 plus
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the absolute value of U2 minus V2, which is the definition of the distance from U to V. Okay, so symmetry is going to check out. And I think I can squeeze the triangle inequality on here. Let's compare U to V but take a detour through W, so that will have coordinates W1 W 2. So the distance from U to V is equal to this. I'm going to run out of room, so instead of rewriting this again, which is in the definition of the function,
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let me go ahead and bring in the coordinates of W. You'll see what I mean. So I'm going to say it's U1 minus V1, but let me bring in—before I write minus V1, let me write minus W1 plus W1 so that I haven't changed anything, and then take off V1. And then likewise, we'll write U2 minus the y component of this other point. So minus W2 plus W2, so it's like adding a fancy version of the number zero, and then minus V2 like that.
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The absolute value function, which is actually the Euclidean metric on the real number line, this has a triangle inequality in its own right. So I can break each of these up and say that this is less than or equal to the absolute value of U1 minus W1 plus the absolute value of W1 minus V1, plus here it's going to be the same idea, absolute value of U2 minus W2
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plus the absolute value of W2 minus V2, which I should have written here. My apologies, probably you saw that when I wrote it. There we go. And if you wanted to, you could bring in another line where you kind of reorder these terms so that this is next to that and this is next to that. But what we have here is the distance from U to W. I'm really struggling with markers right now, I apologize. But these two together are the distance from U to W,
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and these two together are the distance from W to V. Okay, that wraps up the four criteria that we need to check to verify that the taxi cab metric is indeed a proper metric. So R2 with this notion of distance is a metric space. Take a moment, finish this, and then I just want to talk through some visual implications here before we move on to a little result with a proof. Alright, let's make a few visual observations about this metric space. The first thing I want
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to point out is something you may have realized when I was talking about why this is called the taxi cab metric, and that is that there is not a unique path from two different points in the city unless they are on the same street. So as long as I have to make at least one turn, I might have multiple ways to get from, say, this point to this one. So what we could do is drive five blocks to the east and drive four blocks north, or we could drive four blocks north, five blocks east,
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or we could kind of mix and match some east, some north, that sort of thing, as long as we are always moving in the right sense of direction. That's different than R2 equipped with the Euclidean metric. If I had these two points in Euclidean space, there is only one shortest path between them. So we have more variety here on what you might think of as the shortest path. Another geometric discussion worth having is what circles would look like in this
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metric space. Let's draw the circle of radius two centered at the origin. So we'll just make this a circle around the origin in R2 of radius two. Okay, so how could we do that? First, let's stick with this idea that we're working with a city. I could go two blocks east, I could go one block east, one block north, two blocks north, one block west, one block north, west, you know, southwest, south, southeast. Okay, in a city, those are all of the points
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in the metric space which would be distance two from the origin. But that's not actually really what we have here. We have R2, so I would like to be able to actually travel fractional amounts, if you will, or really any amount. So could I go like 1.5, 0.5? Okay, or 0.5, 1.5? Basically, take away the story behind this metric, which we were looking at a city grid. We're not really looking at a city grid. What we have here is the plane. U1, U2, V1, V2, they don't have to be whole
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numbers. That's really what I was trying to say. So we can go any amount up to two blocks east, pair with that some amount north, and fill in this whole straight line segment here. Likewise here, here, and here to conclude that the circle of radius two centered at the origin takes a diamond shape. Okay, so we saw that circles looked a little different in a discrete metric space. Here with this taxi cab metric, our circles look like diamonds.
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Alright, that's the end of our examples. I want to wrap up, though, with an important result connecting norms to metrics. In this last part of the lecture, I want to show you that if you are working in a normed vector space V, there's a function that you write down which is automatically a metric. So we will call that D for metric, and we'll say that this metric is the norm of V minus W. So this
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is a metric defined in terms of the norm that we are getting from the normed vector space. We are going to prove that this metric has the four properties. Here I haven't specified what the norm and the vector space are, which means that if you are working in any kind of normed vector space and you work with the metric, you know it's a metric. You don't need to prove the four properties because we're going to prove them in the most general sense. Before I do that,
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let me just make a couple quick remarks. When we looked at these kinds of computations on Euclidean space, so our norm was like Euclidean magnitude of vectors, and our notion of distance, like the distance formula, we could actually characterize in this way. That was an analogous result to this one. Similarly, when we talked about norms, we had a result that said that if you are working in a real inner product space, from that inner product you can define a norm.
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That's also an analogous result. So inner products induce norms, norms induce metrics. We are going to do the second part of that statement now and prove that this norm induces this metric. It's just a question of checking the four properties. All you need to be a metric space is the set and the metric. The set here is the set of vectors in V, so we get the set from the fact that we're working in this vector space V. The four properties we need to verify—
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As we go through each of these properties, pay attention to how they come so quickly because we know we are working with a valid norm. So when I say valid norm, I mean that this norm has all the properties of a norm that we already covered. For example, notice the distance between any vectors V and W, which is just to kind of restate it, defined as the norm of V minus W, is greater than or equal to zero because we already know that the norm is never negative. So let me write,
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because the norm function has positivity. That was also the first property that we named for norms.
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Okay, let's look at non-degeneracy. I think we can do this as one kind of if-and-only-if chain. I don't think I need to make two arguments to cover the two directions of the if-and-only-if statement. So I used notice here, let me switch to observe. Observe the distance between V and W equals zero if and only if—so this is like the first link in my if-and-only-if chain—if and only if—well, this equals zero. Just going to kind of restate how this metric is defined. But norms are
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non-degenerate, so if the norm of something is zero, then what you took the norm of had to be the origin in the vector space. So let me say that this is true if and only if this quantity in here is, and I'll write zero vector here to distinguish it from scalar zero. So this is like the zero in the vector space, this is just distance zero, they can look different. We're working in a vector space where we have vector addition, additive inverses, things like that, so I can just add W
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to both sides and that addition works. If and only if, because it's well-defined addition, so if I swing W over to the right-hand side, we get equal W, and that concludes our if-and-only-if chain. Alright, just going to write it off to the side here. We have, like, this is non-degenerate. Okay, that's not part of the formal proof, that's kind of a note to myself. Okay, so positivity of the metric was inherited from the positivity of the norm. Similarly,
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non-degeneracy of the metric was inherited from the non-degeneracy of the norm. That's really what it means to say that the norm is inducing the metric. But we have two more properties to check, and I think I would like a little bit more room, so let me erase this and come back and do symmetry and also the triangle inequality. Alright, the reason why I wanted to give myself more room for symmetry,
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which might feel like it's going to be a fast computation, is because the third property of norms is not symmetry, it's multiplicativity. So what I want to do is the full computation to show that symmetry holds for this proposed metric, because each step really relies on the fact that we are working in a normed vector space. So I don't want to rush through this. We have that—let me just start here—the distance from W to V. I want to show that's the same thing
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as the distance from V to W to establish symmetry. By definition, this is the norm of the difference W minus V. Yes, I do just want to flip these, but again, I want to show why I'm able to do that, and it's going to rely on the fact that vector addition is nice in the vector space and the norm has multiplicativity. Okay, what I'm going to do is take this expression and rewrite it with a negative 1 kind of factored
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out front. That's the scalar multiplication in the vector space. So let me write -1 * W + V, and that's just relying on the fact that scalar multiplication works well in a vector space. Now I can use multiplicativity of the norm to bring this negative one in front and take its absolute value. Okay, so that's multiplicativity of the norm. And then here, what I'm left with is -W + V. I'm going to use commutativity of vector addition in the
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vector space to rewrite that as V minus W. Absolute value of negative 1 is one, so we get V minus W. But then, of course, that is the distance from V back to W. Alright, so again, it's tempting to go straight from here to here, but I think to make it clear why that's allowed, you want to do this procedure in here where you highlight things like scalar multiplicativity in a vector space works well, vector addition in the vector space is commutative, the norm has multiplicativity,
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that's where we're really getting the property. So my ability to bring this negative one in front right here is following from the multiplicativity of the norm. Okay, so essentially what we've done is prove that you can do what feels like you should be able to do. So you can swap the order in here because we've proved that that construction works. Okay, let's resume this in a moment with the triangle inequality. Alright, the triangle inequality is actually fast because we're
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working with a norm that has the triangle inequality. So lastly, notice that the distance from V to W is going to be here. I'm going to bring in U. So how about V minus U plus U minus W? The norm has the triangle inequality. That was the fourth property that we covered for norms. So this is less than or equal to V minus U plus U minus W. But now that is the distance from V to U plus the distance from U to W. And that wraps it up.
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So I hope you enjoyed seeing how the four properties of a metric followed from the four properties of the norm. So positivity followed from positivity, non-degeneracy from non-degeneracy, symmetry from multiplicativity, and then here we have the triangle inequality for the metric follows from the triangle inequality for the norm. Summarizing across a few different lectures, we can say the inner product induces a norm and a
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norm induces a metric. But don't try to go the other direction. For example, we take the metric space that we introduced earlier, which consists of my markers with the discrete metric. This is a metric space with the discrete metric, but I'm not going to be able to induce a norm here. This is not going to form a normed vector space. I don't have an origin, I don't have vector addition, I don't have scalar multiplication—struggling to say that word. It's not going to be a vector
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space. I'm also not going to get an inner product here. So the implication there matters: inner product induces norm, norm induces metric. The other remark I want to make before we wrap this up is that while I'm saying that on a normed vector space, this is a valid metric, you don't need to check the four properties. If you have this, you know it's a metric. It's not the only metric,
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so it's not like this is a unique choice to have a metric. In particular, on a normed vector space, I could work with the discrete metric. That would also be a valid choice of a metric. So we're just saying that this is one particular metric that works. This is the induced metric. Alright, that wraps up this lesson on metric spaces. Thank you for your attention.
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