# Supertasks: Doing Infinitely Many Things Speaker: Joel David Hamkins Course: Lectures on Infinity Watch: https://ergo.org/videos/joel-david-hamkins-supertasks-doing-infinitely-many-things Transcript: https://ergo.org/videos/joel-david-hamkins-supertasks-doing-infinitely-many-things/transcript/ YouTube: https://www.youtube.com/watch?v=h02divcjYcc When quoting, name the speaker and link to the watch URL. What happens when you complete infinitely many steps? Joel David Hamkins walks through a series of supertasks that challenge ordinary intuition about infinity. He begins with Thomson's Lamp, a puzzle about toggling a light switch infinitely many times, then connects it to Zeno's paradox to show that supertasks may not be as exotic as they seem. The lecture builds toward deeper puzzles: a deal with the devil, balls placed into and removed from a sack where the final count defies expectation, and a stochastic version where probability-zero events become surprisingly relevant. Hamkins carefully distinguishes between what holds for every individual ball and what holds for all balls collectively, exposing a subtle logical gap. The lecture culminates in the chocolatier's game, where questions about finite versus infinite servings, memory, and the axiom of choice determine whether a glutton can guarantee tasting every flavor. Throughout, Hamkins shows how rigorous thinking about infinity upends common sense. ## What Are Supertasks? Let's talk about supertasks, which are tasks involving infinitely many steps. There's a famous supertask ## Thomson's Lamp: On or Off After Infinity? kind of puzzle called Thompson's Lamp. It's hard to believe, but there are hundreds of papers in the academic literature talking about Thompson's Lamp. But let me tell you about Thompson's Lamp. You're at home, and you're reading in your study in the twilight, but it's getting a little dark, and so you turn on the light. But then it's a little bit too bright, so you switch it off again. But then it's a little bit too dark, so you switch it back on again, and so on. Thompson's Lamp is the supertask. You turn on the light for half a minute, and then you turn it off for a quarter of a minute, and then you turn it on for an eighth of a minute, and off for a sixteenth of a minute, and so on. So like the geometric series that we talked about in another lecture. So therefore, after one minute, you have turned the light on and off infinitely many times. So it's on, off, on, off, on, off, faster and faster, and it's a kind of supertask. And the question that Thompson asked about this is really about the intelligibility of the supertask. What is the state of the lamp after one minute? Is it on or off? And there's all kinds of physics-based objections that one can make to this thought experiment. Let me just draw a little diagram here just so we know what we're talking about. We're going from zero to one, time zero to one, one minute. And the lamp is on for the first half-minute, and then it's off for the half of what remains, which is a quarter. And then it's on, and then it's off. On, off, on, off, on. So it turns on and off infinitely many times. So after one minute, it has turned on and off infinitely many times. And so the question is, at T equal one, is it on or off? Or is that intelligible? Or is it determined? Now, of course, there's all kinds of physics-based objections that you can make to the thought experiment. For example, it's not possible to flick a light switch so rapidly at the end, or electric current flows at a certain rate, and so it wouldn't be able to turn the light on and off in those very narrow moments towards the end. Or, for example, every time it's on, it has to emit photons, but maybe there's only finitely many photons available, and so it couldn't have had infinitely many distinct intervals of being on this. I think all of those objections are quite strong. But they're really beside the point, because really what's at stake isn't whether or not we can actually do a supertask in physical reality, but rather, is it even logically coherent? Is it intelligible to understand a supertask of, a task involving infinitely many steps? Mathematically, for example, we can easily make a kind of step function. If I draw the light, it's on between zero and a half, and then it's off, and then it's on, and then it's off, and then on, off, on, off, on, off, on, off, on, off. And the question is, what is it doing at the limit? At the limit of T equal one? And the point may be mathematically, of course we could have a function that behaved like this and had any value you wanted at T equal one. There's no objection to having it be on at that point or off, and so on. ## Zeno's Paradox Makes Supertasks Real Let's think again about the Zeno situation. Zeno was walking from here to there, or he argued that it's not possible to walk from here to there, and I don't want to do the Zeno paradox in the first manner that we had described it, but rather in the second way where to go from here to there you first go halfway, and then you go halfway of what remains, and halfway of what remains again, and so on. And so when you're walking from here to there, you have done these infinitely many things along the way. You first got halfway, and then you got halfway of what remained, and then halfway of what remained after that, and so on. And the picture looks just like this. In order to walk from here to there, you first had to get to this point, and then to this point, and then to this point, and then to this point, and so on. And so it seems like we can do a supertask. It's perfectly intelligible. If you think that you can walk from here to there, then you should think that you can do infinitely many things if we view each of those things as a separate thing that you have accomplished in walking from here to there. ## Trivial Supertasks Hide in Plain Sight And that suggests a different way of thinking about Thompson's Lamp. Namely, let's turn the lamp and just leave it on for the first minute, and then continue to leave it on for the next half a minute. It's on for the first half-minute, and then it's on for the next quarter-minute, and then it's continuing to be on after that, and on, you leave it on the whole time. So you left the lamp on for one minute altogether, but in doing so, you have done infinitely many things. Namely, you left it on for the first half of that period, and then still again as a second thing you left it on for half of what remained, and then half of what remained after that, and so on. And so that doesn't seem problematic in any way. You just left it on for a minute, and we can view that as an ordinary task which can be thought of as a supertask if you divide up the action into these infinitely many moments. Or we could have left it off, off, off, off, off, off the whole time. That's another kind of trivial supertask. So it seems in at least some cases, we can do a supertask. Let me tell you about a particular supertask that I like quite a lot. It's called the Deal with the Devil. Suppose you've made some really good investments, shrewd investment, and you have infinitely many dollar bills. You're carrying them with you. And let's say the serial numbers are one, three, five, and so on, all the odd numbers. So you have all the odd-numbered dollar bills with you. ## The Deal with the Devil And you enter this underground bar, and there's the Devil sitting there at a table piled high with money. He has all the other bills, the even-numbered bills, zero, two, four, and so on. And the Devil sees you and realizes that he has a particular attachment to your dollar bills, and that he will pay a premium to get your dollar, to get your dollar bills. He's willing to pay you two dollars per dollar just to get your dollar bills. So he says, "Yeah, I'll give you two dollar bills for each one of your dollar bills." And that sounds like, well, maybe it wouldn't matter because if you have two for one, then it seems like okay, you still have infinitely much money, and maybe you're no better off, but it doesn't seem like you would be any worse off, and so what would be the harm in doing this? So the Devil, you say, "What the heck? Let's do it." It doesn't seem like it could harm me. So the Devil makes a contract, and you sign it, and in the contract, it specifies the details of how this exchange is gonna be carried out. And of course, it's gonna be carried out in this supertask manner. So in the first half an hour, you're gonna make one trade. He's gonna give you two dollars, and he's gonna take one dollar bill from you. And then in the next, did I say half hour? In the first half hour? In the next quarter hour, he's going to give you two more dollar bills and take one dollar bill from you. And then in the next eighth and so on, he's gonna give you two dollar bills and take one dollar bill from you, and so on, in the geometric series. And so therefore, after one hour, the whole exchange with all infinitely many trades will be completed. Except actually, he's very fussy about the order in which the bills are going to be exchanged. In the contract, you see now that you're actually undertaking this. He always buys from you your currently lowest numbered bill, and he always pays you with higher numbered bills. So you start the exchange. On the first, you had bills one, three, five, and so on, all the odd-numbered bills, so on the first half hour, he gives you bills, say, two and four, and he takes bill number one from you. Well, it's two for one. And the next round, he gives you six and eight, which are the next two even numbers that he has, and he takes from you bill number two, which he just paid to you. It's your currently lowest numbered bill. So let's make a little graph and see what happens with this. Okay, so here's your bills, and here's the Devil bills. So you have one, three, five, seven, nine, and so on. And the Devil had, I think we said, well, let's just start with two. Okay, two, four, six, eight. So on the first round, he gives you two and four and takes from you bill one. And then in the next quarter hour, he gives you six and eight and takes from you bill two. And on the next round, he had 10 and 12, so he gives you 10 and 12 and takes from you bill three. And on the next round, can you see what's happening? 14, 16. He gives you 14 and 16 and takes from you bill number four. And so we can see that if this process continues, if he always buys from you your currently lowest bill, then as the process continues, your currently lowest bill is gonna just be getting bigger and bigger. He's always paying you with higher numbered bills. And so you can see eventually that every single bill is going to end up in the possession of the Devil, and at the very end, after you've done all the trading, you don't have any money at all. Every single bill was bought by the Devil from you at some stage, and so after all the trades are complete, the Devil has all the money, and you have no money at all. So this is, I like this example because it shows you how an ordinary way of thinking about supertask transactions don't always behave in the same way as finite transactions, and we have to pay attention to the details of how the process is unfolding in this infinitary manner, because the result is sometimes surprising. ## Balls in a Sack: Empty After Infinity? Let me do another one that's something like that. You lost a bet with your friends and the agreement is you have to stand in humiliation holding a big empty sack, a wool sack. And there's a pile of infinitely many billiard balls nearby. And what you have to do is you're going to undertake a certain supertask procedure. At each step, you're going to take two billiard balls from the pile and put them in the sack. And then you're going to take one billiard ball out of the sack and discard it, so that won't be used again. But furthermore, you have to do this faster and faster. So if you do the first step in a half a minute and the next one in a quarter of a minute and then an eighth of a minute and so on, then after one minute, you will have done infinitely many steps. And so you will have completed the supertask in a finite amount of time. And you will be successful or honored or redeemed with your friends if you do it in such a way that the sack is empty at the end. It's empty at the start, but every step you're putting two balls in and only taking one ball out. So maybe it seems impossible because this sack is getting heavier and heavier as you go. After n steps, there's going to be n balls because every step you put two in and take one out, so it has accumulated one extra ball with each step. And so how could it possibly be empty after infinitely many steps? But nevertheless, it is possible for you to be redeemed. And the way that you should do it is let's think of the billiard balls as numbered, like the natural numbers. Say, zero, one, two, three, and so on. So there's ball zero, ball one, ball two. And at every stage, you should take two new balls and put them in, but you should always remove, like the devil, you should always remove the lowest numbered ball from the sack. And that ball will be discarded and not used again. So maybe on the first step you put balls zero and one into the sack and you take out ball zero because that's the lowest number. And then on the next stage, you put balls two and three into the sack and you take out ball one. That's the lowest number. And then in the next stage, you put balls, what did I say? Let's see, zero, one, two, three, and now four, five into the sack, and ball two is coming out of the sack because it's still in there. So at every step, you put two balls in, the next two balls in, and you take out the lowest numbered. So therefore, ball n is getting taken out on the nth step, and then never used again. So therefore, at the very end when you've completed the supertask, the sack must be empty because there can't be any balls in it because ball n was removed from the sack on the nth step and it never got back into the sack. So if there were a ball in the sack, what would it be? It can't be any particular number. All of the balls have been removed from the sack and so therefore the sack, in fact, is empty at the end. So this is another situation where the nature of a supertask, of an infinitary task has this kind of strange effect that simply doesn't occur. And if you were doing only finitely many steps, then because you're getting an extra ball, a net extra ball each time in the sack, then the number of balls in the sack after n steps will be n. But nevertheless, when we arrange the details of it in such a way, in that particular way, then we can make the sack empty at the end. ## Controlling What Remains in the Sack We can also make the sack completely full with infinitely many balls at the end if we wanted, if we always took out, say, the highest numbered ball. So we put in balls zero and one and take out ball one, and then we put in balls two and three and take out ball three, and then we put in balls four and five and take out ball five. So then all the even numbered balls will be in the sack and all the odd numbered balls will be the ones that we remove. So similarly, you can think about, well, how could I arrange so that I had exactly the prime numbers in the sack at the end? Or could I arrange some other pattern to have exactly the multiples of 17 in the sack at the end? And the answer is quite robust. You can actually design a process so that a given target set is exactly what you have at the end. And you can find maybe necessary and sufficient criteria on this target set that you want, to make that happen. There's maybe a way of thinking about this balls in a sack puzzle, a stochastic way, which means thinking of it as a random process. So let's think, what would happen if you put two balls in the sack and then the ball that you selected to take out would be chosen randomly from the balls in the sack? You start with an empty sack, you put two balls in and then pick one of them randomly. And then you put two more balls in, now you have three balls in the sack, and you take one of them randomly. So now there's two. Two more in, take one random and bring it out. ## Random Removal and Vanishing Probability So what is the expectation about what should happen at the end? How many balls would there be? Will there be balls in the sack at the end? Is it likely that there's balls in the sack at the end or not? So let's think about it. Let's think about one of the first balls that goes in the sack. What's the chance that it survives one step? On the first step, two balls go in, and I'm thinking of one of them in particular, and I want to know, how likely is it that it's going to stay in the sack? Well, it has a 50% chance of surviving for one step because there's two balls in the sack and it's chosen randomly, so it's 50/50 chance to survive. Given that it survived one step, what's the chance that it survives the next step? The next step when two more balls go in, so there's three altogether and it's one of the three, and the chance that it survives is two thirds of that, because it's going to survive if the choice is made from one of the other two balls. So we can make a little calculation about this. We're talking about the random choice balls in a sack paradox. We start with an empty sack, we're going to do infinitely many things, in this super task manner so it's all finished in a finite amount of time, we put two balls in at each stage, and then amongst the balls that are currently in the sack, we take one at random and discard it. Initially, the two balls go in, and there's a one-half chance of our favorite ball not being chosen. And then given that it's survived for one step, there's a two-thirds chance that it survives the second step, because two-thirds of the time it will survive for that step, given that it made it that far. And now, what's the chance that it survives on the third picking? Now there's four balls in the sack, as there were two at, you know, we added two, take one, so there's one. Then we put in two, there's three, and we remove one, so now there's two, and now we put in two more, so there's four, and we're going to take one of them so that there's a three-quarter chance that it survives another step, given that it survives the first two steps. So therefore, the chance that it survives for three steps is exactly this product, because there's half chance of surviving the first step, and then given that it's survived that, two-thirds chance of surviving the next one, and then three-quarter chance of surviving after that, and so on. We can see this pattern now. So n over n plus one. So if I've gone n steps, then this product is the chance that the ball will not have been discarded. But I can make this calculation, because I can see the twos cancel, the threes cancel, fours cancel, everybody cancels. And so I just get one over n plus one. So the chance that one of the initial balls survives for n steps is one over n plus one. But as n becomes large, this becomes very tiny, as close to zero as I like. So what that means is that the chance that the ball survives for infinitely many steps is less than this number for every n, which means it must have probability zero. So therefore, what I'm saying is that the chance that the ball is never chosen is probability zero. ## Probability Zero Does Not Mean Impossible It's a probability zero event for it never to be picked. And the same argument works with the later balls that are added, you can argue. If I think about the balls that are in the sack at stage k, then there's a k over k plus one chance of surviving that next round, because this is the number of balls in the sack, and then I've got k plus one over k plus two, and I'm multiplying, again, k plus n over k plus n plus one, and again, we get this cancellation phenomenon happening. And so altogether, it's k over k plus n plus one. And for fixed k, if I make n large, this number goes to zero. So therefore, for any particular ball, at any particular stage, the probability that it survives infinitely many steps is zero. So for every ball, almost surely it will be chosen at some stage. I said almost surely, but actually, the way probabilists use this word, it's a technical word, it has a very particular meaning. It means that the probability of that event is 100%. Which is different from being logically certain, and this is the kind of philosophical point I want to make. In this kind of stochastic reasoning, the difference between a probability zero event and an impossible event, those are not the same thing. Just because something is probability zero doesn't mean that it is logically impossible for it to happen. It could be, it's logically possible, that maybe on the first round one of the balls is red, and maybe that red ball is simply never chosen. That's logically possible. It's very unlikely. It's probability zero. That's what we calculated. The probability of the red ball never being chosen on any step is less than one over n for every n. One over n plus one for every n. So therefore, the probability of that event is zero, even though it's logically possible. So one has to keep in mind the difference between probability zero and impossible are not the same thing. ## Every Ball vs. All Balls: A Subtle Gap But still we can reason with this probabilistic reasoning. It's very likely, for any particular ball, that it's chosen at some stage. And so therefore, what we expect at the end is that the sack should be empty. With probability one, you should expect that the sack is empty, because with probability one, any particular ball is almost surely chosen at some stage. But now if you think about what I just did in that reasoning, what we argued first is that, for every ball, almost surely it's chosen at some stage. But what I said afterwards was, almost surely every ball is chosen. Yes. So that's not quite the same thing. For any particular ball, it's very likely to be chosen at some stage and removed. But that's different from the sack being empty. For the sack being empty, I want to say it's very likely that every ball is chosen at some stage. And so, how do we go from reasoning about a probability calculation for every individual ball versus reasoning about almost surely something is true of every ball. And maybe it's helpful to think about the situation of a dart board. ## Dartboards and Countable Additivity Suppose we're playing darts and I'm throwing darts at a dart board, and maybe I have a uniform distribution of where the dart would land somewhere on the dart board. I'm not paying attention to the case when the dart doesn't land on the dart board, so I'm just looking at the dart is going to land somewhere on the dart board with uniform probability. So that means that the probability that it lands in any particular region is just the ratio of the area of that region to the total area of the dart board. So, for example, the probability that it's on the left side is one-half because the left side and the right side have the same area. They're both one-half of the dart board. And so the uniform probability would mean that with probability one-half, the dart is going to be on the left side. With probability one-half, it's going to be on the right side, or with probability one-half, it's going to be on the bottom half or the top half, and so on and so forth, or it's one-quarter of being in this quadrant, and so on. So the probability of any particular point being hit is zero because a point has zero area, if I make that ratio. The probability of any particular point is zero, so it's very unlikely to hit any particular point. If we think of the dart board as a continuum of points, then the probability of hitting that point exactly is zero. The area of that point is zero. But in that case, what I want to say, as I did with the balls? With the balls we said, "For every ball, almost surely it's going to be removed. Therefore, almost surely, all the balls are removed." But in the dart board case, I'm going to say, "For every point, almost surely the dart doesn't hit that point," but am I willing to say almost surely the dart doesn't hit any particular point? No, because the dart's going to hit some point. I'm going to throw the dart and it's going to hit the board, and it's going to land at that point. And so that means that the dart hit a point which was a probability zero event. Almost surely, there's going to be some probability zero event. There's one way of saying this. It's very likely that very rare things happen. I throw the dart at the dart board and it hits that particular point that it hit, it's very likely going to hit some point, but for any particular point, that's a very rare event. So it's very likely that rare events occur. I can give another proof of this. Suppose I had a coin and I flip it, I'm going to flip it 10 times, and I get heads, heads, tails, heads, whatever the pattern is. I'm going to get some pattern. Almost surely I'm going to get some pattern. But that particular pattern of heads, heads, tails, heads, whatever the pattern is, was very unlikely, one over 2 to the 10. That's a tiny, tiny number. So it's very likely that rare events occur. It's a little bit paradoxical, but it shouldn't be too paradoxical because it's kind of obvious in a way, isn't it? It's very likely that rare things occur because the space of things that could occur is so enormous. But it's very likely that one of them is going to happen. So if we go back to this stochastic process with the balls in a sack, in comparison with the dart board, we're reasoning exactly the same in both cases. For any particular ball, it's very likely that it gets removed, and I want to conclude from that that it's very likely that all the balls are removed. But I don't want to make the same move with the dart board. For any particular point on the dart board, it's very likely that the dart won't land there. But I don't want to say it's very likely that the dart isn't going to land anywhere because it's going to land somewhere very likely. And so what is the reason, how can I be justified in the balls case but not in the dart board case? And the answer has to do with the philosophy of probability and the fact that the number of balls is only a countable infinity, whereas the number of points in the dart board is an uncountable infinity. And it's part of the theory of probability theory that the countable sums of measure zero events, they are still measure zero. The probability theory is countably additive, but it's not uncountably additive. So when we have countably many, when we have a list of countably many probability zero events, then we can say it's probability zero for any of them to happen, whereas we can't make that move in this uncountable space. So to discuss maybe the very subtle differences in these two attitudes, one gets into the philosophy of probability very much, and the justification for why is it that we want our probability measures, and Lebesgue measure, and so on to be countably additive and not to be more than that. Let's do another super task. ## The Chocolatier's Game: Finite Servings And this is what I call the chocolatier's game. The chocolatier's game is a game played with two players. There's the chocolatier, who will be serving up chocolates, exquisite creations on a kind of serving platter. And then the other player is the glutton who will be eating these chocolates as the game proceeds. In the first version of the game, the game has infinitely many rounds, turns. It will go infinitely. And on every round, the chocolatier serves up finitely many new exquisite chocolate creations. But the glutton is allowed to eat only one. And they accumulate on the serving platter. So maybe the glutton serves up 17 chocolates, and the glutton picks one and eats it. And then we get 37 more chocolates on the next round, and the glutton picks one and eats it. And then maybe we get just two more chocolates and the glutton picks one amongst the accumulating chocolates. And the glutton wins after all the stages are completed. And you can think of the process as being finished in a finite amount of time if you wish, although actually, there's nothing at stake in that. We can also just think of the game as taking naturally infinitely many steps and then ask, "Well, what happens after the steps?" And if you like to think of all of that happening in a finite amount of time, that's fine, and you can do that using this geometric series kind of reasoning that we discussed already. But in fact, there's nothing at stake in that decision. And logically, the game proceeds in an infinite succession of turns, and then you look at what happens after all the plays have been made, and it's irrelevant how long that took or whether that took a finite amount of time or an infinite amount of time. Rather, the logic of the game is that there are infinitely many steps and then we look at who won afterwards. The glutton wins if he eats all the chocolate. Every single chocolate that was served must be eaten at some stage. So maybe it's something like the balls in a sack puzzle. And the glutton is allowed to select one of them on offer and eat it. And the chocolates accumulate on the platter. If the chocolatier serves more than one, the glutton can only eat one of them. And so the others are still there for the next time. Maybe they're chosen later, maybe not. On the one hand, it might seem, look, it's impossible for the glutton to eat all the chocolate, because at every stage of the game, the number of chocolates on the serving platter is increasing and it's growing without bound, and it would be absurd for the glutton to be able to eat all of those. But on the other hand, if we use the ideas that we discussed in the deal with the devil or with the balls in a sack, then maybe we can see that the glutton can in fact eat all the chocolate. And here's one way to do it. The glutton is going to be systematic, and the glutton will imagine that the new chocolates coming on the platter, well, the chocolates on the platter will form in his mind a kind of queue, a line. And the new chocolates will always be added mentally at the back of the line, and the glutton will always eat from the front of the line. So this is something like what the devil did when buying always your lowest numbered dollar bill. So the glutton is always going to eat the frontmost chocolate, and the new chocolates coming in, however many there are, will just go at the back. This is the same process that your local restaurant uses when they're doing stock rotation. They get new items coming in that they're going to be preparing the meals with. And the new material always goes at the back of the cupboard, so that you always use the oldest stuff from the front of the cupboard when you're making the meal. In that way, you turn over all the material in time. So that's what the glutton is going to do. New chocolates go to the back, eat from the front. And this is going to be a winning strategy for the glutton, because for any particular chocolate that's ever served, it's going to have its place in line. At that moment, we're going to know exactly what that place is. And therefore, we're going to know exactly which turn that chocolate is going to be eaten on, because there's only finitely many chocolates in front of it. And therefore, at exactly that turn when it comes, that chocolate will be eaten. So for any particular chocolate that's ever served, we know at that moment exactly which turn it will be eaten on, and in particular, it will be eaten. And therefore, the glutton will win by following this strategy. Every single chocolate will be eaten at some stage. So after infinitely many steps, the glutton will have eaten every single chocolate. So the glutton has a winning strategy in the chocolatier's game. ## Infinite Servings and the Zigzag Strategy Now, there's another version of the chocolatier's game, a slightly harder version. Actually, this game, I like it a lot, because it starts out with this easy case that we just talked about, which is quite clear, I think. But we can make it a little bit harder. And in fact, one can make it harder and harder. And in the end, actually, the chocolatier's game becomes quite sophisticated mathematically. I'm just going to be hinting at those deeper things. But we can go at least one step further. Let's think about the form of the chocolatier's game where the chocolatier is allowed to serve infinitely many chocolates on one turn. So each turn, the chocolatier can put down infinitely many chocolates. But the glutton can still eat only one. So the chocolatier puts infinitely many down. The glutton picks one to eat. And then infinitely many more, and the glutton eats just one. And then infinitely many more, and so on. I claim, still the glutton can win. And let's explain how that would happen. First of all, the queuing strategy doesn't work at all now, because if I think of the chocolates on the first round as forming a queue, and then the second-round chocolates go behind those, I'm never going to get to them. I'll be always just eating through the chocolates that were served on round one, and all the countably many stages, I'll never get to the round two chocolates or round three and so on. So we have to think a little bit more imaginatively about it. So what the glutton is going to do is imagine the chocolates on the serving platter filling up an infinite matrix. So the first-round chocolates are placed on the first row. Each one of these is a little chocolate. They're all different. They have exquisite glazing, and maybe some of them have cherries or whatever. So these beautiful chocolates, that's the round one chocolates. And then the round two chocolates will go on a separate row in the glutton's mind. And then the round three chocolates will form their own row here. I haven't told you what the glutton is eating, but let's just understand the matrix. So we're understanding the chocolates being served as filling up this matrix. And now, what the glutton is going to do is eat the chocolates according to a kind of winding path, that looks like this. This winding zigzag path. So on the first turn, the glutton will eat this chocolate. And then on the next turn, the glutton will eat this chocolate. And then on the third turn, the glutton will eat this chocolate, which is available, because this came on the second round already. And then this one, and this one, and this one in the order of this winding path. And this winding path, you can see, it's clear that every chocolate that will ever be served is going to appear on this winding path at some point. And furthermore, every chocolate on the winding path has only finitely many things, because the chocolates that will be eaten ahead of it are basically in this triangular region above and to the left. And therefore, that chocolate will be eaten on the turn which is exactly the number of chocolates that are preceding it on this winding path. So therefore, even though the chocolatier is serving infinitely many chocolates on each turn, nevertheless, the glutton is eating only one, nevertheless, the glutton can systematically eat them in such a way that every chocolate will be eaten at some stage. And so the glutton will win. ## Can the Glutton Win Without Memory? Now that strategy for the glutton requires the glutton to pay attention to the order in which the chocolates are served. But one might want to ask, "Well, does the glutton really need to pay so much attention?" Maybe there's a strategy for the glutton that is going to tell him which chocolates to eat based only on the set of chocolates on offer, regardless of the order in which those chocolates have been served. So this would be called a memory-free strategy, or also it's the difference between a strategy and a tactic in the theory of games. A tactic depends just on the current situation, whereas a strategy is depending on the whole history of play up to that point. And so now the situation becomes quite interesting because, for example, one question that one needs to ask about it is can the chocolatier ever repeat chocolates? We want to know, well, what's on the menu for the chocolatier? Can they make exactly the same chocolate again? And if there were a memory-free strategy, then, for example, if the chocolatier puts two chocolates down, and the glutton chose one of them to eat, and then the chocolatier could simply replace that chocolate with an identical chocolate, and then if the glutton was following a tactic, they would have to make the same choice again. Because it's just depending on what the chocolates are on offer. And so that function would be the same in the same situation, regardless of what had happened before that. So if the glutton is following a tactic and the chocolatier is allowed to serve the identical chocolate again, then the chocolatier can serve two, glutton picks one, and then the chocolatier simply replaces it. And that same one is picked again by the tactic, and then replaced, picked again, replaced, picked again, and so on. And so the other chocolate will never be chosen. So that's a kind of trivial way to see that if the chocolatier is allowed to repeat identical chocolates, then there can't be a memory-free strategy for the glutton. ## Countable vs. Uncountable Creativity Let's not allow that. Let's say the chocolatier must serve a distinct chocolate. He's not allowed to repeat ever. So then that breaks that argument, and we still want to know, is there a winning tactic for the glutton? And now it depends on how creative the chocolatier is. Let's suppose that the list of possible chocolates that the chocolatier could make is, say, infinite. Well, it would have to be infinite, I suppose, but maybe infinite in the manner of the natural numbers. So there's menu item number zero, menu item number one, menu item number two, menu item number three, and so on. I'm not saying that the chocolatier has to serve all those chocolates or in that order, but those are just the possible chocolates that the chocolatier could serve. In that case, the glutton has a winning tactic, which is simply: always eat the chocolate which is the lowest menu item available. So, at any stage, there's finitely many or even infinitely many chocolates on the serving platter, but one of them will be the lowest menu item. And if the glutton always eats the one with the lowest menu item number, then he will eat all the chocolates because any particular chocolate has only finitely many ahead of it on the menu. And so, there couldn't be any chocolate left at infinity because it would have been the lowest one at some point, and therefore, it would have been already eaten. So, if the chocolatier, in other words, if the chocolatier is merely countably creative in the sense that the space of possible chocolates that the chocolatier could serve was only countable, then the glutton has a winning tactic. Well, what about the case where the chocolatier is uncountably creative? Maybe there's a different chocolate for every real number. Like, maybe the width of the glazing or the density of the liqueur in the sherry or something was coming from a real number. So, all of those different real numbers would have their own different, strictly different chocolate type. And then in this case, you can prove that it's actually not possible for the glutton to have a winning tactic. And it's quite interesting, and maybe a little bit mathematically too sophisticated for us right now, so I won't give you the argument. But I'm just telling you that this is a way in which the problem becomes quite sophisticated. Then finally, maybe another thing to mention in that same vein is that there's nearly a winning tactic even in that case. Namely, there's a winning tactic for the glutton in the uncountably creative case, provided that the glutton is also allowed to use the information of the most recently eaten chocolate. So, the glutton, at any stage, is looking at the chocolates on offer, might be infinite, and then also tasting on his tongue the chocolate that he just ate. So, he's allowed to use that information also to determine which one to pick. And then it turns out, in the case where the chocolatier is allowed to serve only finitely many each time, then if the axiom of choice is true, then the glutton has a memory-free strategy, a winning tactic, which also uses this taste of the previous chocolate. And it involves the axiom of choice and well orders and so on, and it's all quite fascinating and wonderful. ## Wrapping Up Supertasks I guess that's all for this lecture. I hope you enjoyed the supertasks and the chocolatier's game. See you next time.