when ···+1+1+1≠1+1+1···

追加: 2026-05-07

[00:00:00]Today, I would like to talk about the ordinal numbers, and we can think of these as some sort of infinite extension, or maybe a better word would be a transfinite extension of the natural numbers.

[00:00:13]And there are two main ways to think about the ordinal numbers. You either think, "Well, everything in this universe is a set." or you think, "Everything in this universe is an ordering." And these are the same approach really down deep. Well, of course they would have to be the same because we're defining the ordinal numbers, and if you define them two different ways, those ways should match.

[00:00:41]And so, I'd like to start by looking at the natural numbers using these two approaches, and we'll see how that will extend to these transfinite natural numbers, in other words, these ordinal numbers.

[00:00:56]And so, generally, I don't think about the number zero as being a natural number, but in this context of ordinals and ordinal arithmetic, we will think about zero as being a natural number.

[00:01:09]So, over here, in this universe or in this realization, where everything is a set, we take the natural number zero simply to be the empty set. Well, that makes sense because how many elements are there in the empty set? Well, there are zero elements in the empty set.

[00:01:27]And then, well, over here in this realization, where everything is order, how should we think about the natural numbers? Well, we should just think of them as like, maybe the empty order. I mean that there is nothing there, and there's no order that you can put on anything there.

[00:01:45]Okay, so now let's move on to the natural number one, and we'll define the natural number one to be the set containing the empty set. Or more appropriately, because we're building this off in some sort of or building this up sort of recursive strategy, we'll say that the number one is the set containing zero. And so, of course, if it's the set containing zero, or deep down it's the set containing the empty set, well, then it has one element. So, that makes sense. Our number one, our ordinal one, should be a set with one element.

[00:02:25]And then, if we go over here to a order, it's no longer an empty order, it's just simply a single element. You can't talk about what's bigger or less than if you only have a single element. Okay. So, now moving on to the number two, thinking of it as a set in this ordinal universe, and motivated by the fact that one had a single element when viewed as a set, zero had zero elements viewed as a set, well, two should probably have two elements when viewed as a set. And what should those two elements be? Well, how about just the two numbers that are less than two? So, that would be zero and one.

[00:03:07]And then, over there in the realization where everything is order, this is where things start to take shape. We would have this order of one being less than two. So, I'm using the labels over here one and two, but really it could be any two elements where you're putting in this order that one is less than the other. Okay. So, now let's move up to three because I think we can see where things are going, and my take when you're looking for some sort of pattern is you should see what the pattern is, and then do a couple more just to be sure. So, let's go over here to three. So, that should be a set containing three elements. It might as well be zero, one, and two.

[00:03:46]And then, over there, it should be an ordering of three elements. And it might as well be one is less than two, which is less than three. We'll use those names.

[00:03:57]Now, let's go dot dot dot down here to an arbitrary element in, and the arbitrary element in in this set realization should have n elements. And based off of what we've seen here, perhaps it should just have all n elements that are less than it.

[00:04:13]So, that would be zero, one, two, ending at n minus one. So, the set containing zero to n minus one is our ordinal n.

[00:04:24]And if we go over here, we could say, "Well, we need some sort of ordering on n elements." And more importantly, it's a total ordering on n elements, if you know something about partial ordering. So, that would be like one less than two, less than three, less than four, all the way up here, less than n minus one, less than n. So, it stops there at n.

[00:04:47]Now, if we were to go down here and think about what would be our first infinite ordinal, well, perhaps it would just be the set of all natural numbers, and that's exactly what it would be.

[00:05:00]That would be our first infinite ordinal, or sometimes said the first limit ordinal.

[00:05:06]And we generally call that as omega. And so, omega will be the set containing zero, one, two, three, so on and so forth.

[00:05:17]So, the important thing here is it has no largest element.

[00:05:22]Now, if we were to go over here to the everything is an order realization, we could think of this as a set containing countably many elements, and those countably many elements are in a linear total order.

[00:05:38]So, in other words, we would have one less than two, less than three, less than four, so on and so forth. So, there's no largest element, and every element has an element which is just smaller than it, and it has a single element that is just smaller than it.

[00:05:57]Okay. So, now we've got this realization of the ordinal numbers up to our first infinite ordinal, and also this one over here where everything is an order up to our first infinite ordinal. Now, in order to get more infinite ordinals, we really need to introduce some ordinal arithmetic. And well, we should probably start with addition because everything else will be built off of the back of addition. Okay, so let's see how addition should be defined. All right, so now we're ready to define addition of ordinals, and we're going to motivate our definition by looking at the following example. So, let's take the ordinal four. So, that's going to be the set containing zero, one, two, and three.

[00:06:42]And I'd like to observe that I can write that as the set containing zero, one, and two union with the set containing three.

[00:06:55]But now, let's observe that this is simply equal to three viewed as a set.

[00:07:00]Remember that three was the set containing zero, one, and two. Union with the set containing three.

[00:07:07]So, this four, which perhaps we should really think of as the number three plus one, can be represented as three union itself.

[00:07:21]And that's the motivation for how to define this successor ordinal, or in other words, starting with an ordinal alpha, we would define alpha plus one to be equal to alpha union the set containing alpha. And if you go back to our definitions, if you will, of the ordinal zero, one, two, three, four up to n, you'll see that whenever we had n plus one, it was simply the set containing or n union the set containing n. So, this matches what we saw with the natural numbers. Okay.

[00:07:56]And then, well, how would you add any two ordinals? Well, there are going to be two cases here.

[00:08:03]So, in the case that you can express the right-hand term in the sum as the successor of something, so in other words, we have alpha plus beta plus one, then we would define this to be alpha plus beta, and then plus one.

[00:08:23]And so, what we're doing here is defining this addition via transfinite recursion.

[00:08:29]And so, we'll define up to alpha plus beta, and then we can define alpha plus beta plus one as simply the plus one of what we got from alpha plus beta.

[00:08:41]And then, if you were to write this as sets, like if you wanted to do that, let's observe that this is simply equal to alpha plus beta union the set containing alpha plus beta.

[00:08:54]Okay, but then, if you think back to how we defined our ordinal omega, there was nothing that you could add one to it and get to omega.

[00:09:06]And that's because omega was something called a limit ordinal.

[00:09:11]And so, we need a case for this when we're working with a limit ordinal as well. And we would do it like this. We would say alpha plus lambda is equal to the union over all beta less than lambda of alpha plus beta. So, we define it in terms of this union.

[00:09:33]Now, I'd like to also point out here that this less than symbol is simply given by being an element of. And so, this means that beta is actually just an element of this lambda here. And so, that's um good to keep in mind that if we're defining our ordinals via this set construction, like we did before.

[00:09:58]>> [snorts] >> Then, the inequality that we get is simply just being an element of. We're creating this big, like, chain of things that are elements of each other. Okay.

[00:10:09]So, anyway, let's uh put down here that lambda is, in fact, a limit ordinal. So, I'll just put limit. And what that means is that there does not exist a beta with lambda equals beta plus one. In other words, it is not the successor or the next ordinal to any ordinal that we have. And now, we're going to give a little proof that addition has some nice properties. We'll actually prove that addition is, in fact, associative. And we'll prove this via what's called transfinite induction.

[00:10:44]And I just like to point out our base case is actually packed away in this defining rule for addition. So, alpha plus beta plus one is equal to alpha plus beta plus one. So, that's our base case, which we are inducting over this last term in the sum.

[00:11:05]Okay. So, let's say we've got an induction hypothesis. This is a strong induction hypothesis because we've already said that our base case is just part of this definition over here. So, let's suppose for all delta less than gamma, we have alpha plus beta plus delta is alpha plus beta plus delta. And now, we're going to have to work in two different cases. One is if we have a limit ordinal, and one is if we do not have a limit ordinal. So, start Let's start with the case when we do not.

[00:11:37]So, let's suppose that gamma is equal to mu plus one for some ordinal mu. So, in other words, gamma is not a limit ordinal. And now, let's look at alpha plus beta and then plus gamma. Let's observe that we can write that as alpha plus beta plus mu plus one.

[00:12:00]Okay, so there we have that. But now, let's observe by our definition over here, this is just simply equal to alpha plus beta plus mu and then plus one.

[00:12:12]But now, what we can do is apply our induction hypothesis to the inside of these parentheses because we have associativity there.

[00:12:20]So, that's going to give us alpha plus beta plus mu and then plus one on the outside.

[00:12:28]But now, we can apply the plus one rule that defines our addition, and that's going to give us alpha plus beta plus mu plus one. And let's see, I think it's got to be parentheses there on the mu plus one. But now, mu plus one was equal to gamma, so we in fact have just finished this all off. We have alpha plus beta plus gamma.

[00:12:54]So, there we have it. We've got addition is associative if we do not have a or limit ordinal.

[00:13:02]Okay, so now, let's look at the case when we do have a limit ordinal.

[00:13:08]And so, that means that we would have alpha plus beta plus gamma is equal to the union over all delta less than gamma of alpha plus beta plus delta. Again, that's by the definition of addition when we have a limit ordinal.

[00:13:29]But now, we can apply our induction hypothesis to the inside, and we'll have the union over all delta less than gamma of alpha plus beta plus gamma.

[00:13:41]Okay, good.

[00:13:43]But now, we can essentially just slide this union on the inside, and that will union this beta plus gamma up to beta plus delta, and leave us with exactly what we need. Sorry, this should have been a delta here. So, that's going to union that up to alpha plus beta plus gamma as needed.

[00:14:05]Okay, so we've got addition being associative. Now, let's check if addition is commutative or not. But to motivate the big picture of checking if addition is commutative or not, I'd like to do an example of addition via orders first, because that'll allow us to illustrate some weird stuff that's going on with our commutativity or, as we'll see, lack of commutativity for addition.

[00:14:32]Okay. So, let's just, as an example, do two plus three.

[00:14:37]And now, as orders, that's like taking this ordering one is less than two and adding to it this ordering one is less than two, which is less than three.

[00:14:50]And I've color-coded my numbers one and two and one and two and three in yellow and pink just to make a point here, and that's that these numbers are you don't They don't have to be natural numbers.

[00:15:04]They are just two elements in order or three elements in order. It's just convenient to have them different colored natural numbers. Okay. So, in this addition via orders model, essentially all we do is remove this plus sign, and we'll put a greater than or less than symbol, depending on which way you're reading it, in between those two.

[00:15:26]So, that's going to give us the following. So, we'll have yellow one is less than yellow two, which is less than pink one, which is less than pink two, which is less than pink three.

[00:15:39]So, it gives us that ordering of five elements.

[00:15:43]But let's observe that we can just rename some things here.

[00:15:48]And I guess maybe you could think of this is, you know, applying some sort of one-to-one function that preserves an ordering, and we would get one is less than two is less than three is less than four is less than five after renaming. So, that would be the ordinal five in the ordinal or in the order representation. And now that we've got our addition of ordinals in our set definition, like this over here, and our intuitive, maybe concatenation of orderings, we can look at addition of commutative those two ways or using those two models.

[00:16:30]So, let's start by looking at omega plus one.

[00:16:35]And let's observe that by our definition over here, omega plus one is simply omega union the set containing omega.

[00:16:45]But let's recall that omega was the set of all natural numbers. So, we could rewrite this as the set containing zero, one, two, three, so on and so forth, all the way up, kind of ending at the end with omega itself. Now, there are infinitely many things between three and omega, so that's important to think of.

[00:17:09]Okay, now, this would be the set theory way of thinking about it. Let's looking at Let's look at the order theory way of thinking about it.

[00:17:17]And so, we could also take omega plus one, and we could view that as this yellow ordering. So, one is less than two is less than three, so on and so forth.

[00:17:30]And then, we are going to add to that this pink ordering, which is just one.

[00:17:37]Now, what we'll do is, like I said before, our addition will just be concatenation of these orderings.

[00:17:44]But the important thing here is that there will be infinitely many things between three and one, even though that pink one will be the largest.

[00:17:55]So, we could write this as one or let's color-code it carefully. So, one less than two less than three, so on and so forth.

[00:18:05]All of those are going to be less than pink one.

[00:18:10]Now, we could rename some things, just like we did over here, and we could rename this ordering one less than two less than three, so on and so forth, and all of those are less than omega plus one.

[00:18:24]And the important thing here is that we have a largest element. So, I'll just say that this is the largest element, kind of obviously.

[00:18:35]And this is also what I'll call a limit point. And a limit point in this ordering is like a limit ordinal. It's something that doesn't have one person just right before it. So, there's no one thing right before this. Now, limit point and limit ordinal are a little bit different because I'm writing omega plus one there instead of omega. Omega is a limit ordinal. But you kind of have to twist your brain into these parallels being just like a little bit off.

[00:19:06]Okay. So, we've got omega plus one in both of these realizations.

[00:19:12]Now, let's look at what's going on with one plus omega.

[00:19:17]So, if we do one plus omega, so since omega is a limit ordinal, that's going to be the union over everything smaller than omega, which I'll call n, and so of one plus n. But look, that's just going to be the union of all natural numbers of one plus n because if n is less than omega, then n is a natural number. That's some nice thing that we can use there.

[00:19:47]And then, I won't prove this, but we do have commutativity of finite ordinals.

[00:19:53]And so, that means that this is the same thing as n + 1.

[00:19:56]But then, as a set, this is the union over all natural numbers of the set containing 0, 1, 2, all the way up to n.

[00:20:06]But if I union all of those sets together, I pretty clearly get the set of all natural numbers, meaning we simply get omega here. So, 1 + omega is omega. So, immediately we can see that omega + 1 and 1 + omega are not the same.

[00:20:24]That being said, I'd like to look at the model or the order model of this as well.

[00:20:30]So, we could have 1 + omega. So, that's like going to be the simple ordering on just one element and then plus the ordering on the natural numbers. So, 1 < 2 < 3, so on and so forth.

[00:20:46]And remember that our addition in this case was just concatenation.

[00:20:51]And so, that means we've got the following setup. We have 1 is less than I should say yellow 1 is less than pink 1, which is less than pink 2, which is less than pink 3, so on and so forth.

[00:21:06]But I think you can see that we can obviously rename those things and we just get exactly what we had before. So, we have 1 < 2 < 3, so on and so forth.

[00:21:18]The important thing here is that there is no largest element.

[00:21:22]So, when we had omega + 1, we did have a largest element. When we have 1 + omega, we get the same thing as omega, no largest element.

[00:21:31]>> [snorts] >> And then, omega + 1 had a limit point, whereas 1 + omega did not have a limit point. And that's actually how you can determine if certain infinite ordinals are equal or not by if they have limit points, how many of them, how they're distributed in the ordinal, as well as do they have a largest element or not.

[00:21:54]Okay, so now that we've looked at this addition of ordinals, I'd like to look at a nice visualization of this addition. So, we can think about omega as this infinite string of dots that's just trailing off to the right.

[00:22:08]And so, I'm having them get dimmer as they go to the right, but you can think about these as just going forever and ever and ever.

[00:22:15]And then, if we look at 1 + omega, that's like putting a single green dot before our omega dots that are trailing off to the right. But now, what we can do is take the smallest blue dot and one at a time recolor it green and slide it over.

[00:22:33]And you'll see that we get really the same picture each time. We have a smallest element, we have no limit points, and we have no largest element.

[00:22:42]And so, this is providing this motivation, this visual motivation that n + omega is just equal to omega, and that's going to be for all natural numbers n.

[00:22:53]So, that means that omega is the same as 1 + omega, which is the same as 2 + omega, which is the same thing as 3 + omega, so on and so forth.

[00:23:02]Now, in contrast, as we saw before, omega + 1 has a largest element.

[00:23:10]And then, omega + 2 is going to have a limit point, and then the point after the limit point will be the largest element. Omega + 3 will have a limit point, a point in the middle, and then a largest element. And so on and so forth.

[00:23:25]And as you can see, each of these is fundamentally different from what we had before. So, omega is different from omega + 1, which is different from omega + 2, and omega + 3, and omega + 4.

[00:23:38]>> [snorts] >> Like as a bit of a summary, omega + n are all distinct. Now, if we put that together with our previous result, we'll see that we get this infinite string of the commutativity of addition not holding. So, 1 + omega is not equal to omega + 1, 2 + omega is not equal to omega + 2, and so on and so forth. I guess the next logical thing to look at would be omega + omega. So, let's just start with omega + 1. So, we've got omega dots, and then bigger than all of those dots, we've got a single green dot, which is the + 1. That's our limit point.

[00:24:17]Then if we just add one at a time. So, omega + 2, omega + 3, omega + 4, and then we just keep going and going and going, we see that with each of these, we have a limit point and then a largest element.

[00:24:32]But then, if we take the limit of that scenario, we could call that omega + omega, and we see we have these two strings of omega elements. We've got a limit point in the middle, if you will, but then we've got no largest element overall. And so, fundamentally, for all natural numbers n, this omega + n is smaller than the omega + omega. And that's because each omega + n had a largest element, but the omega + omega had no largest element, while they both had a single limit point. Or as a bit of a classification, omega + omega was the smallest ordinal with a limit point and no largest element. Now, we did a proof that addition was associative, but I think there's a nice visual explanation of this as well. So, let's look at omega + 3 + omega. So, the omega + 3 can be seen as three dots that are larger than omega dots, and then after that, we've got omega dots trailing off to the right.

[00:25:36]But now, we can change the associativity and then slide the red dots into the green list one at a time, and we'll see that in the end, our calculation just shows that omega + 3 + omega is simply omega + omega.

[00:25:53]To look at something even bigger, let's look at what we would get if we added omega to itself omega times.

[00:26:01]And so, the way to think about this is as taking our chunks of omega many dots and putting them one after another. And then, we can rearrange this into an array. And so, it's important to have the ordering on this array correct. So, you want to read this from left to right and top to bottom in terms of least to greatest.

[00:26:26]And so, we've got this array of omega dots in the both vertical and horizontal position. So, we've done a lot with addition, but some of those things would have been better served with some sort of definition of multiplication. So, let's see what a multiplication of ordinals would look like. Similar to before, multiplication will be defined in this transfinite recursion method.

[00:26:50]So, we'll take alpha * 0 to be 0, and then we'll take alpha * beta + 1 to be equal to alpha * beta + beta. So, essentially, we're defining it in terms of something like the distributive rule.

[00:27:06]Then, well, what if we have a limit ordinal? Well, we're going to define it via a union.

[00:27:11]So, alpha * lambda is going to be equal to the union of all beta < lambda of alpha * beta.

[00:27:18]And so, the intuition we'll use, especially when we're talking about the ordering model of these ordinals, will be alpha * beta is beta copies of alpha put end to end.

[00:27:32]Now, here's a bit of a homework exercise, and you can do this pretty similarly to how we proved that addition was associative. So, first of all, multiplication is associative, and we've got this distributive rule, but we only have distribution from the left. There is actually no a right distributive rule. I'll let you find a nice counterexample for that as well. If you want, maybe post it in the comments.

[00:27:55]There's one that's definitely low-hanging fruit.

[00:27:59]Okay, so let's do a bit of a sanity check that this makes sense on natural numbers before we look at the question of is multiplication commutative. So, let's look at 2 * 3. So, by our definition right here, this is going to be the same thing as 2 * 2 + 2.

[00:28:21]But then, this 2 + 2 will simply be 2 * 1, and then + 2, and then + 2 again. And then, since we have multiplication being associative or addition being associative, it makes sense to just write 2 + 2 + 2.

[00:28:36]But now, I won't go down to the definition of addition, but you can show that 2 + 2 + 2 using our proper definition of addition is 6. So, everything works out in the end.

[00:28:49]So, now, let's look at the order version of this multiplication. So, let's do 2 * 3.

[00:28:57]So, like over here, this is going to be what? Three copies of the order that represents two end to end. So, we've got this yellow 1 is less than yellow 2, and then that's going to be less than magenta 1, which is less than magenta 2, which is going to be less than blue 1, which is less than blue 2. So, see what we did? We just concatenated our ordering for 1 2 together three times.

[00:29:28]But we can pretty clearly just rename this thing, and we would have 1 is less than 2 is less than 3 is less than 4 is less than 5 is less than 6.

[00:29:38]This is really just an ordering on six elements. We're just using these colored ones and twos to stand in for those six elements. Okay, so now let's look at the commutativity or I should say lack of commutativity of multiplication. Okay, so let's show that multiplication is not commutative. So we'll start by looking at two times omega.

[00:30:02]So let's observe by our definition up here since we've got a limit ordinal, that's going to be the union over all natural numbers of two times the natural number. Keeping in mind that omega is simply the set of all natural numbers.

[00:30:18]So like I said, we've got the union over all n and n of two times n. But then writing those as sets, that's going to be the union over all natural numbers of the set containing 0 1 2 ending at 2n minus 1. Because as a set, the ordinal 2n is going to be well, that set right there.

[00:30:40]But I think it's pretty clear that if you take the union of that set, you get all natural numbers, but then as an ordinal, that is omega.

[00:30:49]So we can also look at this via the omega copies of the ordering for 1 2. So I'll write it like this. We're going to do 1 is less than 2, so that's like our ordinal for 2 times omega.

[00:31:06]So let's see. We could start off. We'd have this peach one is less than this peach two, which is less than this blue one, which is less than this blue two, which is less than this magenta one, which is less than this magenta two, which is less than this yellow one, which is less than yellow two, and so on and so forth because we've got omega copies of those.

[00:31:33]But I think it's pretty clear that we can just get rid of all the parentheses and rename things, and we'll see that this is the same thing as the ordering one is less than two is less than three is less than four, so on and so forth, never ending. But that was the ordering representation of omega.

[00:31:52]So there we've got it. We immediately see that two times omega is equal to omega, and we've seen it two different ways.

[00:32:01]Okay, now let's look at omega times two.

[00:32:05]So omega times two, well, let's observe that that's simply equal to omega plus omega by our definition right here of our multiplication of ordinals in terms of what's happening with the precursor, if you will. The precursor of two is one.

[00:32:24]Okay, so let's observe by a definition of addition, that's going to be the union over all natural numbers of omega plus n.

[00:32:37]But then remember that this omega plus n was different from omega. This is in fact equal to the union over all natural numbers >> [snorts] >> of the set containing 0 1 2 all the way up to omega and then omega plus 1 ending at omega plus n minus 1. You can carefully write it down, but I think you can convince yourselves that this set omega plus n will be the set that we've got over there.

[00:33:12]But now if we take that union, let's observe that we essentially get two strings of the natural numbers laid one after another. So in other words, we've got this 0 1 2 3 dot dot dot, and then omega omega plus 1 omega plus 2 dot dot dot. So in our visualization, that looked like omega plus omega from our visualization, which of course it is.

[00:33:42]Now let's look at it in our order ordering representation as well. So or we could think about this as omega times two. But then omega is this ordering one less than two less than three, so on and so forth times two.

[00:34:01]So what we'll do is we'll put two orderings like this back to back.

[00:34:06]But well, let's color code this carefully. We've got yellow one less than yellow two less than yellow three less than yellow four so on and so forth. And then all of that is going to be less than blue one, which is less than blue two, which is less than blue three, which is less than blue four, so on and so forth.

[00:34:27]Now needless to say, we definitely don't get omega in that case either. And let's look at the fundamental difference here.

[00:34:35]Well, neither of them have largest elements, so there's no largest element here, and there's no largest element here.

[00:34:43]But the important thing here is that this one has a limit point of one.

[00:34:49]Or as a set, this has a limit ordinal as part of its set. So there are multiple ways to visualize or think about the difference of 2 plus omega and omega plus 2 seeing that they're most definitely not the same. Highlighting that this multiplication is not commutative within the ordinals. Okay, so let's look at a nice visualization of ordinal multiplication. So in terms of visualization, the way you want to think of alpha times beta is as beta copies of alpha. So 1 times omega is simply omega copies of one. Well, of course that's just going to give you omega many dots laid uh one after each other. And then two times omega will be omega copies of two. So see, we've got these chunks of two dots, but how many chunks of two dots will we have do we have? Well, we have omega number chunks of two dots.

[00:35:46]And so it's pretty clear that you can spread those dots out, and you get exactly the same ordering setup as just omega in the beginning.

[00:35:56]And this is going to continue on for three times omega, four times omega, and in fact all n times omega. So if you have n times omega, that's going to be omega copies of n dots. But again, you can just spread all of this out to get omega in the end. And so what we have is for all natural numbers n, n times omega is simply omega. Now on the other hand, if we look at omega times one, that's like one copy of omega, whereas omega times two is two copies of omega lied one after another. Now importantly, as we pointed out before, even though this has no largest element, it does have one limit point. Then omega times three, well, that doesn't have a largest element, but it has two limit points.

[00:36:46]And then omega times four, no largest element, three limit points. And I think you can kind of see what's going on here. If you have omega times n, it'll pretty clearly have no largest element, but there will be n minus one limit points.

[00:37:04]Providing some really clear evidence that for all natural numbers n, all of the omega time n times n are distinct.

[00:37:11]Setting up another infinitely many failures of the commutativity of multiplication. So two times omega is not omega times two, three times omega is not equal to omega times three, and then so on and so forth. Now I guess the logical next question would be what is omega times omega? So let's go back and revisit omega times five. So there's four limit points, but there's no largest element. And then if you keep going, omega times six, seven, eight, then taking the limit of that picture, we have no largest element, but we have omega many limit points. And so this would be the smallest ordinal where there is no largest element, but there are infinitely many limit point. But of course, we can also think of omega times omega as omega squared, giving us motivation to look at exponentiation of ordinals. So now let's look at exponentiation.

[00:38:09]And if you've internalized the way that we defined addition and multiplication, you can probably guess how to define exponentiation using this recursive strategy that we did before.

[00:38:21]And so namely, alpha to the one should be alpha, and then alpha to the beta plus one should be equal to alpha to the beta times alpha. It's important here that we're multiplying on the right, just by the way. And then if we've got a limit ordinal, alpha to the lambda should be the union of all beta less than lambda of alpha to the beta. Now as a bit of a homework exercise, you can check that these exponentiation properties are satisfied. So we've got alpha to the beta times alpha to the gamma is alpha to the beta plus gamma, and then we've got this rule alpha to the beta to the gamma is alpha to the beta gamma.

[00:39:00]And there might be some missing here, and that's because they don't hold. So maybe explore that if you're interested.

[00:39:08]So let's look at a bit of an example.

[00:39:10]Let's look at two to the omega. So if you know anything about cardinal arithmetic, you might think that two to the omega should be uncountable, and that's because often you say two to the aleph null, the cardinal for countable numbers is the first uncountable cardinal, but that's not going to be the case here.

[00:39:33]So by our definition over here, two to the omega is equal to the union of all natural numbers because that's just the union of all n less than omega, of two to the n. But, that's going to be the union over all natural numbers of the set containing 0 1 2 ending at two to the n and then minus one. But, of course, if you union all of those together, you're just going to get all natural numbers. So, that's just omega.

[00:40:03]So, we've seen that two to the omega is omega using this definition.

[00:40:08]Now, there's another definition that involves the orders that we had before, but that's a little bit more slippery.

[00:40:14]So, I prefer just to do exponentiation with this definition right here.

[00:40:20]Okay, so now let's look at omega squared. So, let's observe that that's simply going to be omega times omega by our definition right here.

[00:40:31]But, then as we saw before, that's going to be the union over all natural numbers of omega times n. Then, you can write that out as the union of the following set. So, it gets kind of gnarly, but it's okay if you just take it one step at a time. So, up here on the first row, I'm going to put 0 1 2 so on and so forth. And down here, I'm going to have omega, omega plus one, omega plus two, so on and so forth.

[00:41:02]And then down here, I'm going to have omega times two, omega times two plus one, omega times two plus two, and so on and so forth. And that's going to end down here at omega times n minus one, omega times n minus one plus two, and then so on and so forth.

[00:41:24]So, I guess we could put this together into the union over all m and n, which are natural numbers, of m times omega plus n. So, it's like in some way some sort of union of these linear polynomials in omega being the variable and m and n coming from natural numbers.

[00:41:50]So, the coefficients are from natural numbers.

[00:41:53]And so, that's essentially just going to be the union of a countable collection of countable sets, making this countable as well.

[00:42:02]And so, two to the omega is countable and omega squared is also countable.

[00:42:07]But, the countability is not really what matters here. It's the ordering that matters. And as we've listed it here, these are ordered from least to greatest, moving like one row at a time.

[00:42:21]All right, so now let's see if we can see a visualization of ordinal exponentiation. So, as we saw before, two to the omega is going to be the union over all n less than omega of two to the n.

[00:42:34]And so, visually, I like to think about this as just taking two to the n dots, maybe one row at a time. So, we've got two to the zero is a single dot, two to the one is two dots, two to the two is four dots, so on and so forth.

[00:42:50]And then, what we do as we union this is we just collapse all of these into a single row. But, of course, if we do a projection or collapse them into a single row, we're going to get all two to the omega, because that's the unioning procedure. But, that's just going to be the same thing as omega many dots.

[00:43:11]But then, this is going to be the same process that we could do for three to the omega to show that that's the same thing as omega. Or, in fact, for all natural numbers n bigger than or equal to two, you can do this exact same process to show that n to the omega is omega.

[00:43:29]Now, powers of omega are a little bit different. And so, omega to the one is clearly just omega.

[00:43:36]Omega squared, you'll have omega copies of omega as we saw before.

[00:43:41]Perhaps it might be nice to visualize this as an omega by omega square, and then use that square to build omega cubed, which will of course be omega copies of omega squared. And then we can count the limit points, and you'll see that we have omega limit points in each of the squares, making a total of omega squared limit points.

[00:44:08]And then you can take all of these squares, and then fuse them together into this crazy cube, this omega by omega by omega cube. And then, omega to the fourth is going to be omega copies of this omega cube. And you can count the number of limit points in each cube, and then the total limit points in the end, you'll have omega cubed limit points. No largest element. And then, well, here's where visualization kind of breaks down, because it would be hard to visualize a four-dimensional cube that is omega size in each direction.

[00:44:45]But, a good takeaway here is that for all natural numbers n, the omega n are all distinct. And then if you want to push it further, the limit of this picture as you go to infinitely many dimensions is what we would call omega to the omega power.

[00:45:03]But now, once we've talked about omega to the omega power, you might be tempted to just go bigger and bigger and bigger.

[00:45:09]In other words, look at power towers of these ordinals.

[00:45:13]And we can do that, and we can measure the size just a little bit at a time.

[00:45:17]So, we'll start down here at the bottom with omega.

[00:45:20]That's less than omega squared, which is less than omega cubed, so on and so forth. All of those are less than omega to the omega.

[00:45:28]And then, you can keep going. That's less than omega to the omega squared, omega to the omega cubed, so on and so forth. Then you hit this omega to the omega to the omega at the top of that.

[00:45:41]And then you can continue this process over and over and over until you get this fixed point. And this fixed point is an ordinal that we call epsilon naught. And it has this crazy property that omega to the epsilon naught is equal to epsilon naught. And you might think that we finally found an uncountable ordinal. Like perhaps epsilon naught is uncountable in size.

[00:46:09]But, we'll see that that's not the case.

[00:46:11]Epsilon naught is in fact still countable. And the fact that epsilon naught is countable really underscores the difference between ordinals and cardinals. Ordinals are all about the arithmetic of infinite arrangements, well, whereas cardinals are all about the arithmetic of infinite sizes.

[00:46:32]So, how can we prove that this epsilon naught is countable? Well, here's a bit of a sketch.

[00:46:38]And so, I'll just prove the following claim first, and then this claim will do most of the work. If alpha is countable, then so is alpha to the omega.

[00:46:54]And so, like I said, this is just like a bit of a sketch. You can fill in the careful details if you want to. But, it goes something like this. So, let's observe that alpha to the omega, so that's going to be the union over all natural numbers n of alpha to the n.

[00:47:12]But, by making some arguments that are pretty similar to what we did before, you can rewrite this as the union over all natural numbers n of the set alpha to the n minus one times c n minus one, all the way down to alpha times c1 plus c0, where the c i's are natural numbers.

[00:47:42]So, in other words, these are just polynomials, where the coefficients come from the natural numbers, and the variable, if you will, is alpha.

[00:47:53]But then, this is a countable union again of countable sets, meaning that it is countable.

[00:48:00]Okay, so this being countable is actually the seed to show that this epsilon naught is countable, because we can express this epsilon naught as again the union of countably many countable sets. And so, let's do that by defining alpha sub zero to be equal to omega.

[00:48:20]We'll define alpha sub one to be omega to the omega. We'll define alpha sub two to be omega to the omega to the omega.

[00:48:29]So, in general, alpha to the n plus one is going to be equal to omega to the power alpha to the n.

[00:48:40]Okay, but then if we observe that our epsilon naught is simply going to be the union over all natural numbers n of this alpha sub n, we have expressed our epsilon naught as the countable union of countable sets, where arguing up here that those alpha sub n are countable.

[00:49:03]Where it follows that these alpha sub n are countable by this claim right here.

[00:49:09]And that's a good place to stop.