 Welcome back to our lecture series, Math 3120, Transition to Advanced Mathematics for Students at Southern Utah University. As usual, be your professor today, Dr. Andrew Missildine. So previously, as we've been talking about functions and particularly about bijections, we brought up the pigeonhole principle. And the pigeonhole principle told us in the language of functions that there cannot exist a bijection between two functions A and B when their cardinalities are different. In particular, if there is a bijection, the cardinalities have to be the same. And therefore, a bijection doesn't exist when their cardinalities are different. And so the question I posed in this video is this process reversible? That is, does there always exist a bijection between two sets of equal cardinality? Well, the short answer is yes, but we cannot explain why that is until we have a proper definition of cardinality. As the current notion of cardinality that we've been using has actually been insufficient. So that leads to our first definition here in this lecture. Suppose we have two sets A and B. We say that A and B have the same cardinality denoted the cardinality of A equals the cardinality of B if there exists a function, a bijection, F that maps from A to B, like so. So I should mention, of course, that the way we've defined this term same cardinality, we've actually defined it so that, yeah, exactly when the two sets have the same cardinality, there's a bijection. So that does answer the question, but sort of like in a cheap shot way there, there exists a bijection because that's the definition of cardinality, all right? But it turns out this notion of cardinality as we just defined is not alien to how we've been approaching things. We say that the cardinality of A is the same as a natural number. That is, we say the cardinality of A is a natural number. Then if there exists a bijection from the set A to the set zero, one, two, three, four, five, up to N minus one. Now you'll notice that in this elementary notion of cardinality we've been using this entire lecture series that this set contains exactly, it contains exactly N elements, right? And so we would have said, oh, this set has N elements inside of it. Now, since there is a bijection from A to the set, they have to have the same cardinality. So we likewise say that A has N elements. If such a bijection between A and this set exists, we say that A is a finite set, okay? Whose cardinality, of course, is this. Now, if A is not a finite set, that is there exists no bijection from A to a set of this form, then we say that A is an infinite set. And we'll denote this by saying that its cardinality is infinity, okay? And so this recaptures the notion of cardinality that we've been using throughout this lecture series. As we've talked about cardinality, we've mostly been fixating on these finite sets, that is sets that can be put in bijection with a set of this form right here. And as we work through combinatorial exercises, which combinatorics is all about computing the cardinality of finite sets, that intuitive notion of cardinality, what didn't disadvantage us whatsoever. We didn't have any notion of function to introduce yet, but that's okay. So we can recapture the notions of cardinality that we were interested in for finite sets, but this then empowers us to be able to talk about cardinalities of infinite sets that we plan to do in this lecture and the remaining lectures for this lecture series here. Now, it might seem obvious, but it's still worth noting here that this notion of cardinality we've just introduced is in fact an equivalence relation. So let's take the relation A is related to B if and only if they have the same cardinality. This is an equivalence relation, okay? So there's three things we need to show. We need to show that this is a reflexive relation, a symmetric relation, and a transitive relation, all right? So let's do reflexivity. So given a set, we wanna show that A is related to A, which would mean that A has the same cardinality of A, which would mean there is a function from A to A. The identity function that we've introduced previously, this is the map, of course, that just takes everything in A and spits back out A, right? This is a bijection from A to A, and therefore A has the same cardinality as A and this shows that the relation is reflexive. Next, we wanna show the symmetric property. So what we wanna show is that if there are two sets, so if there is two sets A and B such that A is related to B, then we wanna show that B is related to A. This is a conditional statement. In order to do that, we will assume the premise and then derive an argument to prove the conclusion there. So for two sets A and B, suppose they're related. That is, suppose they have the same cardinality. By definition of cardinality, there exists a bijection F from A to B. Now this is directional. Functions have directions, and even though we're using an equal sign in here, the relation, we haven't yet proven to be symmetrical. So therefore we need to make sure we go in the right direction. It goes from A to B. That was the definition there. Now since F is a bijection, we've proven that bijections always have inverses, functional inverses there. This is a bijective function that reverses the order of the relation F. It'll now go from B to A. F inverse is a bijection. Therefore, since there exists a bijection from B to A, B has the same cardinality as A and therefore this relation is symmetric. And so the last one we wanna argue is about transitivity. Suppose we have three sets A, B, C, such that A has the same cardinality as B and B has the same cardinality as C. Just because we're using an equal sign here, we should not assume that it's an equal sign. I mean, that's just the symbol we're saying here. Like we've introduced notation. We say these cardinalities are equal because what is a cardinality after all? It's the existence of a bijection. It's curious because the way we've defined cardinality, previously we're like, oh, a set has cardinality. It was a quality of the set, but now the way we're talking about cardinality, cardinality is this relation. It connects two sets together. They have the same cardinality. Now for finite sets, as we mentioned above, if you can connect, if you can relate a function to a specific finite set, then you can talk about its cardinality individually, but not really because that set was deeply connected to the natural number. These equal signs is a notation we introduced, but really it's a relation, right? And we don't necessarily even know what this thing is yet until this theorem is proven, all right? So A has the same cardinality of B. That means there is a bijection F from A to B. B has the same cardinality as C. That means there is a bijection G from B to C. We can compose these two functions together, specifically G composed with F will be a function from A to C. Since these functions are both one to one, so will the composite. Since both of these functions are on two, so will the composite. This is stuff we've shown previously. And therefore the composite of two bijections is a self-unbijection, and this shows that A has the same cardinality of C. This then shows the transitive property, and therefore same cardinality, the relation there, does in fact give us an equivalence relation. And like I mentioned earlier, this recaptures all of our intuitive notion about finite cardinalities that we dealt with when we talked about combinatorial problems previously, but now it empowers us to talk about infinite sets. And so that's what we wanna focus on now. What type of equivalence classes exist among infinite sets? Because each of the finite sets is gonna form its own equivalence class relative to a natural number. The reason we defined finite sets the way we did is that the finite set, each natural number distinguishes a, it distinguishes a equivalence class. It's the representative here. And I should say zero is included in that as well, that the class that's associated to zero, and of course it's just gonna be the empty set. There's only one set whose cardinality is zero, and that's the empty set. For every positive integer, we get things like we described earlier, and those then give us like the canonical representative of that. We can associate to every finite class there, that is the class of finite cardinalities, to a natural number. For infinite numbers, what do you do? Who's the representative? Well, we won't get into that. That leads into conversation about ordinals and cardinal numbers that are beyond the scope of this lecture series, but we can talk about classes of infinite sets with respect to their cardinality. So let's start the natural place to start is with the natural numbers. Who has the same cardinality as the natural numbers? Well, so this first argument we're gonna show here is that the set of natural numbers has the same cardinality as the set of positive integers. Recall that we denote the natural numbers using this blackboard N. N includes zero and all of the positive integers. And then this set here, Z plus we denote the positive integers. So one, two, three, four, zero is not included there. Notice in this statement here that the set of positive integers is in fact a proper subset of the natural numbers because the natural numbers do contain zero and it's not contained here, but every positive integer is contained inside the natural numbers. These are two infinite sets where the positive integers is a proper subset of the natural numbers, but they do have still have the same cardinality. This is a curious fact about infinite sets that a proper subset can actually have the same cardinality. And I wanna make mention that in some and actually many set theory texts, the way they define infinite versus finite, they actually define infinite to be a set which has a bijection with a proper subset. That's, if such a thing exists, you call that an infinite set. And if you're not infinite, you're finite. And that's a little bit different than how we just defined it. And the reason they do it that way is that in those set theory courses, they want to define the natural numbers. And in order to have a set theoretic non-circular definition, you can't use the natural numbers to define finite versus infinite. So you have to define the infinite using a different approach. And that's how they take it proper subsets. If there's a bijection between a proper subset, that's an infinite set. So this type of phenomenon happens all the time. So let's see it for the natural numbers and the positives integers. In order to prove this, we have to find a bijection between the natural numbers and the positive integers. And in this setting, it's super easy. Our bijection F, which will go from the natural numbers to the positive integers, is gonna be the rule where you add one, okay? So given any natural number, plus one to it will give you a positive integer. If you have a positive integer, like two, you just add one to it, you get three. But if you have zero, you add one, you're gonna get one there. So this is in fact a function from the natural numbers to the positive integers. And there's two ways to show that a function is a bijection. One way is to go by the definition. You say that, oh, you prove that it's injective and surjective. That's the definition of bijectivity. But you can also prove that a function is bijective by producing its function inverse. Only bijective functions have inverses. And therefore, if you can come up with the inverse, that makes it bijective. Now, in this example, we probably can come up with the inverse pretty easily. Let G be a map from the positive integers to the natural numbers. And we define this by subtraction by one. And sure enough, when you compose these two functions together, if I do G before F, you're going to subtract one, then add one, the net effect as you did nothing. And likewise, if you do G composed with F, you're gonna add one first, then subtract one, the net effect is as if you did nothing. So F composed with G is the identity on the positive integers. And G composed with F is the identity on the natural numbers. This shows that this function G we just came up with here is in fact the inverse of F. Since F has an inverse, it's a bijection. And that improves the theorem. These two sets have the same cardinality. So again, when you compare this result, we just talked about next to finite sets. It's very, very, very curious because no finite set can be in bijection with a proper subset. But with infinite sets, this can happen all the time. And like I mentioned, you can actually define infinite sets by this property. Every infinite set can do that. Now, if you were like a calculus student, you might deal with some type of dark magic arithmetic like the following. Infinity plus infinity, infinity plus one is equal to infinity. Now for calculus students, they get used to this dark, dark magic right here because doing arithmetic with infinity, as we're discovering in these, in the present lecture and beyond, can be a difficult thing. But they're used to things like, oh, you take limits as X goes to infinity and you might get something like infinity plus one. They simplify that to be infinity. So a calculus student, they get used to a statement like, infinity plus one equals infinity. This kind of justifies that from set's perspective that the cardinality of an infinite set, if you add another element to it, it doesn't change the cardinality whatsoever. And in fact, one can make an induction argument that if you add any finite number to an infinite set, you'll get back the same cardinality as you did before. All right? What if you were to add an infinite amount? All right? Let's consider the set of natural numbers versus the set of integers. So we get positive, negative, and zero. This situation, the natural numbers are now the proper subset of the integers. But the gap between them is not a single number. It's not a finite number. There's infinitely many. If you think about these numbers along the x-axis, right, the natural numbers would be infinite going from zero all the way to the right, but the integers are gonna be infinite in both directions, both to the right and to the left along the natural numbers. But regardless, these two sets have the same cardinality, even though the set difference, right? So in this situation, if you take the set difference between the integers and the natural numbers, the cardinality of this set is still infinite. So even when you remove the proper subset and you still have an infinite amount, these two sets have the same cardinality. Try to breathe that in a little bit here. Now, in order to show that they have the same cardinality, you have to produce a bijection. To find a bijection between the natural numbers and the integers, consider the following map. f of n, which is a natural number, is gonna be equal, or the image is gonna be equal to negative one to the n times two n plus one minus one divided by four. Now, you might wonder like, is this even an integer in the first place? And I should mention that this function right here, it's actually left as an exercise to my students. We have a long list of homework exercises that they are allowed to do. This is there, I'm 3112 in that list there, for which my students and thus I'll leave it as an exercise to the viewer here to prove that this is in fact a bijection. But I'll do some initial calculations for you. If you take your n's and your f of n's right here, looking at some natural numbers, zero, one, two, and three, we'll just do a few here. If you plug in zero, this is gonna give you positive one. You're gonna get one here, one minus one is zero, so that then gives you a zero right here. If you plug in one, this is gonna be a negative right here. Two times one is two plus one is three times negative one is negative three, minus one is negative four, divided by four is negative one, you get like that. When you plug in two, you're gonna get positive one right there, two times two is four, plus one is five, minus one is four, divided by four is one. And as you keep on going through this, so if you plug in three, you're gonna get two. If you plug in four, you're gonna, sorry, when you plug in three, you get negative two. When you plug in four, you get positive two. If you were to plug in five, you're gonna get negative three. When you plug in six, you're gonna get three like so. And so the way this formula simplifies is that whenever you plug in an even number, you're just gonna, the top is just gonna turn into two and because after all, if you have an even number, this disappears one minus one, you just get two N divide. And since the number is even, it's divisible by four. So it does give you an integer. So you're gonna get these things like this, zero, one, two, and three, right? When it's an odd number, this is gonna become negative one. So on the top, you get negative two minus, sorry, negative two N minus two over four. Now, since N is odd, so you have this two K plus one here, you times the negative two, you're gonna get negative four K minus four over four. That's divisible by four. And so then you get the next number you want there. So this does always produce an integer. So that important, it is actually a function that's well-defined here. And you're gonna get this pattern here. Like I said, I'm gonna leave it up to the viewer here to prove that this is in fact a bijection here. And so what this statement tells us is, again, thinking of that calculus student, the integers, like it's infinite in both directions. So in some respect, we're saying that infinity plus infinity is equal to infinity. So even though there's, in some manner of speaking, a lot more integers than our natural numbers. As a set, they have the same cardinality. It's the same infinity. They're both infinite sets, but then they're the same infinity. Now, again, this might not be so surprising because I mean, students at this point might have very different perspectives. They might think this is completely insane because how can the integers have the same size as the natural numbers? Because what I'm saying is there's just as many natural numbers as there are integers, even though when you think of how they're ordered along the real line, it feels like there should be infinitely more. There is, but that's also the same number. Adding infinity didn't actually make the number get bigger. The cardinalities are the same. But on the other extreme, if you were like, well, all infinite sets are infinite in their cardinality, what's the big deal that we're gonna open up your eyes in just a moment? Okay, I wanna do one more example with the natural numbers. So in this proof, I wanna show that the set of natural numbers has the same cardinality as the set of prime numbers, okay? And to motivate this example, I want you to think about the following here. Before we come up with our bijection, think of the following function. We'll call it P from the natural numbers to the natural numbers, like so. And by definition, P sub N is gonna give us the number of primes inside the set, zero, one, two, all the way up to N minus one. And like we mentioned before, when you plug in N equals zero, this set is the empty set right here. But for positive integers, you'll get something that you see right here. So P sub N is gonna count how many primes are inside of that set. So if we do a calculation here, P of zero here, well, the empty set contains nothing, so it has no primes there. P of one, well, if you take that set, it's just the set that contains zero, that's not a prime, so you get zero. P of two, that's giving the set that contains zero and one, there's no primes there, so you get zero. If you look at P of three, well, that's the set that contains zero, one and two, two is a prime. So you do get one element there. P of four, you're gonna get two elements because two and three are now in there. P of five, you still get back two. P of six, you're actually gonna get three now because that set contains two, three and five. P of seven, we get back three. P of eight, you're gonna get back four now because you have two, three, five and seven. So this function will increment every time you find a prime in the previous spot. So like you hit eight, it goes up because the previous number, its predecessor, was a prime number, so that's how this function is evaluating here. And so why is this function of interest, what does it have to do with the problem at hand? Well, in analytic number theory, they would be interested in the following limit, take the limit as n goes to infinity here, take the ratio of P divided by, P of n divided by n in this situation. So if you take the number of primes versus the natural number, so this gives you a ratio that's a percentage, how many natural numbers up to a point r prime? Well, as you take the limit as n goes to infinity, this is gonna turn out to be zero, as in zero percent. As in when you take the set of all natural numbers, there are really no primes in that set. Again, there are primes, they do exist, there's actually infinitely many, but if you were to choose a random natural number, the probability of you choosing a prime is zero percent. That doesn't mean it's impossible, it means it's not predictable. They're so rare, the primes form a set of measures zero inside of the natural numbers. Probabilistically, there are no prime numbers, they do exist, don't get me wrong, I don't want you to misunderstand the probability, but this set, we're gonna call it P, capital P, the set of prime numbers, is essentially it's an infinite invisible subset of the natural numbers. The primes are so sparse that from a probabilistic point of view, you can't find them, not randomly, okay? But yet, with that said, we're gonna show that the natural numbers and the set of prime numbers have the same cardinality. So this is a very important distinction from the example we saw on the previous slide, where we showed that the natural numbers and the integers have the same cardinality. Sure, there's infinitely many more integers than there were natural numbers, so to speak, but with regard to that density that we were talking about a moment ago, basically half of the integers are natural numbers. So if I choose a random integer, about 50-50 chance, I'm gonna pull out a natural number there. So the natural numbers, even though the gap between the natural numbers and the integers, that is the set difference is infinite, still, like the natural numbers is a huge percentage of the integers, but with this regard, this is an extreme here, 0% of the natural numbers are prime numbers, but yet they still have the same cardinality. It's the same infinity, even though it's such a small fraction comparatively, like when you look at the natural numbers, take away the primes, you basically still have all of the natural numbers, using that limit calculation, the limit ratio we were talking about earlier. All right, so now let's get to the point here. Let's come up with a bijection between the natural numbers and the set of primes to prove that they, in fact, do have the same cardinality. And we're gonna find this function recursively. Since my domain is the natural numbers, I can use induction to define this function here. So what we're gonna do is we're gonna set f of zero equal to two. So I'm just gonna take my first natural number, which is zero and assign it to the first prime, which is two. Now, suppose f of n has been defined, then we're gonna define f of n plus one to be the next prime after f of n, okay? That is, we then choose f of n plus one to equal the minimum of the set p, where you take away the image of f, you take away f zero, f one, f two, f three, f four, all the way up to f n, okay? This set, this set right here is a subset of the natural numbers, and therefore the well-ordered principle applies. This set does have a minimum element, so we're gonna define f of n plus one to be that element. Now, oh wait, the well-ordered principle does have a short little caveat. It only applies when you have a non-empty set. How do I know this set is non-empty? Well, that's because it's infinite, right? We used Euler's proof previously that there is infinitely many primes, okay? And as there's infinitely many primes, if I take a finite set of prime numbers and remove it from the set of primes, you still have infinitely many primes, so this is definitely non-empty because it's actually infinite in its cardality. So there always is a next prime because of the infinitude of the prime numbers there, all right? So like I said, this set of images always has the size n plus one, so if you take it away from an infinite set, it's non-empty. The well-ordered principle applies so that this minimum element exists, so this function, so this image is well-defined, and therefore we now have recursively defined a function. So n plus one is gonna give me the, it's gonna give me the next prime after n, the nth prime there, okay? This function is in fact bijective because notice the following here that if you take two images, whenever their inputs are different, so you have two different natural numbers n and m, one is always less than the other since they're different, so let's say that n is less than m. The way we've defined this, this will actually give you f of n is less than f of m, right? Because you pick the nth prime to attach it to n, then the next prime, which will be bigger, is gonna be attached to n plus one, then the next prime, which will be bigger, will be attached to n plus two, all the way up to m, you're always getting bigger at the next step. So whenever the inputs are different, the outputs will have to be different. That makes the function injective. Is it surjective? Absolutely, because to see this, note that p as a subset of the natural numbers, it has a first element, it has a second element, it has a third element, this is a consequence of the well-ordered principle. And so using that recursive function we defined above, one can make an induction argument that every element of p will eventually show up in the image of this function for some element here because if you take the one millionth prime, then it's like, okay, we're gonna grab the first prime, the second prime, the third prime, the fourth prime, by induction, we will eventually get that one millionth prime. More specifically, the nth prime will be the image of n minus one with respect to this function. Because again, zero took the first one, so one takes the second one. So we've offset it by one because we used zero inside the natural numbers there. But in particular, every prime will be grabbed eventually. So the set of primes has the same cardinality as the set of natural numbers, which is a pretty impressive result there. And I want you to make mention, I do want to make mention that if you take this proof and modify it, you can actually show that any subset of the natural numbers, that's infinite, will have the same cardinality as the natural numbers itself because you basically construct a recursive function where you grab the smallest element of that set and then the second smallest and then the second smallest and the next smallest until you get all of them. Okay, and since it's infinite, this process will never end and you have a bijection. I'll leave it up to the viewer to prove that result that every infinite subset of the natural numbers has the same cardinality as the natural numbers itself. All right, we've talked a lot about the natural numbers. So we've talked about the natural numbers. It has the same cardinality as the positive integers, as the primes, as any proper subset has the same cardinality as the integers. We'll talk some more about the cardinality of the natural numbers in the next lecture, lecture 35. But in this one, I do want to say a few things about cardinalities of sets of real numbers, okay? If you take the unit interval, so the real numbers between zero and one, this actually has the same cardinality as the positive x-axis, that is the interval from zero to infinity. Now, when you look at this one right here, I want you to see at first why this statement seems like it might be absurd. You take an interval which has length one, right? So that's a distance. You compare that to a set that has an infinite length. So this interval is infinitely long. So if you're talking about like the lengths of these sets, they're very, very different, right? The two endpoints are exactly one unit apart. These two endpoints are exactly infinitely many feet apart, right? But yet as a set, they still have the same cardinality. So cardinality is measuring how big the set is, but it's measuring not the same thing as you might be thinking. And so we have to make sure we understand what cardinality measures. Cardinality measures how many numbers are inside the set. It doesn't measure something like length or area or volume, which are some other notions of size of a set we might be interested in. Like if we're talking about like the subsets of R2, we might be interested in like, okay, this set right here has an area of two, but like this set right here has an area of four. It's like, it's a bigger one, right? That's not what we're talking about. In fact, one could show that these two rectangles have the exact same cardinality. So when you start talking about length or area or volume, et cetera, this is a notion that an analysis is referred to as measure. I guess you could talk about in geometry as well. I'm not saying that the measure of these sets is the same thing, because the measure here is this is length one, this is infinite length. Those measures are different, but their cardinalities are the same. And one reason we need something like measure, like length, area, volume, hyper volume, et cetera, is because sometimes we need to distinguish between two sets that have the same cardinality, but still it's one's bigger than the other, but with respect to a different notion of big, okay? We have to articulate exactly what we mean. With regard to cardinality, this is just a set theoretic notion. How many numbers are in the set? And in this case, they have the exact same number. That idea of area as a geometric notion. And so if we wanna talk about how sets can be different in size geometrically, then sure we can talk about measure. But in this conversation, we're talking about from a set theoretic point of view, how do these sets compare? And they do in fact have the same cardinality. We can prove this by producing a bijection. So consider the function f from zero one to zero infinity given by the formula f of x equals x over one minus x. All right, so I wanna first convince you that this is in fact a function. So assuming x is a number between zero and one, okay? So in particular, it's positive. So the numerator here is positive. It's also, you also have that x is less than one. If you subtract x from both sides of that inequality, you're gonna get zero is less than one minus x. So that is one minus x is positive as well. So we see that the denominator here is positive. Well, since the numerator is positive and denominator is positive, this ratio will be positive. And as this is a positive integer, it lives inside the integral zero to infinity. So this is a function, it's well-defined as in the image does actually live inside the co-domain like we want. I next claim that this is a bijection because it has an inverse. That's the same trick we did before. We're gonna reverse this around. We're gonna come up with a function whose domain is zero to infinity and whose co-domain is gonna be zero to one right there. So f inverse of x equals x over one plus x, similar reason before by assumption x is positive. So the numerator is positive, but if you add one, you're still gonna be positive. So this gives you a positive number. That's easy to see, it's greater than zero. But notice that x plus one is bigger than x in this situation. So the ratio will be less than one. This object does in fact live inside of the interval zero to one. And then likewise, if you compose these things together, this is a very straightforward algebraic calculation. Feel free to pause the video to double check this. But if you do f compose with what I'm now calling f inverse, this thing will simplify just to be x is the identity and you get a similar calculation in the other direction. So since f has an inverse, it is a bijection. And this shows that the two intervals we talked about have the same cardinality. Likewise, the real line has the same cardinality as the unit interval. So it kind of makes sense now. It's like, okay, if the unit interval can give you half of the real line, then perhaps it can get you all of it, much like we did with the natural numbers as well. This time, we have a much simpler bijection we can throw out here. Take the map r from the real numbers to the positive integers using just the natural exponential g of x equals e to the x. This function is in fact a bijection we've talked about beforehand. The natural log is its inverse. And notice that if we take the composition of g with f inverse, which we had talked about just on the previous slide, this would then be a function. So g maps the real numbers to the positive reels. Okay, that's what g does. And then f inverse that we talked about on the previous slide, it sends the positive reels to the unit interval like so. Since these are both bijections, their composition would also be a bijection. So we found a bijection from the real numbers here and so these two sets have exactly the same cardinality as well. All right, now there's one last thing we wanna talk about in this video before we close it. So we've talked about all these infinite sets that have the same cardinality. In order to give us some closure on this video before we go on to the next lecture next time, we do need to establish the fact that not all infinite sets have the same cardinality. In particular, we will see not in this video, but we will see that the natural numbers have a different cardinality than the real numbers. In fact, what we will show is that the natural numbers have a cardinality strictly less than the real numbers, but we'll make that explicit a little bit more. And the key ingredient here is gonna be the following observation. If you take a set X and you take its power set, the power set is the set of all subsets of X, then the set X and its power set have different cardinalities, which I should mention that if X is an infinite set, then the power set has to also be infinite as well. Because at the very least, the power set contains the collection of all singletons, one element subsets. And as this contains infinitely many elements, there'll be infinitely many singletons. So this thing is at least infinity, which makes an infinite. And so if these two sets are infinite but have different cardinalities, this means there are different infinities. Some infinities are larger than others. Some are different from others. And so that's where we're gonna end this video here. Let's see the proof. We're gonna prove this by contradiction. Suppose for the sake of contradiction that the set and its power set have the same cardinality, which means by definition, there exists a bijection from the set into the power set. And so I'm gonna define a subset of X. We're gonna call it A. And this set A is gonna be like a set of subset of X. It is a subset of X. And it's gonna be all elements of X such that the element X does not belong to the image of itself with respect to F. Because by definition of this function here, by construction, if I take some element inside of X, then F of X is going to be inside of the co-domain, which is the power set, which means F of X, the image here, is a subset of X. Which means it's a possibility that little X belongs to the subset. Now if X doesn't belong to its image, we then put it inside of A, okay? So this does give you a subset of X. So it's an element of the power set. And therefore, since F is a bijection, it is in particular an onto map. And therefore there exists some element A inside the domain that maps onto A, okay? So F of A equals A. Now when it comes to this, because F of A is a set, it's equal to set A, we then have two options. Either little A is not a member of F of A, or A is a member of F of A. One of those two things happens. Now, if it's the first option, if A does not belong to F of A, well remember F of A is equal to capital A. And so if A, if little A does not belong to A by the definition of A, that means that A belongs to capital A, but capital A is equal to F of A. And this gives us a contradiction. We got that A doesn't belong to F of A, but it also does. Well, that's not a possibility. So that's not the case. Well, what if it's the other possibility? What if A does belong to F of A? Well, if A does belong to F of A, since F of A is equal to capital A, by definition of A, we have to conclude that A, little A, does not belong to capital A, but capital, because little A belongs to its image. So it can't be inside of capital A, but capital A is equal to F of A. And so we get another contradiction. There was two possibilities, but both lead to contradiction. This then contradicts the result, which was that there's such a bijection exist. There exists no bijection from X to the power set. In fact, what we've actually shown is that there exists no surjection from X to its power set. But of course, every bijection is a surjection. That's the one we care about right here. So it's not possible for a set to have the same cardinality as its power set. And in the case where X is an infinite set, this means that the cardinality of infinite set is distinct from the cardinality of another infinite set. And we're gonna dig into this more next time. But for now, that then brings us to the end of lecture 34, introducing us to this notion of at least the proper definition of cardinality. If you learned anything about function cardinality, or I should say set cardinality using functions, please like this video, subscribe to the channel to see more videos like this in the future. Share these videos with friends and colleagues who might be interested. And as always, if you do have any questions whatsoever, feel free to post them in the comments below and I'll be glad to answer them as soon as I can.