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0:00
In this video, we are going to introduce our study of continuous functions from one metric space to another. The first thing we are going to do is talk through some terminology. In particular, I want to identify what this notation will mean for us. Then, we will introduce the definition of a functional limit—what does it mean to say the limit of f(x) as x approaches a, etc. We will use that definition to define continuity for such a function. Then, we will do an example proving
0:30
that a vector-valued function of two variables, so a function from R2 to R2, is continuous on its domain. Finally, we will wrap up with a brief look at something called the topological definition of continuity, which we will revisit when we discuss the four characterizations of continuity. Let’s start with the notation, which I will use throughout the rest of our study of functions. The idea here is that I’ve tried to write what it means to have a function from one metric space to
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another as broadly as possible. So here, f is a map from some domain called A (this is our domain) which is a subset of a metric space M. To this metric space, I may associate in a traditional way some metric d. We map potentially to a different metric space, which I’ll typically denote by N, and to this metric space, we may associate the metric ρ.
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Now, it’s possible that our function is just a map from R to R, so just a regular function of the form f(x), but we could look at some more interesting examples than that. Let me talk through domain considerations. The domain A here is the set of points where we will consider our function as defined. It could be a proper subset of the larger metric space M, but even if my function has a natural extension to the rest of M, if we write this notation,
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we are really just looking at how f behaves on A. For example, if I have something like f(x) = x^2, this is just a function from R to R, but if I say that f is a map on the closed bounded interval from 0 to 1 like this, then I really mean that I’m only considering this function on this domain. Moreover, if you write something like this, your function should exist on this domain. This domain would not be appropriate to write for something like g(x) = 1/x because if I was
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trying to write this function and I gave you this domain, that wouldn’t make sense, since the function is not defined at 0. So, the domain of the function is where we will be considering our function, and your function should exist on that domain. If I say f is a map on A, take it for granted that f is defined on all inputs from A. Now, looking at this example I’ve written down, the domain is from 0 to 1. Where are we mapping
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to? This is called the codomain. When we say codomain, we are not suggesting that the function lands on every value in the codomain. Sometimes, I think of this as a larger target space that we are mapping into. To return to this example, the codomain is R because all of my outputs here from the function are real numbers, so f sends the closed bounded interval from [0, 1] to R, but it doesn’t land on every real number. The true set of outputs for my function is not the entire real
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number line; the codomain is this larger target space. The true set of outputs is the image of f, sometimes called the range, and that is denoted f(A). This is like f(x) for all x from the domain. Sorry, my curly bracket looks terrible—so this is the true set of outputs, the true range, the image of our function. It’s where we actually land, and it’s often a proper subset of the codomain, which is just the larger general ambient metric space that we are mapping to.
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There is one more notion that I’d like to define—another piece of terminology—before we move on to the definition of a functional limit, and that is the idea of a preimage. Let me see if I can squeeze it down here. The preimage of a set in the—actually, I’m changing my mind, I’m going to erase this just to give myself a little more room, so let me come back in a moment and I’ll talk through the preimage before we move on to limits and continuity.
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Okay, I wanted to be a little more careful with my definition of a preimage because the notation sometimes confuses students. A preimage of a subset of the codomain—so, I’ll say preimage of some set, which is a subset of the codomain—is defined or written this way: it looks like f-inverse, but this is not an inverse function. This sets up a set; the preimage of S is defined to be all the points from the domain that get mapped to S. So,
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the preimage of a subset of the codomain is like going back to the domain and finding all of the points that land in S. So, this is a subset of A, and it could be empty. Again, let’s take our squaring function f(x) = x^2, and this time I’ll make the domain from -10 to 10. Here, my function is a map from the domain [-10, 10] to the codomain R. If I want to compute the preimage of 4, then I’m saying, "What are all
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the points in the domain that get mapped to 4 in the codomain?" So, what is the preimage of this potential output? It’s the two-element set {-2, 2}. If I say, "What is the preimage of [100, ∞)?"—so, what are all the inputs that map to the closed interval [100, ∞)?—that is also a two-element set: {-10, 10}. The fact that most of the elements in this set don’t get mapped to doesn’t matter; it’s okay. For example, if I take the preimage of—how about 1,000—or
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let me just do, how about [-5, 3]? Okay, so I just changed my mind. Imagine this open interval from [-5, 3]. What points from the domain get mapped to these values? The answer is none of them. Okay, so hopefully that was a good illustration of what the preimage of a subset of the codomain looks like. All right, with this function terminology, we’re ready to go. In a moment, I will come back, and first, we will talk through the idea of a functional limit,
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and then, with that, we will build up our idea of continuity. Here’s the definition of a functional limit, so explaining what it means to take the limit of a function as points in its domain trend to a certain place. Let me talk through this definition, and then I would like for you to pause and compare it to any prior definition of functional limits that you may have seen, for example in real analysis 1.
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So, suppose that f is a map from a domain A, which is a subset of some metric space M, to another metric space, the codomain N. Hiding in the background are the metrics d and ρ for M and N, respectively. Then, let C in the metric space M be an accumulation point of the set A. We say that the limit as x approaches the accumulation point C (and also x is in the domain A) of f(x) equals L (here, L is an element in the codomain N) if,
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for every choice of ε greater than zero that we make, there exists a δ greater than zero such that whenever the distance in the original metric space M between x and C is between zero and δ, it follows that the distance in the codomain metric space N between f(x) and L is less than ε. All right, so what we are defining here is essentially this notation. What does it mean to say that the limit of this function as x approaches C is L? Go ahead and pause,
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like I said, and if you have the ability to quickly look up the definition of a functional limit that you met in real analysis 1, it might be nice to compare these. It should be just about the same, except we could be generalizing the concept that you studied for functions from R to R to general maps from one metric space to another. All right, let’s talk through this definition a little more. I think the first thing worth
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addressing is where C lives. C is this point where we are trending towards with our function. C does not necessarily belong to the domain of the function. You don’t need your function to exist—you don’t need to be able to compute f(C)—in order to potentially have a limit as inputs from the domain approach C. However, by saying that C is an accumulation point of A, that means that we can get arbitrarily close to C with points from A, so this allows us to trend to
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C. Even if C is in the domain A, you should still have the property that it’s an accumulation point, again so that you can get close to it. So, if C is like an isolated point in the domain, there’s no point in computing a limit because you can’t approach C. This language of "accumulation point" here is saying that points from our domain are getting closer and closer to C, and we are trying to see if we can ascertain where the functions are trending to without
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knowing if the function is defined at C, and also not caring what it would be at C. Okay, I jumped ahead right there, but let me go ahead and address what I just said. If you look here, we have this condition that the distance between points x from the domain and C, where we are computing the limit, has to be positive. That’s to say that we don’t ever evaluate the function at C in this definition. We only look at what’s happening near C when we compute a limit.
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Let’s see—so, we’ve got x approaching the accumulation point C. Of course, it’s always in the domain. I will sometimes omit that remark, that x is in A, and kind of take it as implicit that if I’m computing f(x) values, then I must be working in the domain. We will do a proof using this ε-δ language in a moment, but if you’ve written ε-δ proofs, the flow here is the same as what you’ve seen before. Logically, what we imagine is that you choose some small
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number ε first. So, we say, “Okay, imagine ε is greater than zero,” and that might be all that we say about it: let ε be greater than zero. But in our minds, we are imagining that this number ε here is really small—it’s like a very small positive number. With that choice of ε, we then figure out a δ that works. So, this δ here may depend on ε and also often depends on C.
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Okay, so it’s like you’re working towards some point C. You imagine a choice of a positive number ε, and it’s helpful to imagine that ε is very small. Then you figure out, okay, with this desired ε, and for the point C where we’re trending, is there a δ such that for any x, which is not C itself but which is within distance δ of C, can I determine that the value of f(x) is within distance ρ (so, distance in the codomain) of L, the potential limit?
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Again, hopefully, if you’ve studied the ε-δ definition of functional limit before, even if it was just in the context of real analysis 1 for a function from Euclidean space back to Euclidean space (from R to R), this should mirror the thought process that you would have seen in that original definition. A major difference might be this characterization of accumulation points and also using d and ρ, whereas you may have
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had absolute value expressions. So, if you saw in the past something like 0 < |x - C| < delta, then we have |f(x) - L| < epsilon. This is just the statement above but for the specific case of working with the Euclidean metric for both the domain and the codomain. Okay, I am going to step aside for a second, and then I want to erase this so I can put up here the definition of continuity for a function at a point in its domain. We just make a few small
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adjustments to our first definition in order to bring in the idea of being continuous at a point. So, this is the definition of basic continuity for a function at a point in its domain. I’ll talk through this definition and come back to make a couple of remarks about it. Then I will sketch a picture, which I feel is missing from this conversation. We say that f is continuous at a point C in its domain A if one of two things is true. First,
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either C is not an accumulation point, so it’s an isolated point in the domain, and then you’re done. You just stop your assessment right there. Functions are always continuous at their isolated points. That’s not usually what we’re looking at with functional behavior, so the second condition is probably what you’re expecting, and that is that the limit as x approaches C of f(x) exists, meaning this condition is satisfied, and it’s equal to the value the function takes at C. So,
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the limit exists, f(C) exists, and they are equal to each other. Okay, take a moment to kind of read these together and put them in conversation with each other. Then I’ll come back, say a couple of things about this, and then sketch a picture for a function from R to R capturing maybe every little nuance for these two definitions. All right, regular continuity is a point-based definition,
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so notice that we are defining f as continuous at a point in its domain. This is plain, normal continuity. We often say f is continuous, and what we mean by that is that f is continuous on its domain. What we mean by that is that f satisfies this condition for every point in its domain. So, even when we say, “Oh, f is continuous on this closed bounded interval,” we’re saying that f satisfies the point-based definition of continuity for every point in its domain.
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This idea that functions are automatically continuous at their isolated points might be new to you, and it almost feels like a choice that was made. For this example that I came up with, here the domain is the isolated point 1 together with the interval [2, 4], where 2 is in the domain but 4 is not. So why say that the function is continuous at 1? You could try to recreate calculus so that this is not the case, but it will turn out that this isn’t favorable.
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It’s kind of like this choice fits in with everything else about continuity. You don’t have to go into exceptions and special cases when you say that functions are automatically continuous at their isolated points. So, here looking at this graph, our function is continuous at this point in its domain because it’s an isolated point, and that’s the only analysis I need to do at this location in the domain. You could think of it in terms of neighborhoods,
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as in, given this output value, if I want to be near it and near this point in the domain, really, I’m just at that point. So, it’s kind of like you can define a neighborhood around this point where all inputs from the domain in that neighborhood get sent close to this because there’s only one point in the neighborhood. So, if you think in terms of neighborhoods, it also makes sense to decide that functions are continuous at their isolated points. So,
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1 is just a standard that we will always adhere to, and then 2 is like saying, “Do all of this.” Here, the point C is an accumulation point, but it’s also in the domain, so we can actually evaluate what this limit would have to be—it should be f(C) in order for our function to be continuous. So, looking at this picture, there are really three points of interest, maybe four. This point, we’ve already covered that here. If you look at this point in the domain, the functional
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limit exists. So, this point in the domain is an accumulation point; we can approach it, and as we do, as we approach this point, we are tending to this y-value. So, whatever this value is, that is the limit of the function as points in the domain get closer and closer to this point. However, this is not the value of the function at the point, so this function is not continuous right here.
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Here, 4 is not in the domain, so our function is not continuous at 4. It’s not an isolated point; it’s not an accumulation point; it’s not in the domain. It doesn’t make sense to assess continuity outside of the domain, so functions can only be continuous or not continuous in their domain. However, 4 is an accumulation point of the domain, and as we approach 4 from the left, our output values are tending to something, so the limit as x approaches 4 in the domain does exist.
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Maybe the other point of interest would be here. This is just an endpoint—sometimes people wonder if there’s something special about endpoints. There isn’t really because we’re only going to approach in the domain. So here, the only way to approach this point is through the domain from the right, and this function is continuous at this point. So here it’s continuous, here it’s continuous, it’s continuous here, and it’s continuous here. It’s not continuous
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here—that’s the only point in its domain where it is not continuous. And here, it doesn’t make sense to talk about continuity because this isn’t in the domain; however, the functional limit exists. Okay, I think I’ve said enough about these two definitions. Let’s do an example.
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et’s work through a specific example showing that a function is continuous. So, this is not a very complicated function. It is a function of two variables. So here, f is defined as having a vector output, which is (2x, y + 1). So, f(x, y) = (2x, y + 1). This is a function from R2 to R2. So, the domain is all of R2, the codomain is R2. I just want to introduce a little bit of terminology into this. Because the output is a
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vector in R2, we can call this a vector-valued function versus a scalar-valued function. So, if it was a map from R2 to R, then you could say it’s a scalar-valued function. And then, I’m not done writing this, but this is a function of two variables.
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Okay, so it’s a multivariable function. It has x and y as variables, and it maps us to vector quantities. I’m going to prove that this is continuous on all of R2, meaning at any unspecified point (a, b) in the domain, we will prove continuity. I will take you through the scrap work, but I’m going to do it kind of quickly because I assume that this isn’t your first time seeing the mechanics of an epsilon-delta proof. So,
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if you have never done this kind of proof before, I might lose you, in which case I encourage you to revisit epsilon-delta proofs in, say, a real analysis 1 text. Alright, so here’s how I do scrap. And I want to emphasize here that this is what I do on side paper. This is like my work kind of outside of formal writing, so I’m not going to have much sentence structure here or anything like that. This is what I do in order to get to the proof.
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Okay, so I’m going to say, "Well, what does f(x, y) - f(a, b) even look like?" Sorry, I think I said that wrong. What does f(x, y) - f(a, b) look like? And in reality, I probably wouldn’t even write this. I would just jump straight to (2x, y + 1) - (2a, b + 1). The marker is a bit squeaky.
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This is a vector in R2 and another vector in R2. We are computing the norm of their difference, or in other words, we’re computing their Euclidean distance. So that’s going to be the square root of the difference of these squared, plus the difference in y squared. Okay, I’m going to pull the 2 out and square it, so that’s going to be 4 * (x - a)^2, and then here, cancel out the 1 and negative 1, and we’re left with (y - b)^2.
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Do not expand these terms because x is going to get close to a, y is going to get close to b. You would like to see, in these kinds of computations, expressions of the form (x - a) and (y - b). So, these kinds of pieces are actually what you want to see. Don’t expand them. Don’t write x^2 - 2ax + a^2—that would not be productive. Okay, in fact, if you look at this, it almost looks like the distance between just the point (x,
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y) and just the point (a, b). So, if I just computed the Euclidean distance from the point (x, y) to the point (a, b), it would be this expression, if I didn’t have this 4 here. With these kinds of computations, you only want to go up in size. So, I do not want to just take this 4 away because that would actually make me go down in size, and you’re trying to bound things above, above, above. So,
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what I’m going to do is not remove the 4. I’m going to bring a 4 in here and go up in size. Now, since both of these terms have a 4 in common, I can factor the 4 out and take its square root, like that. And as I was saying a moment ago, this square root quantity is exactly the Euclidean distance from the point (x, y) to the point (a, b). If I had chosen (x, y) to be within distance delta of (a, b), so that this square root is less than delta,
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then in my computation—I’m going to do this to the right here—I can now say this is less than 2delta. So, if I’m at some random point (a, b) in the domain, and I say (x, y) is within distance delta of that, we can now say that the Euclidean distance between their respective outputs is less than twice that bound on the input distance—so less than 2delta. But what I would like for it to be is less than epsilon, and this is how you get
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the relationship between delta and epsilon. You take this expression, and you get to some point where you’ve brought in your delta. So, it’s great if you can see exactly this as we did, and that kind of like—the fact that I got exactly 2delta is because these pieces are so nice. So, I was working with nice degree-1 polynomials in each coordinate. Okay, so, since I got 2delta here, got this nice expression in terms of delta,
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then I say, "Well, what if that was less than epsilon? What would that tell me about how delta has to relate to epsilon?" Because remember, in our logical procedure, we imagine epsilon is being chosen, and then delta is computed in terms of epsilon, and potentially also in terms of (a, b). Here, we don’t need to do that. We can just say, "Well, whatever (a, b) is, actually all we need is for delta to be less than epsilon over 2."
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Okay, this is scrap work. I’m going to step aside and erase this, and then I’ll come back and rewrite the proof—meaning, I’ll present the proof as it should be written—and then we will have almost exactly the same computation. I’m tempted not to erase it, but I will. So, I’m going to do the proof from scratch with the knowledge that once I’ve chosen epsilon in my proof, I should then declare delta to satisfy this
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relationship. Before I step away, let me revisit something I brought in a minute ago but didn’t go into too much detail on. Epsilon is fixed—that’s the first thing you pick. Then you do this kind of computation, and you get some relationship with delta and epsilon. The thing I mentioned but didn’t elaborate on is that delta can also depend on (a, b), but delta cannot depend on (x, y). So, it’s okay if delta depends on epsilon—in fact,
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that almost always happens. And it’s okay if delta depends on the anchor point where you are assessing continuity, because we imagine that (a, b) is fixed, and I am determining if my function is continuous at a fixed point in its domain. (x, y) are chosen after delta. So, epsilon first, then delta, then (x, y). So, you can never have delta here depend on (x, y)—the points that are approaching the place in the domain where you are looking for continuity.
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Okay, I think I threw a lot at you right now, so let me step aside, and then I will write this proof, and we will almost be done with this first discussion about continuity. Okay, let’s write the proof. I think that this kind of proof, when you are showing a definition is being met, does not need to be excessively flowery or ornate, so I’m not going to have a huge introduction and a huge conclusion. I’ll get right to the point.
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Okay, the first thing I’ll do is fix where we are computing our computation, where we are assessing continuity in the domain. So, fix (a, b) in R2. That’s a short sentence, so let me continue with "and let epsilon be greater than zero." It’s very important that you name this early in this kind of proof. Don’t start talking about epsilon before you’ve fixed it. So, at this moment, even
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though I didn’t give either of these values, they are not moving around—they are fixed. So, this is a point in R2, and this is now a fixed number. It will not change from this moment forwards. Choose delta in (0, epsilon / 2), that open interval. This is my style of choosing delta, so I want delta to be greater than zero and less than or equal to epsilon over 2. Any delta in this interval will work. So, this is what I do. Sometimes,
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other people who like to do analysis for fun will say, "Choose delta equal to epsilon over 2." They will have a specific choice of delta, and that’s fine too—that’s just a different style. However, even though my choice of delta is unspecified, it will work if it’s anything between zero and epsilon over 2. From this moment forwards, delta is now fixed. Okay, so delta has been chosen. Now, whenever the Euclidean distance from (x, y) to (a, b) is less than delta, it follows
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that the Euclidean distance between their outputs—and this is going to look exactly like my scrap computation—just save room, I’m going to jump straight to the square root. Hope that’s okay. But if I had a little bit more room, I would write out exactly every step from the scrap. I’m just going to go straight to sqrt((2(x - a))^2 + (y - b)^2). Okay, and this is equal to sqrt(4(x - a)^2 + (y - b)^2), and that is less than or equal to...
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I hate to do this, but I’m going to again skip another line, just in the interest of space, but it should follow exactly the scrap work. I regret it, but I’m going to go from here to bringing in that 4 to now saying 2 times the Euclidean distance from (x, y) to (a, b). Okay, in general, I don’t like to skip too many algebraic steps as I work through this kind of computation. It is followable from here to here to here,
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but if my board were like two feet longer, I would put in everything we did in the scrap. This is less than delta. So, we’re assuming that this is true and showing what would follow as a consequence. So, this is true, so this is less than 2delta, but my choice of delta is less than epsilon / 2, and here I’m just going to write all to the right that that is epsilon, keeping in mind that I would normally like to put that down, but I’m out of room.
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So, as I said, I don’t think that these proofs should be too wordy. They should be complete sentences, they should show all the work, and they should have every logical necessity in the definition—meaning, you know, a point where you are assessing continuity, an epsilon which is fixed at a positive number, a specific choice of delta, the precision that we are choosing now to work with points from the domain within distance delta of our anchor point. You need that.
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Do not skip any of these clauses, like especially this one, but don’t skip them. Put them all there because we are showing the definition as being satisfied, and you really need your proof to follow the logical train of thought from the definition, but you don’t need much more. So, you know, I said everything I needed to, and I didn’t say much more—and arguably, I didn’t quite say enough right here. So, once I get to the end and I’m saying, "Hey,
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look, if the points in the domain are within distance delta of each other, the outputs in the codomain are within distance epsilon," I usually just wrap up by saying, "as desired," meaning that we have now satisfied the definition as it’s stated. So, it’s not a mystery now what the conclusion is. Okay, so this function is continuous at any point in R2. So, probably, if I had a little bit more room, I would say, "Therefore, this function is
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continuous on R2," because this is just a random unspecified point. I mean, it’s specific here, but we’ve now shown any point in R2 is a point of continuity for this function. This is, again, I hope, not your first epsilon-delta proof because if it was, I think you should probably look at a few more examples. I will also mention that this function, as I alluded to when we were working through the scrap work, has polynomial coordinates. In fact,
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degree-1 polynomial coordinates, which meant that the algebra simplified really nicely. Whenever you’re working with a polynomial expression and you’re doing this kind of continuity computation, you will see these kinds of pieces emerge. There’s a way to get to something very clean at the end. I also maybe want to mention that there are results like, if a vector-valued function is continuous in its coordinates, then it’s continuous. So,
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if I looked at this function as somebody who’s been doing this for a while, I wouldn’t need to do this work to know that it was continuous because I would say, "Okay, 2x is a polynomial—that’s continuous. y + 1 is a polynomial—that’s continuous." So, this is really just an example of what it looks like to verify that the definition of continuity is being met. Okay, we will have a couple more examples of this down the road. So,
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what I would like to do now is step aside so that I can erase this, and then I want to wrap up our introduction of continuity with something that may be brand new. It might not remind you of anything from real analysis 1, and that is the topological definition of continuity. Let's wrap up our first look at continuity with a brief discussion about the topological definition of continuity, which is phrased in terms of open sets. Okay, I just want to present this to you
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and give you the main idea. We will revisit this notion when we look at the four characterizations of continuity. So what this definition says is that f is continuous on its domain if, for every open subset of the co-domain (so here U is an open set in the co-domain), when you look at the pre-image (so remember that idea of figuring out all of the points in the domain that get mapped to U—that's the pre-image of U), that is open relative to A. So this idea of being "open
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relative to" is what I want to give you a working understanding of for the rest of this video. So, I have two approaches to explaining this, and I'll start with this kind of visual approach, where we imagine that A (so here are two functions over basically the same domain, A, to this closed bounded interval) and they are mapping to R. So, the co-domain, like the y-axis here, is R. This one is continuous; this one is not. Okay, my first attempt at looking at the idea of
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computing a pre-image and assessing whether it's open relative to the domain is to imagine that the domain is actually the entire metric space. So, normally for these closed bounded intervals, you would imagine them as subsets of R, but I haven't sketched R here. So notice there's no x-axis, there's just the domain. Okay, so if our function is not a map from a proper subset of a metric
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space to another one but a map on, say, the entire metric space, so here imagine this is the entire metric space we're mapping from, what does it look like to compute the pre-image of an open set? Okay, let's start with this example here on the right, where the function is continuous. Let me sketch a couple of open U's and let you take a moment to try to visualize, just on this domain, what the pre-image of each U looks like. So, here's U1 to that little open interval,
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and then I want to take you to—this is going to be a little bit annoying, but let me do—make a little adjustment here, not sure I'm going to be able to fix it. Okay, there we go. So, U2 is a little open interval here, and then, being a little bit annoying, this might be tricky, but let's do U3 like this. Okay, U3 might be a bit complicated, but especially focus on U1 and U2. Take a moment to visualize their pre-images here in the domain, and keep in mind that the domain is
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really like the entire metric space, so this is not a subset of anything, like a proper subset. Okay, I think U1 is the easiest, so we'll start there. Who maps to U1? Nobody. So, the pre-image (let's call this function f) of U1 is empty, which is open. So that's fine—the pre-image of U1 is an open set. What about U2? Well, nobody maps here, and then this endpoint maps to U2 up to about here. So, I changed this function so that we really just have to focus on this
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left side. So if I map backwards, the pre-image includes this left point and goes to about there.
40:28
I'll say that this interval is A, B, so that the pre-image of U2 is A, and then let's call this point C, like that. And then U3—let me just sketch it on the picture—I'll do it in pink. Okay, so let's see who maps to U3—we get that like that. Okay, actually, not that bad. So let's see, it's from about here to there, together with there to about there, so the union of these two open intervals is the pre-image of U3.
41:18
Okay, I think before we go forwards with the definition, let me give you just one example over on the left. So, over here to the left, figure out what the pre-image of this neighborhood U4 looks like. All right, for U4 over here on the left, see from there, and then this bottom half of U4 is empty because of the jump discontinuity, so we hit it with this point, then nothing to the right. So we have this kind of interval.
41:56
Okay, if the domain is the entire metric space, then you would say: Is the pre-image of an open set open? So, if this is the entire metric space and I don't acknowledge the existence of any numbers to the left or right of this interval, are these pre-images open? Well, this one is, because this is just an open set. So if it's open, we're good. Likewise, this union of two open intervals is also open, so this one's good. You might hesitate for a moment looking at the pre-image
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of U2 because it doesn't look like a traditionally open set, but if you do not acknowledge anything to the left of A, so that this is the full metric space, then A is an interior point of this pre-image. Any neighborhood around A would only go to the right because there is no such thing as going to the left. So, A, while it looks like it's on the edge, it's still an interior point of the set, which is the pre-image of U2. So this is open in this metric space.
43:07
On the other hand, if I look here—then, you know, if my closed bounded interval from A to B is the metric space—and I haven't said it yet, but I probably should have: we're working with the Euclidean metric here, so imagining the Euclidean real number line but I'm chopping off everything that's outside of these two domains—if I look here, this is not open relative to this metric space, and it's not open in a very traditional way. If I take this edge right here,
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it's not an interior point, because seeing as it's here inside of the metric space, not like A, which is kind of special, a neighborhood around it would travel to the right and pick up points that do not belong to this open interval. It's like now I have to go back and kind of adjust, so... Like, think of A as the entire metric space. If A were the entire metric space, would the pre-image of any open set be open? This is one way to think about it. So,
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you know, closed bounded intervals may be a subset of the real number line, but if you just focus on the closed bounded interval and that was a metric space, would this be open in that metric space? And the answer is yes. So, this is open relative to its
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set. That's kind of how I like to think about it the most. But let me give you another working definition, and one that's actually quite useful because it gives you a characterization that you can write down, and that is: a pre-image is open relative to the domain if it can be written as the intersection of an open set and the domain. I think I'll just say that out loud and maybe add it here. So, if I take each of these pre-images,
44:54
I could write them as the intersection of an open set in R—sorry, let me say that better—okay, if I now say this domain is a proper subset of a larger metric space (maybe I'm treating the closed bounded interval from A to B as a subset of the Euclidean line), so this is kind of the more normal visualization of such a function, in this situation this set no longer looks open because there's a larger metric space outside of our domain. However,
45:27
the definition of "open relative to A" is that we can take the pre-image and write it (in any case for a continuous function) as the intersection of a traditionally open set and the domain. Here that's easy. I would just write it as the empty set intersect A. Here, I would just write that as this open set intersect A. So, let me call this W1 and W2. So, if it's already open, it's no problem—just write it as itself intersect A. And in this kind of special situation, where it's open
46:04
relative to the domain as a metric space, but when I embed it in the larger domain it doesn't look open, I can write it as an open interval intersect A. Let's go from 0 to C in the picture. Okay, so the topological definition of continuity is that the pre-image of any open subset of the co-domain is open relative to the domain. The two ways you can think about that are: If the domain was the whole starting metric space,
46:35
it would just be open, or, if the domain is a proper subset, then any pre-image can be written as the intersection of an open set and the domain. So, here this is from 0 to C. All right, that's just a look. I just wanted to put this out here so that when I bring this back up when we talk about the four characterizations of continuity, you know what it is, because I think if you've never seen this before, it's a little bit
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unexpected at first glance. So, when we look at the four characterizations of continuity, I didn't want you to be studying those four characterizations and learning this at the same time. So, you've seen this, you have some time to think about it, and then we will talk about the four main ways to characterize a continuous function. Two of them we've now seen in this video. One is the Epsilon-Delta definition; it's just your
47:22
traditional analysis definition of continuity. Another is this one. And then we will have two others—one with sequences and one which is actually a lot like this one but with closed sets. Okay, let's wrap up this discussion here. We have a lot more to talk about with continuous functions. So, thank you for your attention.
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