 Throughout this last chapter of our lecture series, we keep on having these analogies. Group theory is to Simi lattice theory, as lattice theory is to ring theory, as rings with unity is to bounded lattices, and finally as fields are to Boolean algebras. Boolean algebras are the most highly structured objects in the lattice setting, and they're kind of like fields when you make that nice comparison. Now another thing we studied when we studied ring theory is the idea of a unique factorization. So it begs the question, do lattices or Boolean algebras have unique factorization? Well if a Boolean algebra is quote unquote atomic, then it does in fact have unique factorization. That's what we want to do right here and use that actually to characterize all atomic Boolean algebras up to isomorphism. In particular, since every finite Boolean algebra will be atomic, this will classify all finite Boolean algebras up to isomorphism. So what is an atom? For a Boolean algebra to be atomic, what is an atom? So let B be a Boolean algebra. We say an element A inside the Boolean algebra is an atom if for every element X in the Boolean algebra, if X happens to be less than or equal to the atom, it's either because X is the atom itself or X is equal to 0, the minimum element of the Boolean algebra. So atoms are exactly the minimum non-zero elements of a Boolean algebra or equivalently an atom, an A, that is an atom so that whenever you have the statement X join Y equals A, it implies that either X was equal to A or X is equal to 0. So atoms in a Boolean algebra are kind of like irreducible elements. That is any factorization that results in giving you an atom, either you took the atom itself or you took a unit, right? I mean, this is the zero is of course the identity with respect to join. So atoms just be clear, atoms are the irreducible elements, so to speak, for the join operation. This is not with regard to meat. We're factoring things uniquely using joins. That's what our goal is going to be. Now much like a ring theory, you don't necessarily expect irreducibles to always exist. Certainly, there are settings for which we can guarantee they exist like a unique factorization domain. You always have irreducibles and in fact irreducibles and prime numbers are exactly the same thing. You know, if you have a no theory in ring, you can guarantee they exist. There are sometimes they do and sometimes they don't. The same can also happen amongst lattices, even on Boolean algebras. You don't always have atoms. If atoms basically always exist, we can say that the Boolean algebra is atomic. In particular, a Boolean algebra is atomic if for all elements inside of your Boolean algebra, there exists some atom A that lives inside the Boolean algebra. It's an atom such that A is less than or equal to x. And I guess we should specify here that x is not zero. There's no atoms below the zero element because it's the minimal element, but if you take a non-zero element, a Boolean algebra is atomic if it has an atom that is less than it. Again, there's analogies to how we did unique factorization on rings. That's the same idea there. First of all, all finite Boolean algebras are going to be atomic. They have atoms because basically since your Boolean algebras finite, you have that minimal element, which is zero. Then we're going to take the set of all elements that are basically one step above zero. There's nothing between zero and that element. Since the set's finite, if you look at the order, you can't have an infinite descent. So eventually, if you're descending down any element, you're eventually going to have to put a place that is right before zero and you get an atom for finite algebras, finite Boolean algebras in that situation. Now, of course, there do exist infinite atomic Boolean algebras. The best example, of course, is going to be the power set of a set for which if x is an infinite set, that means its power set will likewise be infinite. Soon enough, its cardinality will be probably larger than the set x, but I digress in that situation. If we have an infinite set x, its power set will be an infinite set as well. In that situation, the single tens, I can very easily describe them, the single tens are going to be the atoms of that Boolean algebra. Therefore, a power set algebra is always going to be an atomic Boolean algebra. There do exist, of course, infinite non-atomic Boolean algebras, although that takes us beyond where I want to go in this lecture. It's going to go beyond the scope of our course. We are essentially at the end, our very last lecture here. So instead, what I want to do is prove, you know, classify what a Boolean atomic algebra looks like. And so that's what our theorem is going to do that. Now, this theorem is going to be stated in terms of finite of Boolean algebras. And that's mostly just because that's how the statement is phrased in the book. But honestly, I could erase the word finite, and I could just say an atomic Boolean algebra, and then let x be the set of atoms. And then we're going to then argue that b is isomorphic to the power set of x as a Boolean algebra. And in particular, the order of any atomic Boolean algebra is necessarily going to be 2 to the cardinality of x. And that's because that's the cardinality of this thing right here. So the last statement is a very quick corollary. We want to prove that these things are isomorphic. Now, what does it mean for Boolean algebras to be isomorphic? What does it mean for lattices to be isomorphic? What does it mean for semi-lattices to be isomorphic? We've never talked about those things. Now, the notion of a homomorphism, I hope by this point in the lecture series, is much straightforward. We want to map between, like if you want a homomorphism between semi-lattices, you want to map between two semi-lattices so that the binary operation is preserved. In a lattice, we want to prove that a homomorphism between lattices, that it's a map which preserves the two operations of meet and join. If you're looking at a homomorphism between Boolean algebras, we need to preserve the two binary operations of meet and join. But we also need that complements map to complements. So, phi of x prime is equal to phi of x prime. And I will leave it as an exercise for the viewer here to prove that if you have a homomorphism that preserves meet and joins, then it necessarily makes the map an increasing map. That is, it preserves the order of the partial order set. And then as a consequence, it'll preserve complements as well. So, if we have a homomorphism between Boolean algebras that preserves the meet and join operations, then it'll automatically cover complements as well. So, I don't need to cover that one. I just need to come up with a, well, for an isomorphism, that's going to be a bijective homomorphism, of course. And so, what we need to do is prove there exists a bijective join and meet preserving map between this Boolean algebras and that Boolean algebras. And then the two will be isomorphic. And so, with regard to the theory of Boolean algebras, those two objects are one and the same thing. There's no reason to distinguish between them. Now, the proof is going to be based upon the following. We are going to prove that in an atomic Boolean algebra, every element has a unique factorization, the so-called atomic factorization. This is analogous to a prime factorization in a unique factorization domain. This factorization, of course, will be unique up to reordering. Because every element is idempotent, we don't have to worry about repeated atoms because repeats make no difference when your operations are idempotent. And we also don't have to worry about associates in this situation because there are no units in a Boolean algebra. The algebraic structure doesn't work that way. So, the first thing we're going to do is we're going to prove that every element in an atomic Boolean algebra has a unique atomic factorization. So, suppose that x is some generic element of the Boolean algebra. And suppose it has something, something less than it, which, of course, it could be that a could be itself. It could be zero. Most likely, a is going to be an atom because after all, since this is an atomic Boolean algebra, every element has an atom less than or equal to it other than zero. But for this next statement, a doesn't have to be an atom. This is true for any situation where a is less than or equal to x. So, x, of course, is equal to x meet one because one is the identity of meat. By the complement axiom, one is equal to a join a complement for which then we can distribute x across the join right there. So, we get that x meet a join x meet a prime is equal to that as well. For which, because a is less than or equal to x, by definition, that means that x join a is equal to a. So, remember that this symbol right here means that a join x is equal to x and it means that a meet x is equal to a. So, when we take x meet a right here, that's equal to a. And so, we then have the following statement. Now, I claim that Boolean algebras are essentially the same thing as just power sets with unions and intersections. So, I want you to think about what this statement we were saying if we were in the set theoretic setting. So, let's say that we have some set x. I'm claiming that if a is a subset of x, then we can actually decompose x as a union x take away a. Like so, so you have, so a, excuse me, the set x can be decomposed into a subset and to its set complement. That observation, that decomposition works for any element inside of a Boolean algebra. If you have x and something less than it, you can decompose x into the a part and the not a part. That's all that thing is saying, okay? So, next, I want us to suppose we have two distinct atoms that belong to an element x. Now, clearly because they're two distinct atoms, if we take a meet b, that has to equal zero, right? When it comes to an atom, what's gonna happen here is that if a, you know, a meet b, you want something smaller than a and smaller than b, which the only thing smaller than a is a and b, the only thing smaller than, excuse me, the only thing smaller than a is a and zero, the only thing smaller than b is b and zero. And so then the greatest lower bound has to be zero. So the intersection of atoms is always equal to zero. All right, so next what I wanna do is make the following observation. If you take your atom b, or again, we are assuming b is an atom in this situation, this exercise is gonna be very similar to the one we just did here. Well, b is equal to b meet one, for which by compliments, one is a join a compliment. Then we can distribute b, like I said, this is very similar. You're going to get b meet a join b meet a compliment. And as we observed a moment ago, a meet b is equal to zero, you get zero and adding zero or, excuse me, joining zero to anything doesn't change anything. So we end up with b equaling a meet a prime. And so from that statement, we can then make the inference about the inequality here. b is less than or equal to a prime, right? So in particular, b is also less than or equal to b intersect a prime, because b is less than equal to b and b, as we just observed is less than or equal to a prime. And so b will be less than or equal to their greatest lower bound, which would be b intersect a prime in that situation. All right, so now we're gonna get to the heart of our observation about atomic decomposition. So we have these two atoms, a and b. So by the very first statement that we proved in this proof here, right? So just as a reminder, what we said here is that, whoops, if a is less than or equal to x, then we can decompose x as a union the compliment. So x meet a compliment, like so, okay? And so honestly in Boolean Algebra, sometimes people introduce the notation that x minus a by definition is x meet a compliment, much like we do in set theory. Although in set theories, often use this more of slant minus sign, that's fine. That's fine, you can use that if you want to. So this is just saying that, okay, x decomposes into a union x minus a, okay? So you have that statement right there, that's the first statement we did. But we just observed that, we just observed that b is gonna be less than a prime right here and b is also less than x. So in particular, b is going to be less than or equal to x meet a prime, right? These are the things we just observed from before. And therefore this statement applies to that one right there. If we make that statement there, I'll write it again, if b is less than or equal to x meet a prime, then we get that x meet a prime can b decompose into the b part union the not b part. So we end up with x meet a prime meet b prime. I don't have to worry about parentheses in that last case, because meet is an associative operation. And so this right here can be substituted with this right here. So we get b join a meet a prime meet b prime like so. And again, I don't need parentheses here because join is an associative operation in that situation for which then we're going to re-associate this like so. Let me use a different color to emphasize that because we have x meet a prime meet b prime. So if we re-associate there, you get x meet a prime meet b prime, but by De Morgan's law, a compliment meet b compliment is a join b compliment. So making that substitution right here, we get our final statement. So that x with these two different atoms, x can be written as the a part plus the b part plus the not a b part, all right? And so by induction, we can continue this process for each atom a one a two all the way up to a n. Each of those atoms belong to x, a i is less than or equal to x. And by similar reason as we had up here, a i does not belong to the a one a two a three up to a minus one part. So by induction, we can continue this process. We end up with something like x meet, and then you're gonna have a bunch of things, a join of all of these elements up to n, like so. And then we did not right there. So by induction, you can keep on doing this, all right? This is of course where the, this is where finite makes things a lot easier. This process has to eventually terminate for an infinite atomic Boolean algebra. That's actually part of the definition. I wasn't very explicit about the definition here because this after all that focuses on the finite case. But for an atomic Boolean algebra, you do have the assumption that every element has finitely many atoms contained inside of it. Okay? Which again, I'm not gonna go through all of that. I don't wanna worry too much about infinite atomic Boolean algebras. Finite ones is gonna be sufficient for our purposes. So eventually this process is gonna have to terminate so that eventually there's no more atoms. And so this part basically becomes, this join is just going to die off with x. So this last part just vanishes off. So basically what I'm saying is by induction, we then get to this decomposition into atoms, the so-called atomic factorization. And up to reordering, we claim that this thing is unique. That is you don't have different atoms. You can't factor it one way with one atoms and factor it a different way with different atoms. And so let's look at that. Let's suppose we have two factorizations of the element x where you have the atomic factorization a one, join a two, join all the way up to a n. And then on the right b one, join b two all the way up to b m like so. You have these two different factorizations. Well, if the second factorization doesn't involve a one then look what happens when you take a, well, I mean a one, a two, a i doesn't matter. We could without the laws, generally assume it's a one, whatever else deal with a i since that's on the screen. If the second factorization didn't involve a i, right? Then take the meat of a i with x. Well, since that shows up in one of these things here by the distributive properties and all the other axioms of a Boolean algebra, you can argue that this thing is not gonna equal x jk. It's gonna equal a i, okay? But when you consider the factorization on the right, you're gonna get a i meet all of these b's. And so this is, let me show you the details that we did over here that you omitted them. You're gonna use the distributive law so you distribute the a i. So you get a i meet b one and then you're gonna get that a bunch of times a i meet b two, a i meet b three, et cetera, all the way through b m, right? And now since these are each atoms, none of those, well, most of those atoms, they're different from a i because there's no repeats here. After all, it's ident potent. If there's any repeats, we can consolidate them together, suck them together there. So we can, if a i doesn't show up in this factorization, then each of these meats is gonna equal zero all the way through. And therefore, if you take zero, join zero, join zero, et cetera, you're just gonna end up with a zero. But in conversely, if you do this with the a's, right, a n, all of these things are gonna be zero except there is an a i that shows up. And so this simplifies just to be a i. A i, of course, is not equal to zero. We get a contradiction. And so that tells us that a i is an atom that appears in this atomic factorization. And going through this, we then get a unique factorization. So Boolean algebras have unique factorization in the category of lattices. These are unique atomic factorizations. So Boolean algebras kind of like a UFD with regard to lattice theory, which after all fields are UFD, those two that matches up with our analog here. All right, so now we're in a position where we can define the isomorphism between B and the power set here. What we're gonna do is we're gonna map an element x to the set, a one, a two, all the way up to a n. Remember, x here is the set of atoms of the Boolean algebra. And so this right here is a set of atoms where these atoms, a one, a two, up to a n, these are the atoms that come from the atomic factorization of x. So if a i is an atom of x, we put it inside the set. And so that's what we map to. This clearly is going to be a subset of x. So it is a map into the power set of x. We wanna now show that it's homomorphic. Let x and y be two elements of the Boolean algebra. And suppose their atomic factorizations are the following. x can be factored into n many atoms, call them a one through a n. y can be factored into m many atoms, call them b one through b m. And so then if we take the image of this map under the join of x and y, this would then, well, if you take x with it substituted its factorization to the same thing for y, we get the following here. You're gonna get x one join a, excuse me, a one join a two all the way to a n, join b one, b two, b m, like so. For which this, since this is a factorization of atoms, it'll map to the set a one, a two up to a n, and b one, b two up to b m. For which that set naturally can be decomposed into a union for which you can put all the a's together in one set, the b's together in the second set. There could be repetition in these sets. I mean, there could be overlapping atoms between x and y. That's not important in this. That doesn't frustrate anything here. Clearly this set is the image of x because those are the atoms of x. And this set is the image of y because those are the atoms of y. So we see that this map very naturally preserves the join operation. And that's mostly because we defined it using the atomic factorization, which comes from how to decompose an element using joins. The meet operation is gonna be a little bit more challenging because we have it that it's join homomorphic. Why is it meet homomorphic? Well, we're gonna do the same basic thing, although this one's gonna be a little bit more convoluted but we'll handle it just fine, no problem. Now, if you take phi of x, meet y. Well, we have the factorization for x. We have the factorization for y. Since our Boolean-Alderist distributive, it's kind of like a foil type thing. You have to do all the possible combinations. It's like the foil method, which is a consequence of the distributive property. So when you multiply this out, you're gonna have a bunch of different, you're gonna have a join of a bunch of meets. And those meets are gonna look like some A meets some B and you allow the indices i and j to run from one to n and one to m respectfully, right? But these are atoms. These meets will either equal zero for which inside of a join a zero doesn't do anything or it's equal to A i, right? I mean, and that would happen only if A i equals B j. So whenever A i equals B j, you retain the A i but if they are distinct elements, their meet is zero and the join, since it's the x zero is the identity of join, it'll just vanish and so you don't have them anymore. So this sum, this join simplifies down to be whenever we have a match, right? For which then as a set, this is now a decomposition into atoms. It's an atomic factorization. And so that means this will map to the set when we grab all of the A i's that appear in the set. Well, A i appears inside that decomposition exactly when A i equals B j for some i and some j, like so. All right, now this set, every set can be decomposed into a union of single tens for which those single tens occur only when A i equals B j. The same index set is in play here for which these single tens can be represented as intersections of A i intersect B j because if A i and B j are the same element, the intersection of single tens will just be the single 10 A i. But if A i and B j are different elements, then their intersection will be just an empty set. And if you throw an empty set into a union, it doesn't enlarge the union whatsoever. So these two unions are in fact the same thing, okay? For which intersections and unions are distributive. After all, it is a Boolean algebra. So we can factor this thing, reverse foil it into the following. We have a union of all of the A's intersect the union of all of the B's, which again, if you go backwards, if you foil this thing out by the distributive property, you get something like this. For which if you take the union of all of these single tens of A's, that's just the set of A's. If you take similarly the union of all these single tens of B's, that's just the set of B's. And if we take the, if we take, sorry, that's a typo right there, this should be an intersection symbol because it's the same symbol we have right there. I don't know why it turned into a union. So then we get the set of A's, which that's the image of X and we get the set of B's, which is the image of Y intersected together. And so therefore, voila, we then turned our meat into an intersection, which is what we expected to be. This then proves that fee is meat homomorphic. And as we already talked about compliments, compliments are preserved automatically by Boolean homomorphisms. So since we have a bijective Boolean homomorphism, this gives a Boolean isomorphism. So every atomic Boolean algebra is isomorphic to a power set with unions and intersections. And which admittedly we didn't do the infinite atomic ones, I confess, we kind of talked about it. So if you don't wanna believe that since we didn't apply the proof, just settle with the fact that every finite a Boolean algebra is, every finite Boolean algebra is in fact isomorphic to a power set. And as such, the order of this Boolean algebra is necessarily equal to two raised to, well, the number of atoms that are in the system. Oops, number of atoms. And that classifies all finite Boolean algebras.