 Welcome back to our lecture series Math 31-20, Transition to Advanced Mathematics for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. Lecture 36 is the last lecture in our series on Transition to Advanced Mathematics, but definitely is a goodie. So we need to finish up some loose hands and plot holes that we've left inside of this topics of cardinality of infinite sets. So first of all, we've mentioned that all sets with the same cardinality as the natural numbers are exactly what we refer to as these countable infinite sets. We mentioned previously that countable infinite sets are the smallest of all of the infinite sets. It's a countable infinity, that is, a if not, is the smallest infinity. What does it actually mean to be the smallest infinity? We know what it means for cardinality to have the same size, but what does it mean for one cardinality to be larger than another? That sounds like a partial order, doesn't it? Let's define it properly. So imagine we have two sets, A and B. We say that the cardinality of A is less than or equal to the cardinality of B, and we denote this as the cardinality of A is less than or equal to cardinality of B. Which if you stop there for a moment, when A and B had finite cardinalities, this is just the usual ordering of the natural numbers, no big deal. But when you have infinite sets, what does it mean for an infinity to be less than or equal to another infinity? I mean, they're not natural numbers anymore, they're not real numbers. There's a new type of order in town here. So what does it mean for a cardinality less than or equal to another cardinality? Well, finite are infinite. That means there exists some injection F from A to B. An injection here, of course, being an injective function of one-to-one function. I'm using a symbol that sure looks like a partial order. It sounds like a partial order. It tastes like a partial order. Is this a partial order? Well, of course it is. We're going to prove that in just a second. I mean, the name seems to suggest it's obviously a partial order. Now, in order to be a partial order, remember, it needs to be a reflexive relation, an anti-symmetric relation, and a transitive relation. Now, proving that it's reflexive and transitive will be identical to how we showed that having the same cardinality was an equivalence relation, which equivalence relations are also symmetric. Well, they are symmetric, but they're reflexive and transitive, is what I meant to say there. So that's going to be basically the exact same argument as before. But unlike a equivalence relation, which is symmetric, partial orders are anti-symmetric. And as I've hinted towards in the past, when we talk about partial orders, anti-symmetry is one of the hardest of the three axioms to prove. And in fact, the way we do that in this case is we're going to invoke the Cantor-Bernstein-Schroeder theorem, which says that if there is an injection F from A to B and there's an injection G from B to A, there actually exists a bijection H from A to B. So the presence of injective maps in both directions gives us the existence of a bijective map. And the idea is, if you have these two sets, A and B, basically you take the map here. And then since you take the map back and you kind of go back and forth taking unions of these infinite sequences, and you can construct a bijective map, we're not going to provide the details of that proof here. In fact, it's a long proof. And as we kind of finish this lecture series, I don't necessarily want to focus on this challenging proof. If you want to see a proof of it, you can take a look at the proof inside the book. It's in section 14.4 of the book of proof there. But this is going to be the key ingredient for proving anti-symmetry. So let's now look at the proof of that result here. So let's define a relation between sets. We say that a set is less than or equal to another set. A is less than or equal to B, if and only if the cardinality of A is less than or equal to cardinality of B. So basically we're saying that this relation we defined on the previous slide is in fact a partial order, okay? So it's got to be reflexive, anti-symmetric and transitive. Well, reflexivity is just like we did with same cardinality. Given any set, there exists the identity map from A to A, which is an injection. I mean, it's actually a bijection, but every bijection by definition is one to one. So that's an injection. And so this shows that A has the same cardinality or has less than or equal to cardinality as A, right? This is actually why we say less than or equal to because they could actually have the same cardinality. I mean, if there was an injection, there could possibly be a bijection. In fact, the injection that showed this might actually be a bijection that shows this, okay? So it is reflexive, okay? Then we get to anti-symmetric. Well, basically the contour of Bernstein-Schroder theorem gives us exactly what we want here. Suppose we have two sets such that A's cardinality is less than or equal to B's and B's cardinality is less than or equal to A. Then by the contour of Bernstein-Schroder theorem, since, because this implies there's an injection from A to B, this implies there's an injection from A or B to A, that's by the definition. And since there's injective maps in both directions, the theorem gives us there's a bijection. So F, there exists some bijection. F, that goes from A to B. Well, since there's a bijection, that means they have the same cardinality. And that's what we mean by anti-symmetric in the situation. If the inequality goes in both direction and actually forces an equality here. So they have the same cardinality. I'm not saying that the same set, but they have the same cardinalities, okay? And then transitivity, exact same arguments we have before. If we have three sets, A, B, and C, such that the cardinality of A is less than or equal to cardinality of B and the cardinality of B is less than or equal to cardinality of C. Well, by definition, this first one implies there is an injective map from A to B. The second statement implies that there's an injective map from B to C. And when you compose those together, G composed with F, this then gives you a map from A to C. And the composition of one to one's maps is likewise one to one. So that tells us that the cardinality of A is less than or equal to the cardinality of C. And that then gives us that this is a transitive relation, thus proving the result. So using injective maps, we can then form these comparisons between cardinalities. You can say when one cardinality is larger than another. So let's get back to the original question. Why is countably infinite the smallest infinity? Well, if you have an infinite set A, then it turns out that the cardinality of the natural numbers, which is alif-naught, is less than or equal to the cardinality of A. And this statement right here is exactly what we're trying to show here, is that alif-naught is less than or equal to every infinite cardinal. So we wanna prove this statement. This will argue that the countably infinite sets are the smallest infinite sets. Clearly finite sets are smaller, but alif-naught is the smallest infinite cardinal. Now, in order to show this statement, we have to construct a one-to-one function. And given that the domain is the natural numbers, we can construct this function recursively. We have the induction principle for the natural numbers, which allows us to do recursion. That's how we're gonna do it here. Now, A by assumption is an infinite set. And so in particular, it's not empty. There's something inside of A. So I don't care which elements you want, choose something inside of A. There's a lot of options, but choose something. And we're gonna call that element A0. And we're gonna define our function so that f of zero equals A0. That element you just picked, then becomes the image of zero, okay? Next, now that we've defined the image of zero, then we're gonna construct a new set. We're gonna call that new set A1. A1 by definition will be A, take away A0, the element we just assigned to F to zero there. Now, because A is an infinite set, if infinite set minus an element, is still gonna be an infinite set. And as such, it's in fact non-empty. That's what I actually care about here. It's non-empty, A1 is not empty because A is infinite. Taking away one element can't take away everything. But A1 is non-empty. As such, since A1 is a non-empty set, there's something inside of A1. And therefore, pick your favorite element inside of A1. We're gonna call that one little A1 right here, okay? Then we're gonna define our function F such that F of one is then evaluates to be A1, right? So F of one is equal to A1. So then we've now defined F of zero to be A0, we've defined F1 to be A1. And so imagine we keep on doing this, all right? So we keep on doing this recursively. Suppose that we have chosen N elements from A, call them A0, A1, A2, A3 all the way up to A and minus one. And we've done this in such a way that none of the As are the same. So A1 is distinct from Aj whenever the indices are different. And this is how we were doing it before. So A0 is any element in A, then A1 was something in A that wasn't A0. A2 would be something that's in A that's not A0 or A1. A3 we selected to be something that's in A that wasn't A0, A1, A2. And since A is an infinite set, we haven't exhausted all the elements of A yet. So I've always been able to choose elements AI and AJ so that they're distinct from each other. And then every time we selected one of these elements AI, we assign that to be the image of I under this map F that we're building. So F of I equals AI. And we do this for all of the Is up to N. We haven't done N yet. And then also for the sake of simplicity here, we also have introduced this sequence of sets AI, which AI is gonna be AI minus one, take away the little element AI. So in terms of sets, we started off with A, which basically A was just A0. We didn't take away anything. Then sitting inside of that is A1, sitting inside of that is A2, sitting inside of that is A3. And we've done this all the way up so far to A in minus one. So all of these sets, and each time we go from one set to the other, we removed an element. So as we went from A0 to A1, we removed little A0. As we went from A1 to A2, we removed little A1. As we go from A2 to A3, we removed little A3, et cetera, et cetera. So we've done that up to the end stage. That is, we've done the first N numbers, zero through N minus one, all right? And so that's our inductive hypothesis right here. So now with all of these elements, these N elements we've decided, now let's construct the set capital A sub N. This will be capital A sub N minus one. We're gonna take away the very last image we decided upon, which was A sub N minus one. Now be aware that since A was an infinite set, to form A N, we have removed N elements from an infinite set. A N will still be infinite. And in particular, since it's infinite, it's not empty. There's something in there. As such, choose your favorite element among the set A N that hasn't already been captured, call that little A N, and we're gonna define the functional relationship F of N to be A sub N. And so this is then our inductive step. So we've done a base case, inductive hypothesis, inductive step. So then we can scream induction at this moment. By induction, this function has been recursively find F as a map from the natural numbers to A, such that every element in the domain has a unique element in A that it's been assigned to. So N will map to A sub N, right? And we've also done this in a manner such that we know that AI doesn't equal AJ. That was part of our induction that we did above. And as these are the functional images, since F of one, excuse me, since F of I is equal to AI, and since F of J is equal to AJ, so long as I and J are different, then the AIs and AJs will be different. Therefore the images F of I and F of J will be different. So since different input necessarily have different output, this makes our function into a one to one function. And therefore we've now proven that the natural numbers have a cardinality less than or equal to the cardinality of any infinite set. So the above theorems, particularly the one we just have proven, justifies our claim that N is the smallest infinity. Now, if N is the smallest infinity, it sort of begs the question, is there a next smallest infinity? Who comes after ALIF naught? Well, I mean, using the subscript notation from before, it would make sense to ask about ALIF one. So ALIF one would then be the next infinity after ALIF zero. We know there's other infinities, right? We've proven the existence of uncountable sets. So who's ALIF one? Now, it turns out answering that question is actually a very difficult question. We know that ALIF one would have to be uncountable because countable infinities are ALIF naught and there's nothing smaller for an infinite cardinal than ALIF naught there. So okay, ALIF one is an uncountable set. That's good. And we do know that ALIF one exists. That I can say, right? The so-called ZFC axioms of set theory guarantee the existence of ALIF naught. So we have the Zermelo-Frankl choice axioms here. So Zermelo and Frankl were mathematicians who helped develop the modern day framework of set theory. I mean, contour, I should say. Gayle contour, of course, is one of the largest names in modern day set theory. There would be modern set theory without contour here. And of course, we also have some of the names from the previous theorem we saw. Like in addition to contour, there's Bernstein, Schroeder, Guttel would of course be another one to mention. There's lots, lots of mathematicians that I won't name here right now. But the ZF and the ZFC set theory axioms are named after those two mathematicians. This third symbol, C, is short for choice because in the original formation of ZF set theory, there was not the axiom of choice. The axiom of choice basically allows us to make choices that we just used in the previous proof. Remember I said things like, pick your favorite element in this infinite set. It turns out without the axiom of choice, there exist models of ZF set theory for which that might not be possible. And so the choice axiom is, it gives the existence of a so-called choice function that's the rigorous meaning there. But it allows us to do lots of things like pick a random element of an infinite set, guarantee that infinite Cartesian products are non-empty. It allows us to form, to define maximal elements in various partial orders. We can put a well order on any set as if we can extend the well ordering principle to any set like the real numbers. There does exist an ordering of the real numbers such that every non-empty subset has a minimal element. It's not the standard ordering of the real numbers but such a thing exists because of the axiom of choice. Now, the axiom of choice has a lot of controversy that I'm not gonna really go into inside of this lecture. There was a reason it wasn't originally included. Its importance wasn't obvious at the beginning and people sort of caught on that we needed choice. But also there's some, I guess there are some controversies, there are some oddities that come about because of choice. The axiom of choice often leads to very many non-constructive existence statements like can we construct a Hamill basis for the real numbers as a vector space over the rational numbers, things like that. There's the Bonak-Tarski paradox which allows us to cut a sphere into, I should say a ball, a unit ball inside of R3. We can cut that into five pieces and reassemble it in such a way that we have exactly two balls now of each having the same volume as the original one. So sort of like a miraculous creation of a new ball. That kind of gives some people some odd feelings, right? I mean, if you think of it from cardinality, the two sets have the same cardinality. So the fact that you can form two balls is not too surprising from a cardinality point of view. But the fact that this can be done in a measure preserving way, you sort of created a second ball out of nothing, that can be disconcerting to some people. And so there's a lot of controversy that at least used to exist about the axiom of choice. But honestly, the benefits of the axiom of choice definitely outweigh the consequences, the bad consequences that the mathematical community has almost unanimously accepted the axiom of choice as a standard well-accepted axiom of set theory. So we use it freely throughout this lesson. We've also used it in some of the previous lessons. I didn't draw attention to it, but 10 brownie points to anyone who can find our reference to the axiom of choice in the previous lecture there. It's a nice little Easter egg that I'm sure you could find if you want to go snooping for it. But honestly, if we want to make any comparisons about infinite cardinals, we need the axiom of choice. And that's exactly what we're doing now. I mean, that's the title of our section, comparing cardinals, cardinalities here. So we need the axiom of choice, okay? So the axiom of choice guarantees the existence of a next infinity, the so-called ALIF-1. But who is it? Do we even have a candidate for who it would be? Well, it turns out there is one candidate and it's this friend right here. We're gonna call the cardinality of the real numbers the continuum. And we get this name, of course, from calculus here that if we think of like an interval, like the unit interval zero to one, along the x-axis, the idea of the continuum is that as we go along this line, there are no gaps or breaks of numbers whatsoever. The continuum is kind of like the opposite of the integers which are discrete as you go between integers, there's always gaps between them. So like there's nothing between zero and one. There's no integers between zero and two. There's no integers between two and three. This is sort of the important thing, like nothing between zero and one, nothing between one and two, but nothing between two and three inside of the integer realm. That's what makes them discrete. The real numbers on the other hand have the opposite effect that we, that's the continuous realm. Continuous as in like, there's no breaks. We often care about continuous functions inside of calculus. And as such, the cardinality of the real numbers is then dubbed the continuum. It is a cardinality that's strictly larger than the accountability infinite. We haven't yet proven that, though we've hinted around it. Like we did previously show that the real numbers has the same cardinality as an infinite segment of it. So like if you take the segment zero to infinity, that's the same cardinality. We've also shown that it's the same cardinality as a interval of finite lengths, like the unit interval right here. So these are all at the same cardinalities. And so that leads to that, but we never actually compared this to the natural numbers. How do they compare? Well, we're gonna see in just a second that these sets are actually bigger than the natural numbers. And it actually gives us a very natural candidate for this ale of one. It's the next infinity, maybe. We'll get to more of that in just a second. So before we do that, I wanna present the very famous Cantor's diagonalization argument here. Admittedly, we actually have another way we're gonna prove that the real numbers are uncountable, but this is such a classic result appropriate to what we're talking about right now. I kind of have to mention it. So what we wanna do is count the set of binary sequences. So binary sequences, this would be things like zero, zero, zero, one, one, zero, zero, one, one, one, zero. Keep on going. So the sequence consists of only zeros and ones. For convenience, we're gonna denote this set as two to the N. And the idea here of two to the N to give some motivations, like if you consider the set zero, one, that's a set of size two. And a infinite sequence is like an infinite Cartesian product, because you can think of a sequence as an ordered list of infinite length. And so as we've done things, like if you have a set, a squared gives you the Cartesian product of ordered pairs, a cube gives you the Cartesian product of ordered triples, ordered quadruples, ordered quintuples, et cetera. We could say like A to infinity, but now we know that there's different infinities so that could be ambiguous. So we might say that, oh, okay, A to the N is then the set of countably infinite ordered pairs, which of course is exactly what a sequence is. And so that's where this notation is coming from. So two to the N is the set of binary sequences. And this set of course is going to be uncountable. And we're gonna see this with a very classic argument due to Cantor here. All right, and this is gonna be approved by contradiction. So let's suppose that two to the N is in fact a countable set. That is the collection of binary sequences could be ordered, that is it could be listed. We can enumerate all of them. So there's an A zero, there's an A one, there's an A two and A three. Every binary sequence has an index associated to a natural number. Now for each of these sequences, let's look at one somewhere in this list here. You take this binary sequence AI, it shows up in the sequence of sequences, but in itself is a sequence. So AI is a list of numbers zero and one. So we'll call the first one there AI zero, AI one, AI two, AI three, AI four, et cetera, et cetera, et cetera. So now we have a sequence of sequences. So we can list them and actually form this type of two-dimensional array, like an infinite matrix. Let's consider some things like this before. So here is a zero, here's a one, here's a two, here's a three, here's a four, and then continue on. Since there's cannibly many of these things, I can list all of the sequences and then actually write all the terms by term. I'm gonna get this infinite matrix here. So now the reason this argument has its name is I want you to consider the diagonal of this array here. We're gonna construct a sequence, a binary sequence we're gonna call it B by the following rule. The nth position of B is going to be the negation of the nth position of the nth sequence in this sequence here. So that is looking along this, we're going to change the nn position of this matrix. So that is to say that if the nn position was a zero, define bn to be one. If the nn position was a one, we'll define bn to be zero. So just toggle between zero and one, like so. I claim that this sequence, this binary sequence B does not live inside of our list anywhere. It doesn't. So in fact, B does not equal a n for any of the a n's that we have previously mentioned. And why is that? Well, if B was equal to one of the a n's, what that would mean is that term by term, they would always agree with each other. So there's zero positions would agree, their first positions would agree, their second positions, their third positions, their fourth positions, their 5,280th positions would all agree with each other. But when you look at the zero sequence, b and a zero, zero, sorry, b and a zero disagree with each other because they disagree in the zero position. If a zero, zero was one, b by definition would then be zero. Or if a zero, zero was zero, then b zero would actually be one. I chose b zero to be the opposite of a zero, zero. So we get that b does not equal a zero, okay? What about the next one on the list? Well, I also claim that b does not equal a one because if b equaled a one, that means that b one would have to equal a one one. But by construction, if a one one was zero, then b one would be one. But if a one one was one, then by construction b zero would be one. So these actually disagree with each other, okay? And this then happens as we go down the list, b two disagrees with a two two. So b is not a two. And likewise, b three disagrees with a three three. Therefore b doesn't equal a three. b four doesn't match a four four. So b doesn't equal a four. And we can do this for every term in the sequence so that there never is a position where a four, a n could equal b. So b never equals a n. So b is not in this list. And this contradicts the assumption that the binary sequences are countable because if they were countable, this list would contain every binary sequence. But I've now constructed a binary sequence that's not in the list. I missed something. And it's not like I can just add it back in because if I added it like in the zero position, it's like, whoops, I forgot one. Let's put it in the list somewhere. I could then repeat this argument, construct a new element that's still not in the list. And therefore, since we found a contradiction here, we have to then conclude that our original assumption was bad, that the set of binary sequences is not countable. Therefore, since it's infinite, it has to be uncountable. So we've now established that the cardinality of the natural numbers that is a-lif zero is strictly less than the cardinality of binary sequences, okay? Now let's look at a very similar result. We're gonna argue that the set of binary sequences, again, are set two to the end, has the same cardinality as the power set of the natural numbers. So in particular, this should be the power set of the natural numbers there, which be aware just so you know, we've proven previously that if you take the cardinality of any set, this will always be strictly less than the cardinality of its power set. We've shown specifically that these things are not equal to each other. With finite sets, it's obvious that they get bigger, but it's not too difficult to argue here that this in fact holds, because given any set X, you can map into the power set by the rule that little X will map to its single 10. This would be a one-to-one map. And so we do get that the cardinality of X is less than or equal to the cardinality of the power set. If you also apply the fact that these cardinalities are distinct, we get that every set is strictly smaller than its power set. So the power set of the natural numbers has to be an uncountable set because it's bigger than the natural numbers. And so we're gonna show that these things are actually equal. And in fact, this would then provide a second proof that two to the end is uncountable because it's the same cardinality as an uncountable set that we already know is uncountable. I'm not gonna go through all the details of this one because in some respect, it gives us something we already know, but I really do want this connection right here. I'm gonna give you the bijection. So this bijection is the following. So we're gonna go from two to the N to the power set of N given by the following rule that if you have some binary sequence A, then, so again, A is some sequence consisting of A0, A1, A2, A3 all the way down. Okay, that's what A is, it's a binary sequence. Using A, we can construct a subset of the natural numbers by the following rule. F of A contains the element N if F of N is equal to a one, all right? If F of N is equal to one, then N belongs to the set F of A. If F of N is equal to zero, then N doesn't belong to F of A. And this then constructs a subset of the natural numbers. You can think of this as an infinite decision tree that whenever you go along your binary sequence, you include the element if the number was one and you exclude the element if the number was, if the number was zero. So if we take a sequence like zero, zero, one, one, one, zero, one, and let's go from there. What this means is it's like, okay, we're building our set F of A, this is our sequence here A, F of A. Well, we don't include zero, we don't include one. We don't include zero, we don't include one. We do include two, three, and four. Two, three, four. We don't include five, we do include six. So we then get six, and then we keep on going. I mean, I only have the first initial segment here. If this thing ever converges to repeating zeros, this would give us a finite set, but most of the time you'd probably expect this to be an infinite set here. So this gives us a map from binary sequences into subsets of the natural numbers, using the sort of like infinite decision tree here. I'm gonna leave it as an exercise to the viewer here to prove that this is in fact a bijection. We have to show that it's one to one and onto, or conversely, you could just produce a inverse to this map, which wouldn't be too difficult to do. All right, so that then gets us to the one that I want to talk about, the set of real numbers. So let's come back to the continuum here. The set of real numbers has the same cardinality as the power set of the natural numbers. And how do we see that? Well, how we're gonna prove this is we're gonna show that the real numbers have the same cardinality as two to the n. We have just shown that two to the n has the same cardinality as the power set of the natural numbers. And so as this is a transitive relation, if this is the continuum, then this has to be the continuum as well. Now, how are we gonna show that these two sets have the same cardinality? Well, we're actually gonna come up with not a bijection. We're gonna come up with an injection. Bijections are possible, but a lot harder to describe here. Let's first come with it up with an injective map from two to the n to r. And how we're gonna do that is the following way. Again, f of a, we have to come up with an image for some binary sequence. This binary sequence, we can think of it as a set. It's an ordered set. So there's some a0, a1, a2, a3, continue on. Using this binary sequence, we can actually form an infinite series, like from calculus two here. So we're gonna take the series where n equals zero ranges from zero to infinity. So we grab every term of the sequence. We're gonna come up with the fractions for which the numerator is a sub n and the denominator is 10 to the n plus one, okay? So this is actually given us a number that is less than, it's gonna be somewhere between zero and one, just so you're aware. I'll leave it up again as an exercise to the viewer here to prove that this is an injective map, that no two different sequences have the same evaluation, right? Because this series is a convergent series, okay? I'll let you kind of prove that. Again, that's a calculus two argument there. But the thing is, if these were all ones, this would be a geometric series. And but sometimes there's zeros gonna show up in there. So by the comparison test, this would likewise be convergent and assets. This does give you a real number. It might be irrational, it could be rational, probably irrational, but this does give you a well-defined map. We just need to prove that it's in fact one-to-one, that different sequences will give you different series, okay? Then in the reverse direction, because the fact that this gives us a one-to-one map shows us that the cardinality of binary sequences is less than or equal to the continuum. Now to reverse the direction, consider the following maps. Take any bijection from the real numbers to the unit interval. We have already considered such. We did some of those two lectures ago. So use any one of those, okay? A bijection from the real numbers to the unit interval. Then we're gonna construct an injection from the unit interval into two to the end by the following rule. That what we're gonna do is essentially try to reverse the process we did a moment ago, that we are gonna take the numbers in the unit interval. Again, there's some arguments to be provided here that I'm not providing the details for. But every number in the unit interval zero to one can be written in a decimal expansion. That's why I'm using base 10 here. That if you take any number between zero and one, the whole number part's gonna be zero. Then just look at the decimal expansion. But we wanna do this in binary, binary here. So I guess I actually misprinted here. This should be a two. We don't wanna do the decimal expansion. We wanna do the binary expansion so that these coefficients are either zero or one. Whoopsie about that one. So we wanna do this so that we get the following here. Now, there are some things I should make mention here is this function even well-defined. There is an issue that when you have repeating decimals or in this case, repeating binaries, you actually could get different representations which potentially could make different things. So you have to deal with that. Like basically what I'm saying is like the issue is if you have the decimal expansion, we'll do decimal expansions. If you have 0.99999 forever onward, right? So this is 0.9 repeated. This is actually the number one itself. And likewise, if you did like 0.09 repeated, this is actually one-tenth, right? So there are different ways of representing a repeated decimal or a rationals in this case. So you have to be a little bit careful about that. So again, I'm hiding some of the details in this argument here. But avoiding these issues, there are ways you can find a unique representation for these numbers to make this into a well-defined map. In particular, however you choose it, different numbers will have different power series representations. The same power series cannot represent two different numbers unless they're equal to each other. So however you decide to handle numbers which have multiple power series representation, it doesn't matter. If you can make this thing into something that's well-defined, it will then be injective. This map is injective. And then if you compose an injection with a bijection, that itself will be injective. This gives us the containment in the other direction so that the continuum is less than or equal to the cardinality of the binary sequences. Therefore, these are both the cardinality of the continuum here. So I do wanna make a comment here that we now know that this set, which is the same cardinality as the power set of the natural numbers, which has the same cardinality as the reals. These are all continuum cardinalities. We note that the continuum is actually, it's larger than ale if not, but it's actually really, really, really big. I want you to think about like the following situation here, the following thought experiment. Using only straight line segments, we could create an entire alphabet of so-called block letters. And this would be in a manner like similar to like eight-bit video games. So we could say something like here's an A, right? And then although that comma needs to not be curved, so like a slant there, we could do a B, and that kind of looks like an eight, but we could fix that if you wanted to put a little bit more angle to it, something like that. We could do a C, we could do a D, we could do an E, we could do an F, G wouldn't be so difficult, something like that. And we can keep on going, right? We could do the whole alphabet. We did capital letters, we could also do lowercase, but in particular, we could make block letters that then resemble the English alphabet. I mean, people have been doing this with low graphics all the time. I mentioned like eight-bit video games. I love those as kid, but they have these blocky-like letters. Each letter, when you draw it in the plane, like think of that A for a moment, and you can think of it in terms of, in terms of like the X and Y axis here, so the X axis, the Y axis. You could think of this A, and I'll use a different color to illustrate here. This A, you can think of as a subset of the plane here. Now, the plane we usually think of as R2, the fact that we use straight lines only means that each of these can be represented as a line with a rational slope and a rational intercept. We really could visualize this not as a subset of the real plane. We could think of this as a subset of Q2. That is, what if we only allow rational coordinates? We could still create this alphabet in that scenario using rational coordinates. That is, we don't allow any points with irrational points. So we could visualize the letter A as a subset of Q2. We could do that for every letter, B, C, D, E, F, G, et cetera. And then if we start to put the letters next to each other, we can start to make words, right? So if you put the letters A and D together, you start to spell a word like and. With respect to a set, it's like you took this set, union that set, union that set. And so you can concatenate letters together using unions and make a larger subset of the plane. Well, if you then have words, you can repeat this process to make sentences. That if you take unions of words, then you can form a sentence. Throw in punctuations there as well. Then if you have sentences, you can take unions of sentences to make paragraphs. The paragraph we can also identify with a subset of Q2. Every paragraph that's ever been written, we can identify with a subset of Q2. And then if you can make paragraphs, then you can take unions of paragraphs and we can start writing documents. We can start writing books. We can start writing novels. And so in this manner, every book ever written could be identified with a subset of Q2. And since it's a subset of Q2, any book you ever wanted, we could think of it as an element of the power set of Q2. So I mean, this will include things like The Hobbit, Harry Potter and the Sorcerer's Stone, the Bible, any book you could think of is identifiable as a subset of Q2 and therefore it could be identified with an element of PQ2. That's a very interesting thought. But it even goes beyond that. This doesn't just include every book that's ever written. You could identify any possible book as an element here. So any fan fiction that's ever been written, let me know you're in a real book, right? When I was in high school, I liked to write these Dragon Ball Z fan fiction. Those are also inside of this set. Every head canon you've ever come up with, you're like, oh, I didn't like the ending to Star Wars Episode VIII. I'm gonna change it so that this happens instead. That also, if it was written down, is a book and that would be part of the set. Your autobiography is inside of this set, right? I mean, if you're watching this and I don't care who you are, whoever is watching this video, your biography is inside of the set. Even though it's not complete yet, because you're still alive and haven't finished your life yet, it's still in there. And in addition to your biography, there's an infinitude of alternative biographies of you or more specifically of your variants that are living in alternate dimensions that exist inside the multiverse. Every one of those possible biographies is inside of this set. PQ2 is huge. It has all of that information. All the information plus a bunch of bunch of gibberish, right? All of the plays that monkeys have tried to type that didn't turn out very good. And there are a bunch of gibberish there inside this set as well, of course, right? This is a huge set. And remember that Q2 is the Cartesian product of Q times Q, which these are countable sets. The Cartesian product of two countable sets is likewise countable. And so in fact, Q2 has the same cardinality as N. Therefore, this set right here has the same cardinality as P of N. So there exists a bijection from PQ2 to PN, all right? So all these books that exist here could be identified here. And likewise, P sub N could be identified with the natural of the real numbers, excuse me, because there's a bijection from this set to that set. So any book here if it follows through that bijection then is a real number. There is some number along the x-axis that literally has your name on it. Hope you can find it someday, right? So let's kind of summarize what we have discovered so far. Because we've been talking about the countable infinite sets. We've talked about the continuum here, but this is just the tip of the iceberg. Consider the following sequence of infinities. We have countable infinite, this is a list of zero. Then we have the cardinality of its power set, which is larger. This is the continuum as we have just discovered. But then you could take the power set of the continuum. You could take that power set of the power set of the continuum. You could take the power set of the power set of the continuum, et cetera, et cetera. And then you can keep on going, taking power sets and power sets and power sets. And in particular, every time you take another power set, you take another set, that is you get a new infinity. It's larger than the previous ones. So if you progress this down this sequence, you're gonna get infinitely many infinities, countably infinite to be specific, but you have countably infinite distinct infinities. That's a lot of infinities there. There's infinitely many infinities. But it doesn't stop there because if you took all of these sets listed in this sequence, what will happen if you take the union of all of them? That you can argue will give you a bigger set with a larger cardinality than any of the ones in the list. And then that set, we'll call it, we'll call it, I don't know, X here. That set then has a power set, which then has a power set, which then has a power set. And each time I take another power set, I get something with a strictly larger cardinality, right? So then what happens if you take the union of this set, of all of these sets, I should say, this is gonna give you another sequence, a new set that's bigger, for which you could take its power set, its power set, its power set, you could go on. Then you take the union of that sequence, you're gonna get a new set and you can start taking power sets all over again. And then you can keep on doing this. In which case then you get an infinite sequence of infinite sequences of sets, all of which have larger cardinalities. Each and every time the cardinality got bigger. But then once you've completed this, like this two-dimensional array of sets, you could then take the union of all of these, and that gives you a new set, for which then you can take its power set, its power set that gives you a sequence, then you take the union of that sequence, you can then get a new sequence going, then another sequence, then another sequence, take a union, you get a second two-dimensional array that's bigger than everything. You can then repeat this process to make another two-dimensional infinite array, and then another, and then another, then another. So you can create infinitely many two-dimensional infinite arrays. If you take the union of all of those two-dimensional infinite arrays, you then are going to get a three-dimensional infinite array. I'm not gonna even try to draw that. For which then with that array, you can take the union of it, that gives you a bigger set for which you can then start taking power sets, power sets, repeat this process. You can then create, like there's a lot of infinities floating around here, but you can take another three, you get another three-dimensional infinite array, and then another, repeat recursively. You can then get another three-dimensional array, another one, another one, another one, take the union of all of those, you get a four-dimensional array. Repeat this process, you get another four-dimensional array, another four-dimensional array, another one, until you get infinitely many four-dimensional arrays. Taking the union, you get a five-dimensional array, repeating that process, and getting infinitely many five-dimensional arrays. You can take a union and get six-dimensionals, right? And then keep on going, right? until we've now created infinite dimensional array of infinities, all these different infinities, right? And this is still just the tip of the iceberg right here. Are these examples in Pogmaw? This is still only given as countably many distinct infinities, right? In the end, the moral of the story is that you're gonna be able to get, we can keep on going, keep on going, keep on going. There's gonna be uncountably many infinities. I mean, there are, I can't, I mean, honestly, I can't even tell you how many infinities there are because there's not a word to describe how big this process can get. I mean, if anyone feels dizzy like I do right now, I totally get you, right? I mean, because this process never ends. It really gives new meaning to buzz year, buzz light years, a catchphrase to infinity and beyond. But let's get back to the original question we asked ourselves. What about ALF one? Where does ALF one sit inside of this list? Does it sit anywhere? Did I, did I miss it? Well, one natural candidate is that it's the real numbers that we had mentioned before that the so-called continuum hypothesis states that the continuum is the smallest infinity larger than countable infinity. That is the continuum is ALF one. Now, this is listed as a conjecture because the statement is not, it has never been proven true nor has it been proven false technically. Let me kind of explain what that means in just a second. So look, is it true or false? Well, there's two compelling arguments why the continuum hypothesis would be true or false. In defense of the true, no one has ever proven there's a set A that sits strictly between the natural numbers and the real numbers, such that the cardinalities are strict. No set has ever been discovered, okay? And as such, I mean, and some of the greatest mathematical minds have worked on this problem, right? And so the fact that no one's found anything does seem to give a little doubt of its existence. Now that itself is not a proof, right? I mean, it could be that it still exists but just no one found it, right? But no one seemed to be able to find anything that sits between the two. So maybe, maybe it doesn't exist. On the other hand, as we've been talking about in this lesson, the cardinality of the continuum is huge, huge, huge, huge, right? Every document ever that could be written is inside of this set, right? And as such, it feels like it's too big so that there's nothing in between, right? There's gotta be something in between, right? How? I mean, we haven't found it but it feels like maybe there ought to be, right? How could it not? It's so stinking huge. Now, what's the actual answer to the continuum hypothesis? The continuum hypothesis is true and it's actually false. Wait, wait, wait, wait. How can it be true and false? What does that even mean? Well, to be more specific, the truth value is dependent on the model of set theory that you're using. There exist consistent models of set theory where the continuum hypothesis holds. It is a theorem of that theory. There exist other consistent models of set theory for which the continuum hypothesis is actually false and its negation is a theorem of that theory. It comes down to basically the ZFC axioms that we had talked about before, okay? The ZFC axioms tell us what you can or cannot do with a set. And it turns out that the ZFC axioms are insufficient as a logical system to prove or disprove the continuum hypothesis. The proper logical word here is that the continuum hypothesis is independent of the ZFC axioms. There exist models for which the continuum hypothesis holds and some where the continuum hypothesis fails. And so this curious topic that we end our lecture series with is a fun one because it kind of shows us what's in front of us. The mathematical community at one point is gonna have to decide is the continuum hypothesis an axiom of our set theory or not? Because I say there are some models of set theory for which it holds and some that where it doesn't. Which model of set theory have we been using in this lecture series? What model of set theory does the mathematical use at large? I mean, other than like very specialized set theorist. What set theory model are we using? We're not exactly sure. In some regard we have to figure that out. In some regard we're gonna actually decide whether the continuum hypothesis holds or not. And there are huge implications based upon whichever decision. And so with that, with this very curious ending to our lecture series I want to illustrate with the continuum hypothesis that now as we reach the end of lecture 36 we've now completed our transition from elementary mathematics you could say naive mathematics to advanced mathematics. Now that does not mean that we are done. No, no, no, no, no, no. We are nowhere near the end of our mathematical journey. This right here is only the beginning. It's the end of this lecture series but the beginning of the study of mathematics. So I do appreciate everyone who's been participating in this lecture series. Participate all the comments that people have shared here. If you do have any comments in the future of course feel free to post them on this video or any videos in this lecture series whatsoever. And I'll gladly answer them as soon as I can. Like these videos if you learned anything from them. Subscribe to the channel if you wanna see more things like this in the future. There are many other mathematical courses you can find here on the channel. Feel free to take a look at any of them. And I hope that you have an enjoyable mathematical journey and that this truly is just the beginning for you. Not the end.