 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 Misseldine. In lecture 35, we're going to talk some more about cardinalities of infinite sets. We've talked about this in lecture 34, and we discovered looking at power sets of infinite sets, for example, that not all infinities are created equal. That is, there are some that are bigger than others. In this lecture, we're actually going to explore the smallest type of infinity, so-called countably infinite sets. Let's get to the titular topic for this lesson here. Let A be a set. We say that A is countably infinite, or some people use the word denumerable here, if the cardinality of A is the same as the cardinality of the natural numbers. In that situation, we denote this that A is countably infinite by saying the cardinality of A is equal to A-lif-naught. This symbol right here is actually a Hebrew letter here. It's pronounced A-lif, and it's kind of like Alpha, but it's the Hebrew version of an A, I believe. If I misspeak it all, I'm not the expert of the Hebrew alphabet or anything like that. But A-lif-naught is the symbol we used to denote the cardinality of countably infinite sets. These are sets that have the same cardinality as the natural numbers. We say that a set is countable if it's either finite or is countably infinite. And if a set is not countable, we say it is uncountable. Now, I do have to caution you that in different mathematical texts, they can use these words slightly different. For example, when someone says denumerable, they might use that as a synonym to countable as we've defined right here, and some people might use the word countable to mean countably infinite and not finite. So you have to be very careful what these words mean. In our lecture series, countable means that it's finite, are countably infinite, and countably infinite means you have the same cardinality as the natural numbers. Now, clearly the natural numbers have the same cardinality as themselves, so the natural numbers are a countably infinite set. In the previous lesson, lesson 34, we showed that the positive integers have the same cardinality as the natural numbers. We came up with a bijection in that situation between the two, and therefore the positive integers gives us a countably infinite set. We also showed that the set of integers, positive, negative, or zero has the same cardinality as the natural numbers, so this is also a countably infinite set. We also looked at the set of prime numbers and showed that they are countably infinite. In fact, we made an argument with regard to the primes there, that if you take any subset, or any infinite subset of the natural numbers, this will likewise be a countably infinite set. That also applies to the integers as well. If you take any infinite subset of the integers, you will get a countably infinite set. So in particular, every subset of the natural numbers is countable. Likewise, we can actually argue that any subset of a countable set is countable, but we'll get to more of that in a little bit. Now, we also talked about in the previous lesson that if you have a set X, that the power set, that is the set of subsets of X, has a different cardinality than the set itself. So if you apply that to the natural numbers, the natural numbers, which of course is countably infinite, it does not have the same cardinality as the power set of natural numbers, which the power set of natural numbers itself has to be an infinite set, and therefore, since this is an infinite set, it's not finite, and it doesn't have the same cardinality as the natural numbers. So this is an example of an uncountable set. So the power set of the natural numbers is an uncountable set. And in the very next lesson, we'll talk some more about this set. In this lecture, I wanna explore countable sets. For example, why do we call them countable sets in the first place? So a set A is countably infinite if and only if there exist elements, that is to say that the elements of the set can be arranged in an non-repeating sequence. So there's some A0, A1, A2, A3, A4, A5, et cetera. And every term inside of the set shows up in this sequence once and only once. That makes it a countably infinite set. These two notions are exactly the same thing. Now, be aware that if you have a finite set, okay, so you have some finite set, we'll call it B for a moment. Now, because it's finite, there exist elements in there, we'll call them like A, B, C, D, et cetera, right? It's finite, so this does stop after a while. And so you, of those finite elements, pick one of them, and we can basically label that the 0th element, B0. Then grab another element that you haven't grabbed already, call that one B1. Grab another one B2, B3, et cetera, until you get to the end of your list, say BN or something like that. If your set's finite, we can then we can associate each of these elements with a natural number. Because after all, if it's finite, there exists a bijection from B into the set, zero, one, two, all the way up to N. The way I've written it, I'm assuming that the cardinality of B is N plus one. No big deal there. But because there's a bijection between the set, the finite set to some set like this, you can then use the inverse to provide a labeling, like, oh, there's a B0 element, a B1, a B2. You can list each of these elements. And so you can count them. There's a first element, a second element, a third element. And if you want to start counting at zero, that's fine too, no big deal. We're saying that for countably infinite sets, we can do the same thing. And therefore, a set is countable if it's finite or countably infinite. And so the moral of the story here is that a set is countable if you can list every element for natural numbers. And then for the infinite case, it never stops, mind you. But that's fine. So let's prove this. This is an infinite only of statements. So we have to prove it in both directions. So let's first assume that A is a countably infinite set. Well, if it's countably infinite, by definition, there exists a bijection, which we'll call it F, from the natural numbers into A. And that bijection, if we call it F, then we can define the following term. I'm gonna define A sub N to be the image of N under this map F. And therefore, this is an element of A, all right? Since the map is surjective, every element of A appears in this list of numbers. It appears in this sequence. And so of course, we can write this A0, A1, A2, A3, A4, et cetera, et cetera. This is just, of course, F of zero, F of one, F of two, F of three, et cetera. That's all that this is. These terms are just the images of the function listed in the order that the natural numbers have, zero, one, two, three, four. Since this map is surjective, every element of A will appear in this list somewhere, because that's what it means to be surjective. Likewise, because this function is injective, it's one to one, this sequence has no repetitions, because if there were two numbers that were the same, that means we would have two images that were the same, and that would violate the injectivity of the function. So because there's a bijection between the natural numbers in our set, we can create an infinite sequence using the elements of the set A with no repetitions and everyone shows up exactly once, okay? So that gives us the first direction. Now we wanna go the other direction. Suppose that the set A is equal to an infinite non-repeating sequence. So every element in A can be listed in this sequence with no repetitions whatsoever. We then are gonna define a function. We're gonna define a function G, which goes from A to the natural numbers, such that G of A sub N will maps on to N, okay? Now, since this sequence is non-repeating, that means the element A will never show up twice inside this list. So this is a well-defined map. Now like if you had an element, call it X, that was like A1 and it was A2, then what is G of X? Is it one or two? That would be a problem. But since there's no repetition in the sequence, this is well-defined, like if X shows up in the 12th spot, then G of X is equal to 12. It's a function, all right? It's well-defined. And it's also not too difficult to see that this function is a bijection. It has an inverse, or you can send this thing back like so. And so this is a bijection. So our function, excuse me, our set, since the function is a bijection, is then countably infinite. There's a bijection going back in the other direction. And so this explains why we call these sets countable. A set is countable if you can list all of them in order, right? And it doesn't matter what the order is, because this function kind of places an order on it. It doesn't matter what that order is, but every element can be listed. You can count all of them. That's where this word denumerable comes from. You can enumerate every element inside of the set there. I also wanna make mention that in this theorem, we talked about non-repeating sequences. If you have a sequence where everything is listed and there are repetitions, you can just remove the repetitions, and therefore every sequence can be restricted to a sub-sequence, which is non-repeating elements. If again, there are any repetitions, just throw them out. Now sure, when you restrict it, maybe it's no longer a infinite sequence, maybe it's just a finite list, that's fine. That still will be a countable set in that situation. So the moral of the story of theorem 14-4 here is that if you can list the elements of the set, that makes it countable. And in particular for countable infinite sets, that would be an infinite sequence in that situation. So with that in mind, I wanna present to us some corollaries for which I'm not gonna prove any of these four corollaries. I'm gonna leave these as exercise to the viewer here, exercises for my students. And honestly, we can prove each and every one of these using this listing condition that they're countable sets there. So I'll give you some hints on how you prove these corollaries here. Corollaries because these are consequences of the previous theorem we just proved. So imagine we have a subset. So A is a subset of B. If B is countable, then likewise A is countable. And how could you argue this? Well, since B is countable, you can list each element here. Since A is a subset of B, and you can list every element of B inside of a sequence, you can then think of A as a subsequence of that sequence. So you just remove all the elements that are in B that are not in A. That'll give you a subsequence. And now since you can list every element in A, that shows it's countable by the previous theorem. So I kind of talked about the proof. I'll leave it up to you to kind of write the specific details of that. The next corollary, let X be a partition on a set A. If the set which is being partitioned is countable, then the partition is countable. If you have a countable set and you place a partition on it, there are finitely many equivalence classes on that. So another way of saying that is that if you take an equivalence relation on a countable set, the number of equivalence classes is itself countable. I'll leave it to the viewer to check that. Let's get some functions into play here. So if F is a function from A to B, and if the co-domain is countable and the function is injective, then A is likewise countable. So if the co-domain is countable and the function is one to one, then the domain has to be countable as well. And then the dual argument there is that our last corollary, if F is again a function from A to B, this time the domain is countable and the function is surjective, that is the functions onto, then the co-domain has to be countable as well. And so again, I'll leave it up to the viewer here to prove all these results, but these are very simple results that come from the theorem we just proved. If you can list all the elements, that makes it a countable set. We kind of talked about how you do this one. I'll leave it up to you to kind of argue this one here. Now I mentioned that if you have this one, you can use that to prove this one. And if you have this one, you can use that to prove this one here. But I'll let you fill in the details of those. We're gonna look at some harder ones here. So I wanna talk about how countable sets interact with the operations of sets that we've already been talking about. Like for example, if you have two countable sets, what about their union? And it is a fact that the union of two countable sets will be countable itself. Now I don't need to say much about intersections because if you're like, oh, A and B are countable, what about A intersect B? Well, if you take A intersect anything, if A is countable, it's a subset of A, which was countable by assumption. So by the previous corollary we had just talked about, all subsets of countable sets are countable. So intersections are too easy. Like if you intersect two countable sets, you're gonna get something countable. Honestly, if you intersect a countable set with anything, you'll get something countable. Unions on the other hand, a little bit harder because you're getting something bigger. How do I know that bigger thing is still countable? Like what if you take two sets and you make it so big that you can't count it anymore, right? Well, let's look at the proof of this thing. Suppose A and B are countable. Then that means our sets A and B can be expressed as sequences. Now those sequences could be infinite or finite. I don't care. We get two sequences. So all the elements of A will look like A0, A1, A2, keep on going. If it's finite, this will eventually terminate. If it's infinite, it'll go on forever, not a problem. Do the same thing for B. Every element of B can be listed. B0, B1, B2, B3, B4 keep on going, okay? Then in order to show that A union B is countable, all we have to do is argue that the elements of A union B can themselves be listed. And we're gonna list every element of A union B using the following strategy. We're gonna do A0, B0, A1, B1, A2, B2. And then we're gonna continue to alternate between the two elements. The next one would be A3, then B3, then A4, then B4, then A5, then B5, et cetera, et cetera. So we're gonna mesh the two sequences together interlocking them like so. Now, if any of the sequences are finite, then you can stop alternating. Once you've exhausted all of the elements from A, then just continue to list elements for B forever onward. That's how we're gonna handle that. Or same thing, if B is finite, then once you've reached the end of B, you can stop your zipper. Because that's the idea. We're trying to zipper these two sequences together. Now, if one of them has more teeth than the other, then just stop the zipper process at that moment. If they're both finite, this will be just fine. Now, we're taking a union. It's very possible that the elements of A do overlap with the elements of B. There could be an element that's in A and B. Like so. As such, this sequence very likely could have repetitions inside of it. But if there's any repeated elements, just remove them after the first appearance. So we have this sequence. If there's any repetitions like, oh, A0 is equal to B2. Which is like, oh, just remove B2 from the list. We will then get a non-repeating sequence of elements. And every element of A union B will be in this list. Because if it was in A, it showed up here. Thus, it'll show up there. If it was in B, it'll show up here. And thus, we'll also show up there. And as such, every element of A or B will show up eventually. And so we've now constructed a non-repeating sequence of elements from A union B. Therefore, A union B is a countable set. Now, if we apply induction here, we can actually show that an arbitrary union, that is a finite union of countable sets, is likewise countable. So this is a corollary to the previous theorem we just talked about. So A1, A2, up to AN, these are each countable sets for some natural number N. Then the union where K ranges from one to N of A sub K, this is itself a countable set. And we can prove this by induction. So for induction, we have to do the base case. So suppose N equals one. A note that if you take the union of K equals one to one of AK, this is just A1 itself. And A1 is a countable set. Now, I guess technically the base case should start off with zero because that's the first natural number there. And so in this situation, you're taking the union where you have nothing inside of it. An empty union is actually the empty set itself. And as such, the empty set, it's a finite set. It's countable, so that would still work. Some people find this a little disconcerting, but we should be very comfortable with these trivial zero cases. So really the base case should be N equals zero, but if you're uncomfortable with that because like an empty union, oh no, then take N equals one and you can do your base case there, not a big deal. An empty union would be the empty set because the empty set acts like the identity for unions. Okay, so we get our base case. So then let's take our inductive hypothesis. So suppose that the union where you take K equals one to M or M is some positive integer here to take the union, the M-fold union of AK that this is a countable set, okay? So then that's our inductive hypothesis. Assume, no, not assume. Consider the union where this time we go from K equals one to M plus one, you take the union of all those sets A sub K. Well, so the only difference between this union and this union is we've now joined on one more set. So we can actually decompose this union into the M-fold union where we get A1 through AM and then you just unite to it A sub M plus one. Well, AM plus one is countable by assumption. Now is the assumption of the sets A1, A2, A3 up to A, whatever, AN, these are all countable sets. So that's a countable set. By our inductive hypothesis, this is a countable set and therefore the union of two countable sets is countable. That's what we proved previously. So theorem 1429, which is just above on the slide shows that the union of two countable sets is countable. So therefore this is countable and therefore the result follows by induction. If you take a finite union of countable sets, that is itself countable. Same thing with intersections, but that argument's much easier. If you take the intersection of A sub K where K ranges from one to N, I don't need induction there because this is a subset of A sub K. We'll just say it's the subset of A1, right? Which A1 is countable. So this intersection is countable as well, right? No big deal, not a problem there. Intersections are much easier because subsets of countable sets are themselves countable. All right, what if we take an infinite union of countable sets? That is, suppose we have an infinite sequence, A0, A1, A2, A3, A4, A5, et cetera, et cetera. We have a countable set for each of the natural numbers. Each of these A sub Is is a countable set, okay? So then if we take the union where K equals zero to infinity of all of these sets A sub K, if each of the sets A K is countable, then this countable union will also be countable. And I do say this is countable union. If you take a union where, you take the union of the A sub X's where X is an element of the real numbers, we're gonna see not yet, but the real numbers is an uncountable set. If you take an uncountable union of countable sets, this might not be countable. Not countable. Well, maybe. It's hard to say. I mean, like if you take the natural numbers and you unite that even on uncountable amount of times because the overlap is the whole thing, you can still be countable. It's like, but the thing is if you take an uncountable union of things, because when I say an uncountable union, I mean the index set of the union is itself an uncountable set, you might not be countable anymore, right? But with regard to the countable unions of countable sets, this is in fact always going to be a countable thing. So we've already taken care of the finite case. So to show that a countable union of countable sets is countable, we only have to deal with the union be countably infinite. So there are countably infinite, there's countably infinite many sets inside of this union. If each of the sets is countable, this is going to be a countable set, all right? And the way we're going to do this is that we have to come up with a sequence. We're going to list every element in the set in an infinite sequence, okay? Now, since AI is a countable set, so AI is just one of the sets inside of this list here, since it is a countable set, again, it could be infinite or finite, I don't care, each of the elements inside of AI can be listed as a sequence, which we're going to use a double subscript to keep track of things here. So the zero with element of AI, we're going to call AI zero, the first element we'll call it AI one, the second element we'll call it AI two, et cetera, AI three, AI four, AI five, AI six, et cetera, et cetera, et cetera. So every element here gets that. And again, if there's any, if it's finite, this might stop after a while, not a big deal. Now, every element in the countable union could be arranged not as a sequence, but actually as a two-dimensional array, kind of like an infinite matrix over here, for which the first row of the matrix is the first set, AI zero, the second row of the matrix is the second set AI one, the third row here will be AI two, the next row would be AI three, et cetera, and et cetera. Okay, so that's how we've done, we've arranged it here. And now every element will show up somewhere in this array because every element in AI zero shows up in the first row, every element of AI one shows up in the next row, every element of AI two shows up in the next row. So every element will show up eventually inside of this thing. So be aware that this two-dimensional infinite matrix, well, it's potentially infinite matrix, some of these rows might terminate after a while and the whole thing might terminate, I suppose, although we are assuming infinite rows in this situation, the columns might stop after a while. Anyways, every element inside this countable union will show up somewhere in this array. So how do you turn this two-dimensional array into a one-dimensional sequence? Well, we can accomplish that using what we call the snake method. We're gonna have a little sneaky snake that's gonna slither through this entire diagram. To use in a calculus analogy, what we are talking about is a space-filling curve. But instead of having some type of continuous topology, we're actually talking about a discrete array. So it's much easier to get a space-filling curve, but that's exactly what we're trying to do here. We're gonna make a sequence by snaking through all of it. So what you're gonna do is you're gonna start with the upper corner, the A-00 spot. Then following this snake, you're gonna go to the A-1 spot, A-1, 1-0, 2-0, 2-1, 2-2, 1-2, 0-2, 0-3, 1-3, 2-3, 3-3, 3-2, 3-1, 3-0, 4-0, 4-1, 4-2, 4-3, 4-4, 3-4, 2-4, 1-4, 0-4, whoo! You get all those on the screen. Now this continues on potentially forever and ever and ever. So the next one is when you end, because you have these layers, right? So basically what you're doing is like, there's this layer. There's this one, there's this one, there's this one, there's this one, there's this one right here. So you have these layers, these L-shaped objects inside of the array. And so then what you do is you just transverse each of these layers, each of these L's. And once you've finished the layer, you just jump to the next layer and go over. Then you jump over, do the next one, then jump over, do the next one. So what we do is we jump here and then we do this layer. Then we jump down and go to the next one, like so, and you keep on doing this over and over and over again, going towards infinity, okay? So basically when you end, one of your coordinates is gonna be a zero. If the second coordinate is zero, then you're just gonna increase the first coordinate by one. But if you end with the first coordinate being zero, then you're just gonna increase the second coordinate by one. So like I said, whenever you end a layer, you end with a zero. Either the first or second coordinate is a zero. Just increment the non-zero coordinate to the next natural number and then continue going through the layers. This snake method will grab everything in the two-dimensional array. And so this then turns everything into a sequence, right? So that sequence will look like what I read earlier. A zero, zero, zero, one, one, one, zero, two, zero, two, one, two, two, one, two, zero, two, zero, three, one, three, et cetera, et cetera, et cetera. You're gonna go all the way through. You're gonna grab everything, all right? Now, if there reaches a point where there isn't a number because the list was finite, that's okay. Just skip over that spot. No big deal. If you ever reach a point where you get a repetition, like this number and this number were the same thing, the first time you find it, add it to the list. The second time you find it, just skip over it. You don't use it. Like I said, and if it's finite, if there isn't a spot, if there's a gap, just skip over until you find the next number, okay? This strategy will construct a non-repeating sequence, infinite sequence here, I guess with repetitions it might be finite, but whatever. You are gonna get a non-repeating list of numbers, finite or infinite, that'll then, every element inside the union falls inside of that list somewhere. And therefore, by the theorem we've proven, this is a countable set because we can list it all. And again, by similar reasoning, if you take an infinite intersection, this of course is a subset of a one, so it'll be countable. With intersections, you can take any intersection you want. If you take an uncountable intersection of countable sets, this will still be countable because they all sit inside of a one and subsets of countable sets are countable. So intersections are no problems whatsoever. If you take finite or countably infinite unions of countable sets, that does give you a countable set. If you take an uncountable union of countable sets, then all bets are off. I mean, anything could happen in that situation. It could be countable maybe, but most likely it'd be uncountable. Another operation that we wanna talk about here is Cartesian products. This gets a little bit more difficult, but sure enough, if you have two countable sets, the Cartesian product of those two countable sets is likewise countable. The argument is actually very similar to the SNCC method we used on the previous theorem. So suppose we have two countable sets. They could be infinite, they could be finite, it doesn't matter. Since they're countable, the elements of A can be listed, A0, A1, A2. The elements of B can be listed as well, B0, B1, B2, et cetera. Then with these two listed in mind, we can construct a two-dimensional array that contains every element of A cross B. Because A cross B is ordered pairs of elements. So you take an element from A and you pair it with an element of B. You're gonna use the elements of A to give you your first coordinate. You're gonna take the elements of B to be your second coordinate. And so likewise, you construct this very large, most likely infinite matrix again, for which you have the A0, B0, A0, B1, A0, B2, A0, B3, A0, B4, A1, B0, A1, B1, A1, B2, A1, B3, A1, B4, A2, B0, A2, B1, A2, B2. You get the idea, right? Using the elements, the index of elements from A as your first coordinate and index of elements from B for your second coordinate, we can then come up with another two-dimensional infinite matrix that keeps track of all of the elements here. Every element inside of this set shows up in this matrix somewhere. And so using the snake method again, you can slither your way through this entire process and everyone's gonna get hit. If there ever is a repeat, that actually isn't gonna happen here because this list is non-repeating, we can assume that this list is non-repeating. So none of the older pairs are ever gonna repeat itself. Now, if this is a finite list, there might be a point where it stops, because if there's any gaps, the snake will just jump over it. And there might be a point where there's nothing left because maybe both sets are finite, that's fine. If it terminates, that's perfectly fine too. This shows us that the Cartesian product of two-countable sets is countable, again, using the snake method. Using induction, we can show that the finite product of countable sets is likewise countable. So if you have countable sets A1, A2 up to An, and N is a positive integer right here, then the product K1 up to N of the AKs is likewise countable. This is the exact same induction argument we did before. The base case, when N equals one, you're taking a Cartesian product of just one set, that'll just give you back A1 itself for which that was countable by assumption. So that gives us the base case. Again, your base case is really zero in which case if you take the product of nothing, you give back the empty set, that of course is also a countable set. So whichever base case you decide to use, it's gonna be just fine, no big deal. People don't typically talk about empty products, but it's relevant to mention there. So then your inductive hypothesis, assume you take the M-fold product, Cartesian product of countable sets, assume that's countable, then look at the M plus one-fold Cartesian product. Well, the M plus one-fold Cartesian product is the M-fold Cartesian product times the next function there, or the next set, excuse me, M sub M plus one. By assumption, A M plus one is countable. By our inductive hypothesis, the M-fold Cartesian product is countable. And then we just previously shown that the Cartesian product of two countable sets is itself countable. And therefore this gives us that the M plus one-fold Cartesian product is countable. So any finite Cartesian product of countable sets is countable. So if I took, for example, the set, zero one cross the natural numbers, this is a countable set. It's countable, because you took a countable set cross a countable set. You could even take, for example, N cross N, this would likewise be a countable set. And you can come up with more exotic ones as well. On the other hand though, you have to be very cautious. This is about as far as we can go here, because unlike the theorem we saw above about countable unions, the corresponding theorem for Cartesian products is false. That is to say that if you take the Cartesian product, where K ranges from one to infinity of A sub K, this set is uncountable. Now with the unions, if you take a countable union of countable sets, that is countable. But the countable product of countable set is actually guaranteed to be uncountable. And this is an example we're gonna explore more in the next lecture here. We're actually gonna look at the set, take the set of binary sequences. That is to say we're gonna take the, if you have a binary sequence, you know, this would be something like zero one one one zero zero one zero one one zero one et cetera, right? So this is an infinite list of numbers that consist of only zero and ones. If you take the set that contains all of these things, this set we are gonna see is actually uncountable. It's an uncountable set. And this set of course is actually an example of such a thing here. This set X is actually equal to the infinite Cartesian product where K goes from one to infinity here of the set zero one. So you can identify a sequence as an infinite Cartesian product because after all, what is a sequence? It's an infinite list order does matter. So if a Cartesian product gives you ordered pairs, a three-fold Cartesian product gives you ordered triples, a four-fold Cartesian product gives you ordered quadruples. And so an infinite, accountability infinite Cartesian product is actually gonna give you sequences. And so if you take the infinite product of the set zero one, you just do this over and over and over again, right? So in some respect, this is the set zero one to the infinite power. More precisely, this is the alof not power there. This actually gives you an uncountable set. We're actually gonna show that this set has the same cardinality as the power set of the natural numbers. But again, we'll talk about that more next time. Before we end this video, there is one other important example I wanna mention, the rational numbers. We've talked about the integers, the integers are countable. The natural numbers are, of course, are countable by definition, but the rational numbers are likewise a countable set. That one's a little bit hard to believe because like if you look at it in terms of the geometry of the real line, right? Okay, we got the integers, I can see that. So you have like zero one, but between any two integers, there's a gap. There's not any integers between zero and one. That's what makes these things discrete. On the other hand, with the rational numbers, between any two rational numbers, there's another rational number and another one and another one. So honestly, between any two integers, there's infinitely many rational numbers, right? So the terminology we use here is that the rational numbers are dense. That between any two, there's always another one. So given any like any interval of rational numbers, there's infinitely many rational numbers inside of that set. You can't do that with integers. Integers have gaps, they are discrete. But the rational numbers, on the other hand, are dense. So in some regard, one can make the argument that there are more rational numbers than integers. But that's actually a falsehood. At least when we're talking about cardinality. Like we had mentioned before, if you're talking about like measure or geometry, maybe there's a geometric notion and analytic notion or a topological notion of size. But if we're only talking about set theory, we're talking about cardinality. And the cardinality of the rational numbers is the exact same as the natural numbers. This is a countable set. And the way we're gonna do this is the following. We're gonna construct a function F where F is a function from the rational numbers to Z cross Z. I want you to be aware that Z cross Z is a countable set. How do I know that? Well, Z is a countable set. We came up with a specific bijection between the natural numbers and the integers. And as we've just have seen, the Cartesian product of two countable sets is countable. So this is a countable set. Now we have a function between a set, the rational numbers with a countable set. Well, what's the formula here? Where if you take any rational number X, I want you to take F of X to be the ordered pair P comma Q where X is equal to P comma Q and P comma Q is the lowest terms. So for example, if X equals one half, you're gonna map this over to one comma two. If X equals three, you're gonna map this over to three comma one. If X equals zero, you're gonna map this over to zero comma one. If X equals negative three halves, you map this over to negative three comma two. You always assume the denominator is gonna be positive. You can always make that assumption. This is at lowest terms. There's no common divisors between the two terms there. That is the numerator and denominator have a GCD of one. Such a number always exists and this is unique for every rational number. So if you use the lowest terms representation, this is well-defined as a function. So this is a function. Is it one to one? Well, if F of X equals F of Y, that means their image is the same. So there's some P and Q. Well, if F of X is P comma Q, that means X as a rational number is equal to the fraction P comma P over Q. But likewise, if F of Y has the same image, that means Y as a fraction, as a rational number is equal to the fraction P over Q since they're the same fraction, they're the same rational number. So that's injective. And I should mention that one of the corollaries we mentioned earlier in this lesson said that if you have a map, let's look at this again. If you have a map from a function to a countable set that is injective, that corollary guarantees that this thing is, the domain is countable as well. So because we can inject, we can embed the rational numbers inside of a countable set that makes themselves countable. Now, you could also prove the rational numbers is countable directly using a snake argument that we did before. And if you want to, I'll leave that up to the viewer to do that. But this gives us some examples of some very important countable sets and operations that you can do with countable sets to retain the accountability of the set. And our next lesson, which will actually be the final lesson for this lecture series number 36, we'll talk about uncountable sets and some of the consequences to set theory and advanced mathematics. But that brings us to the end of lecture 35. Thanks for watching. If you learned anything about countable sets in this lecture, please like the video, subscribe to the channel to see more videos like this in the future. Refer this video to friends or colleagues who might be interested in this as well. And as always, if you have any questions, feel free to post them in the comments below and I'll be glad to answer them as soon as I can.