# An Extensive Categorisation of Infinite Paradoxes

In­fini­ties are one of the most com­plex and con­found­ing top­ics in math­e­mat­ics and they lead to an ab­surd num­ber of para­doxes. How­ever, many of the para­doxes turn out to be vari­a­tions on the same theme once you dig into what is ac­tu­ally hap­pen­ing. I will provide in­for­mal hints on how sur­real num­bers could help us solve some of these para­doxes, al­though the fo­cus on this post is pri­mar­ily cat­e­gori­sa­tion, so please don’t mis­take these for for­mal proofs. I’m also aware that sim­ply not­ing that a for­mal­i­sa­tion pro­vides a satis­fac­tory solu­tion doesn’t philo­soph­i­cally jus­tify its use, but this is also not the fo­cus of this post. I plan to up­date this post to in­clude new in­finite para­doxes that I learn about, in­clud­ing ones that are posted in the com­ments.

In­fini­tar­ian Paral­y­sis: Sup­pose there are an in­finite num­ber of peo­ple and they are happy so there is in­finite util­ity. A man punches 100 peo­ple and de­stroys 1000 util­ity. He then ar­gues that he hasn’t done any­thing wrong as there was an in­finite amount util­ity be­fore and that there is still an in­finite util­ity af­ter. What is wrong with this ar­gu­ment?

If we use car­di­nal num­bers, then we can’t make such a dis­tinc­tion. How­ever, if we use sur­real num­bers, then n+1 is differ­ent from n ev­ery when n is in­finite.

Para­dox of the Gods: “A man walks a mile from a point α. But there is an in­finity of gods each of whom, un­known to the oth­ers, in­tends to ob­struct him. One of them will raise a bar­rier to stop his fur­ther ad­vance if he reaches the half-mile point, a sec­ond if he reaches the quar­ter-mile point, a third if he goes one-eighth of a mile, and so on ad in­fini­tum. So he can­not even get started, be­cause how­ever short a dis­tance he trav­els he will already have been stopped by a bar­rier. But in that case no bar­rier will rise, so that there is noth­ing to stop him set­ting off. He has been forced to stay where he is by the mere un­fulfilled in­ten­tions of the gods”

Let n be a sur­real num­ber rep­re­sent­ing the num­ber of Gods. The man can move: 1/​2^n dis­tance be­fore he is stopped.

Two En­velopes Para­dox: Sup­pose that there are two sealed en­velopes with one hav­ing twice as much money in it as the other such that you can’t see how much is in ei­ther. Hav­ing picked one, should you switch?

If your cur­rent en­velope has x, then there is a 50% chance of re­ceiv­ing 2x and a 50% chance of re­ceiv­ing x/​2, which then cre­ates an ex­pected value of 5x/​4. But then ac­cord­ing to this logic we should also want to switch again, but then we’d get back to our origi­nal en­velope.

This para­dox is un­der­stood to be due to treat­ing a con­di­tional prob­a­bil­ity as an un­con­di­tional prob­a­bil­ity. The ex­pected value calcu­la­tion should be 1/​2[EV(B|A>B) + EV(B|A<B)] = 1/​2[EV(A/​2|A>B) + EV(2A|A<B)] = 14 EV(A|A>B) + EV(A|A<B)

How­ever, if we have a uniform dis­tri­bu­tion over the pos­i­tive re­als then ar­guably EV(A|A>B) = EV(A|A<B) in which case the para­dox re­oc­curs. Such a prior is typ­i­cally called an im­proper prior as it has an in­finite in­te­gral so it can’t be nor­mal­ised. How­ever, even if we can’t prac­ti­cally work with such a prior, these pri­ors are still rele­vant as it can pro­duce a use­ful pos­te­rior prior. For this rea­son, it seems worth­while un­der­stand­ing how this para­dox is pro­duced in this case

Part 2:

Sup­pose we uniformly con­sider all num­bers be­tween 1/​n and n for a sur­real n and we dou­ble this to gen­er­ate a sec­ond num­ber be­tween 2/​n and 2n. We can then flip a coin to de­ter­mine the or­der. This now cre­ates a situ­a­tion where EV(A|A>B) = n+1/​n and EV(A|A<B) = (n+1/​n)/​2. So these ex­pected val­ues are un­equal and we avoid the para­dox.

Sphere of Suffer­ing: Sup­pose we have an in­finite uni­verse with peo­ple in a 3-di­men­sional grid. In world 1, ev­ery­one is ini­tially suffer­ing, ex­cept for a per­son at the ori­gin and by time t the hap­piness spreads out to peo­ple within a dis­tance of t from the ori­gin. In world 2, ev­ery­one is ini­tially happy, ex­cept for a per­son at the ori­gin who is suffer­ing and this spreads out. Which world is bet­ter? In the first world, no mat­ter how much time passes, more peo­ple will be suffer­ing than happy, but ev­ery­one be­comes happy af­ter some finite time and re­mains that way for the rest of time.

With sur­real num­bers, we can use l to rep­re­sent how far the grid ex­tents in any par­tic­u­lar di­rec­tion and q to rep­re­sent how many time steps there are. We should then be able to figure out whether q is long enough that the ma­jor­ity of in­di­vi­d­ual time slices are happy or suffer­ing.

Sup­pose we have an in­finite set of finite num­bers (for ex­am­ple the nat­u­ral num­bers or the re­als), for any finite num­ber x, there ex­ists an­other finite num­ber 2x. Non-for­mally, we can say that each finite num­ber is in the first half; in­deed the first 1/​x for any in­te­ger x. Similarly, for any num­ber x, there’s always an­other num­ber x+n for ar­bi­trary n.

The canon­i­cal ex­am­ple is the Hilbert Ho­tel. If we have an in­finite num­ber of rooms la­bel­led 1, 2, 3… all of which are full, we fit an ex­tra per­son in by shift­ing ev­ery­one up one room num­ber. Similarly, we can fit an in­finite num­ber of ad­di­tional peo­ple in the ho­tel by send­ing each per­son in room x to room 2x to free up all of the odd rooms. This kind of trick only works be­cause any finite-in­dexed per­son isn’t in the last n rooms or in the last-half of rooms.

If we model the sizes of sets with car­di­nal num­bers, then all countable sets have the same size. In con­trast, if we were to say that there were n rooms where n is a sur­real num­ber, then n would be the last room and any­one in a room > n/​2 would be in the last half of rooms. So by mod­el­ling these situ­a­tions in this man­ner pre­vents last half para­doxes, but ar­guably in­sist­ing on the ex­is­tence of rooms with in­finite in­dexes changes the prob­lem. After all, in the origi­nal Hilbert Ho­tel, all rooms have a finite nat­u­ral num­ber, but the to­tal num­ber of rooms are in­finite.

Gal­ileo’s Para­dox: Similar to the Hilbert Ho­tel, but notes that ev­ery nat­u­ral num­ber has a square, but the that only some are squares, so the num­ber of squares must be both at least the num­ber of in­te­gers and less than the num­ber of in­te­gers.

Ba­con’s Puz­zle: Sup­pose that there is an in­finitely many peo­ple in a line each with ei­ther a black hat or a white hat. Each per­son can see the hats of ev­ery­one in front of them, but they can’t see their own hat. Sup­pose that ev­ery­one was able to get to­gether be­fore­hand to co-or­di­nate on a strat­egy. How would they be able to en­sure that only a finite num­ber of peo­ple would guess the colour of their hat in­cor­rectly? This pos­si­bil­ity seems para­dox­i­cal as each per­son should only have a 50% chance of guess­ing their hat cor­rectly.

We can di­vide each in­finite se­quence into equiv­alence classes where two se­quences are equiv­a­lent if they stop differ­ing af­ter a finite num­ber of places. Let’s sup­pose that we choose one rep­re­sen­ta­tive from each equiv­alence class (this re­quires the ax­iom of choice). If each per­son guesses their hat as per this rep­re­sen­ta­tive, af­ter a finite num­ber of places, each per­son’s guess will be cor­rect.

This para­dox is a re­sult of the rep­re­sen­ta­tive never last differ­ing from the ac­tu­ally cho­sen se­quence in the last half of the line due to this in­dex hav­ing to be finite.

Trumped: Don­ald Trump is re­peat­edly offered two days in heaven for one day in hell. Since heaven is as good as hell is bad, Trump de­cides that this is a good deal and ac­cepts. After com­plet­ing his first day in hell, God comes back and offers him the deal again. Each day he ac­cepts, but the re­sult is that Trump is never let into heaven.

Let n be a sur­real num­ber rep­re­sent­ing how many fu­ture time steps there are. Trump should ac­cept the deal as long as the day num­ber is less than n/​3

St Peters­berg Para­dox: A coin is tossed un­til it comes up heads and if it comes up heads on the nth toss, then you win 2^n dol­lars and the game ends. What is the fair price for play­ing this game? The ex­pected value is in­finite, but if you pay in­finity, then it is im­pos­si­ble for you to win money.

If we model this with sur­re­als, then sim­ply stat­ing that there is po­ten­tially an in­finite num­ber of tosses is un­defined. Let’s sup­pose first that there is a sur­real num­ber n that bounds the max­i­mum num­ber of tosses. Then your ex­pected value is $n and if we were to pay$n/​2, then we’d ex­pect to make $n/​2 profit. Now, set­ting a limit on the num­ber of tosses, even an in­finite limit is prob­a­bly against the spirit of the prob­lem, but the point re­mains, that pay­ing in­finity to play this game isn’t un­rea­son­able as it first sounds and the game effec­tively re­duces to Pas­cal’s Wager. Trou­ble in St. Peters­berg: Sup­pose we have a coin and we toss it un­til we re­ceive a tails and then stop. We are offered the fol­low­ing deals: • Offer 1: Lose$1 if it never lands on tails, gain $3 if it lands on tails on toss 1 • Offer 2: Lose$4 if if lands on tails on toss 1, gain $9 if it lands on tails on toss 2 • Offer 2: Lose$10 if if lands on tails on toss 2, gain $13 if it lands on tails on toss 3 And so on, when we calcu­late the losses by dou­bling and adding two and the gains by dou­bling and adding 3. Each deal has a pos­i­tive ex­pected value, so we should ac­cept all the deals, but then we ex­pect to lose in each pos­si­bil­ity. In the finite case, there is a large amount of gain in the fi­nal offer and this isn’t coun­tered by a loss in a sub­se­quence bet, which is how the EV ends up be­ing pos­i­tive de­spite be­ing nega­tive in ev­ery other case. The para­dox oc­curs for in­fini­ties be­cause there is no last n with an un­miti­gated gain, but if we say that we are offered n deals where n is a sur­real num­ber, then deal n will have this prop­erty. Dice Room Mur­ders: Sup­pose that a se­rial kil­ler takes a man hostage. They then roll a ten-sided dice and re­lease them if it comes up 10, kil­ling them oth­er­wise. If they were kil­led, then the se­rial kil­ler re­peats tak­ing twice as many hostages each time un­til they are re­leased. It seems that each per­son taken hostage should have a 110 chance of sur­viv­ing, but there are always more peo­ple who sur­vive than die. There’s an in­finites­i­mal chance that the dice never comes up a 10. In­clud­ing this pos­si­bil­ity, ap­prox­i­mately 50% of peo­ple should sur­vive, ex­clud­ing it the prob­a­bil­ity is 110. This is a re­sult of a bi­ased sam­ple. To see a finite ver­sion, imag­ine that there are two peo­ple. One tosses a coin. If it’s not heads, the other tosses a coin too. The per­son-flip in­stances will be 5050 heads/​tails, but if you av­er­age over peo­ple be­fore av­er­ag­ing over pos­si­bil­ities, you’ll ex­pect most of the flips to be heads. How you in­ter­pret this de­pends on your per­spec­tive on SSA and SIA. In any case, noth­ing to do with in­fini­ties. Ross-Lit­tle­wood Para­dox: Sup­pose we have a line and at each time pe­riod we add 10 balls to the end and then re­move the first ball re­main­ing. How many balls re­main at the end if we perform this an in­finite num­ber of times? Our first in­tu­ition would be to an­swer in­finite as 9 balls are added in each time pe­riod. On the other hand, the nth ball added is re­moved at time 10n, so ar­guably all balls are re­moved. Clearly, we can see that this para­dox is a re­sult of all balls be­ing in the first 110. Soc­cer Teams: Sup­pose that there are two soc­cer teams. Sup­pose one team has play­ers with abil­ities …-3,-2,-1,0,1,2,3… Now lets sup­pose a sec­ond team started off equally as good, but all the play­ers trained un­til they raised their abil­ity lev­els by one. It seems that the sec­ond team should be bet­ter as each player has strictly im­proved, but it also seems like it should be equally good as both teams have the same dis­tri­bu­tion. With sur­real num­bers, sup­pose that the teams origi­nally have play­ers be­tween -n and n in abil­ity. After the play­ers have trained, their abil­ities end up be­ing be­tween -n+1 and n+1. So the dis­tri­bu­tion ends up be­ing differ­ent af­ter all. Pos­i­tive Soc­cer Teams: Imag­ine that the skill level of your soc­cer team is 1,2,3,4… You im­prove ev­ery­one’s score by two which should im­prove the team, how­ever your skill lev­els are now 3,4… Since your skill lev­els are now a sub­set of the origi­nal it could be ar­gued that the origi­nal team is bet­ter since it has two ad­di­tion (weakly pos­i­tive) play­ers as­sist­ing. Can God Pick an In­te­ger at Ran­dom? - Sup­pose that there are an in­finite amount of planets la­bel­led 1,2,3… and God has cho­sen ex­actly one. If you bet that God didn’t pick that planet, you lose$2 if God ac­tu­ally chose it and you re­ceive $1/​2^n. On each planet, there is a 1/​∞ chance of it be­ing cho­sen and a (∞-1)/​∞ chance of it not be­ing cho­sen. So there is an in­finites­i­mal ex­pected loss and a finite ex­pected gain, given a pos­i­tive ex­pected value. This sug­gests we should bet on each planet, but then we lose$2 on one planet win less than \$1 from all the other planets.

Let sur­real n be the num­ber of planets. We ex­pect to win on each finite val­ued planet, but for sur­real-val­ued planets, our ex­pected gain be­comes not only in­finites­i­mal, but smaller than our ex­pected loss.

Banach-Tarski Para­dox: This para­dox in­volves de­scribing a way in which a ball can be di­vided up into five sets that can then be re­assem­bled into two iden­ti­cal balls. How does this make any sense? This might not ap­pear like an in­finite para­dox as first, but this be­comes ap­par­ent once you dig into the de­tails.

(Other var­i­ants in­clude the Haus­dorff para­dox, Sier­pin­ski-Mazurk­iewicz Para­dox and Von Neu­mann Para­dox)

Let’s first ex­plain how the proof of this para­dox works. We di­vide up the sphere by us­ing the free group of rank 2 to cre­ate equiv­alence classes of points. If you don’t know group the­ory, you can sim­ply think of this as com­bi­na­tions of where an el­e­ment is not al­lowed to be next to its in­verse, plus the spe­cial el­e­ment 1. We can then think of this purely in terms of these se­quences.

Let S rep­re­sent all such se­quences and S[a] rep­re­sent a se­quence start­ing with a. Then .

Fur­ther

In other words, not­ing that there is a bi­jec­tion be­tween all se­quences and all se­quences start­ing with any sym­bol s, we can write

Break­ing down similarly, we get:

In other words, al­most pre­cisely two copies of it­self apart from 1.

How­ever, in­stead of just con­sid­er­ing the se­quences of in­finite length, it might be helpful to as­sign them a sur­real length n. Then S[a] con­sists of “a” plus a string of n-1 char­ac­ters. So S[aa] isn’t ac­tu­ally con­gru­ent to S[a] as the former has n-2 ad­di­tion char­ac­ters and the later n-1. This time in­stead of the para­dox rely­ing on there be­ing that no finite num­bers are in the last half, it’s rely­ing on there be­ing no finite length strings that can’t have ei­ther “a” or “b” prepended in front of them which is prac­ti­cally equiv­a­lent if we think of these strings as rep­re­sent­ing num­bers in qua­ter­nary.

The Headache: Imag­ine that peo­ple live for 80 years. In one world each per­son has a headache for the first month of their life and are happy the rest, in the other, each per­son is happy for the first month, but has a headache af­ter that. Fur­ther as­sume that the pop­u­la­tion triples at the end of each month. Which world is bet­ter? In the first, peo­ple live the ma­jor­ity of their life headache free, but in the sec­ond, the ma­jor­ity of peo­ple at any time are headache free.

If we say that the world runs for t timesteps where t is a sur­real num­ber, then the peo­ple in the last timesteps don’t get to live all of their lives, so it’s bet­ter to choose the world where peo­ple only have a headache for the first month.

The Magic Dart­board: Imag­ine that we have a dart­board where each point is col­ored ei­ther black or white. It is pos­si­ble to con­struct a dart­board where all but mea­sure 0 of each ver­ti­cal line is black and all but mea­sure 0 of each hori­zon­tal line is white. This means that that we should ex­pect any par­tic­u­lar point to be black with prob­a­bil­ity 0 and white with prob­a­bil­ity 0, but it has to be some color.

One way of con­struct­ing this situ­a­tion is to first start with a bi­jec­tion f from [0, 1] onto the countable or­di­nals which is known to ex­ist. We let the black points be those ones where f(x) < f(y). So given any f(y) there are only a countable num­ber of or­di­nals less than it, so only a countable num­ber of x that are black. This means that the mea­sure of black points in that line must be 0, and by sym­me­try we can get the same re­sult for any hori­zon­tal line.

We only know that there will be a countable num­ber of black x for each hori­zon­tal line be­cause f(x) will always be in the first 1/​n of the or­di­nals for ar­bi­trary n. If on the other hand we al­lowed f(x) to be say in the last half of countable or­di­nals, then for that x we would get the ma­jor­ity of points be­ing black. This is dis­tinct from the other para­doxes in this sec­tion as for this ar­gu­ment to be cor­rect, this the­o­rem would have to be wrong. If we were to bite this bul­let, it would sug­gest any other proof us­ing similar tech­niques might also be wrong. I haven’t in­ves­ti­gated enough to con­clude whether this would be a rea­son­able thing to do, but it could have all kinds of con­se­quences.

The canon­i­cal ex­am­ple is Thom­son’s Lamp. Sup­pose we have a lamp that is turned on at t=-1, off at t=-1/​2, on at t=1/​4, ect. At t=0, will the lamp be on or off?

With sur­real num­bers, this ques­tion will de­pend on whether the num­ber of times that the switch is pressed is rep­re­sented by an odd or even Om­rific num­ber, which will de­pend in a rel­a­tively com­plex man­ner on how we define the re­als.

Grandi’s Series: What is the sum of 1-1+1-1...?

Us­ing sur­real num­bers, we can as­sign a length n to the se­ries as merely say­ing that it is in­finite lacks re­s­olute. The sum then de­pends on whether n is even or odd.

This is one class of para­doxes that sur­real num­bers don’t help with as sur­re­als don’t have a largest finite num­ber or a small­est in­fnity.

Satan’s Ap­ple: Satan has cut a deli­cious ap­ple into in­finitely many pieces. Eve can take as many pieces as she likes, but if she takes in­finitely many pieces she will be kicked out of par­adise and this will out­weigh the ap­ple. For any finite num­ber i, it seems like she should take that ap­ple piece, but then she will end up tak­ing in­finitely many pieces.

I find this para­dox the most trou­bling for at­tempts to for­mal­ise ac­tual in­fini­ties. If we ac­tu­ally had in­finitely many pieces, then we should be able to paint all finitely num­bered pieces red and all in­finitely num­bered pieces blue, but any finite num­ber plus one is finite and any in­finite num­ber minus one is in­finite, so it doesn’t seem like we can have a red and blue piece next to each other. But then, what does the bound­ary look like.

Th­ese “para­doxes” may point to in­ter­est­ing math­e­mat­i­cal phe­nomenon, but are so eas­ily re­solved that they hardly de­serve to be called para­doxes.

Gabriel’s Horn: Con­sider ro­tat­ing 1/​x around the x-axis. This can be proven to have finite vol­ume, but in­finite sur­face area. So it can’t con­tain enough paint to paint its sur­face.

It only can’t paint its sur­face if we as­sume a fixed finite thick­ness of paint. As x ap­proaches in­finity the size of the cross-sec­tion of the horn ap­proaches 0, so past a cer­tain point, this would make the paint thicker than the horn.

Miscellaneous

Ber­trand Para­dox: Sup­pose we have an equilat­eral tri­an­gle in­scribed in a cir­cle. If we choose a chord at ran­dom, what is the prob­a­bil­ity that the length of chord is longer than a side of the tri­an­gle.

There are at least three differ­ent meth­ods that give differ­ent re­sults:

• Pick­ing two ran­dom end points gives a prob­a­bil­ity of 13

• Pick­ing a ran­dom ra­dius then a ran­dom point on that ra­dius gives prob­a­bil­ity of 12

• Pick­ing a ran­dom point and us­ing it as a mid­point gives 14

Now some of these method will count di­ame­ters mul­ti­ple times, but even af­ter these are ex­cluded, we still ob­tain the same prob­a­bil­ities.

We need to bite the bul­let here and say that all of these prob­a­bil­ities are valid. It’s just that we can’t just choose a “ran­dom chord” with­out spec­i­fy­ing this more pre­cisely. In other words, there isn’t just a sin­gle set of chords, but mul­ti­ple that can be defined in differ­ent ways.

Zeno’s Para­doxes: There are tech­ni­cally mul­ti­ple para­doxes, but let’s go with the Dich­tomy Para­dox. Be­fore you can walk a dis­tance, you must go half way. But be­fore you can get halfway, you must get a quar­ter-way and be­fore that an eight of the way. So mov­ing a finite dis­tance re­quires an in­finite num­ber of tasks to be com­plete which is im­pos­si­ble.

It’s gen­er­ally con­sider un­con­tro­ver­sial and bor­ing these days that in­finite se­quences can con­verge. But more in­ter­est­ing, this para­dox seems to be a re­sult of claiming an in­finite amount of time in­ter­vals to di­verge, whilst al­low­ing an in­finite num­ber of space in­ter­vals to con­verge, which is a ma­jor in­con­sis­tency.

Skolem’s Para­dox: Any countable ax­iomi­sa­tion of set the­ory has a countable model ac­cord­ing to the Löwen­heim–Skolem the­o­rem, but Can­tor’s The­o­rem proves that there must be an un­countable set. It seems like this con­fu­sion arises from mix­ing up whether we want to know if there ex­ists a set that con­tains un­countably many el­e­ments or if the set con­tains un­countably many el­e­ments in the model (the cor­re­spond­ing defi­ni­tion of mem­ber­ship in the model only refers to el­e­ments in the model). So at a high level, there doesn’t seem to be very much in­ter­est­ing here, but I haven’t dug enough into the philo­soph­i­cal dis­cus­sion to ver­ify that it isn’t ac­tu­ally rele­vant.

• Sur­real num­bers are use­less for all of these para­doxes.

In­fini­tar­ian paral­y­sis: Us­ing sur­real-val­ued util­ities cre­ates more in­fini­tar­ian paral­y­sis than it solves, I think. You’ll never take an op­por­tu­nity to in­crease util­ity by be­cause it will always have higher ex­pected util­ity to fo­cus all of your at­ten­tion on try­ing to find ways to in­crease util­ity by , since there’s some (how­ever small) prob­a­bil­ity that such efforts would suc­ceed, so the ex­pected util­ity of fo­cus­ing your efforts on look­ing for ways to in­crease util­ity by will have ex­pected util­ity , which is higher than . I think a bet­ter solu­tion would be to note that for any per­son, a nonzero frac­tion of peo­ple are close enough to iden­ti­cal to that per­son that they will make the same de­ci­sions, so any de­ci­sion that any­one makes af­fects a nonzero frac­tion of peo­ple. Mea­sure the­ory is prob­a­bly a bet­ter frame­work than sur­real num­bers for for­mal­iz­ing what is meant by “frac­tion” here.

Para­dox of the gods: The in­tro­duc­tion of sur­real num­bers solves noth­ing. Why wouldn’t he be able to ad­vance more than miles if no gods erect any bar­ri­ers un­til he ad­vances miles for some finite ?

Two-en­velopes para­dox: it doesn’t make sense to model your un­cer­tainty over how much money is in the first en­velope with a uniform sur­real-val­ued prob­a­bil­ity dis­tri­bu­tion on for an in­finite sur­real , be­cause then the prob­a­bil­ity that there is a finite amount of money in the en­velope is in­finites­i­mal, but we’re try­ing to model the situ­a­tion in which we know there’s a finite amount of money in the en­velope and just have no idea which finite amount.

Sphere of suffer­ing: Sur­real num­bers are not the right tool for mea­sur­ing the vol­ume of Eu­clidean space or the du­ra­tion of for­ever.

Hilbert ho­tel: As you men­tioned, us­ing sur­re­als in the way you pro­pose changes the prob­lem.

Trumped, Trou­ble in St. Peters­burg, Soc­cer teams, Can God choose an in­te­ger at ran­dom?, The Headache: Us­ing sur­re­als in the way you pro­pose in each of these changes the prob­lems in ex­actly the same way it does for the Hilbert ho­tel.

St. Peters­burg para­dox: If you pay in­finity dol­lars to play the game, then you lose in­finity dol­lars with prob­a­bil­ity 1. Doesn’t sound like a great deal.

Banach-Tarski Para­dox: The free group only con­sists of se­quences of finite length.

The Magic Dart­board: First, a nit­pick: that proof re­lies on the con­tinuum hy­poth­e­sis, which is in­de­pen­dent of ZFC. Aside from that, the proof is cor­rect, which means any re­s­olu­tion along the lines you’re imag­in­ing that im­ply that no magic dart­boards ex­ist is go­ing to im­ply that the con­tinuum hy­poth­e­sis is false. Worse, the fact that for any countable or­di­nal, there are countably many smaller countable or­di­nals and un­countably many larger countable or­di­nals fol­lows from very min­i­mal math­e­mat­i­cal as­sump­tions, and is of­ten used in de­scrip­tive set the­ory with­out bring­ing in the con­tinuum hy­poth­e­sis at all, so if you start try­ing to change math to make sense of “the sec­ond half of the countable or­di­nals”, you’re go­ing to have a bad time.

Par­ity para­doxes: The lengths of the se­quences in­volved here are the or­di­nal , not a sur­real num­ber. You might ob­ject that there is also a sur­real num­ber called , but this is differ­ent from the or­di­nal . Arith­metic op­er­a­tions act differ­ently on or­di­nals than they do on the copies of those or­di­nals in the sur­real num­bers, so there’s no rea­son­able sense in which the sur­re­als con­tain the or­di­nals. Ex­am­ple: if you add an­other el­e­ment to the be­gin­ning of ei­ther se­quence (i.e. flip the switch at , or add a at the be­gin­ning of the sum, re­spec­tively), then you’ve added one thing, so the sur­real num­ber should in­crease by , but the or­der-type is un­changed, so the or­di­nal re­mains the same.

• Thanks for your feed­back. I’ll note that these are only in­for­mal hints/​thoughts on how sur­real num­bers could help us here and that I’ll be pro­vid­ing a more de­vel­oped ver­sion of some of these thoughts in a fu­ture post.

In­fini­tar­ian paral­y­sis: I con­sider Pas­cal’s Mug­ging to be its own seper­ate prob­lem. In­deed Pas­cal’s Mug­ging type is­sues are already pre­sent with the more stan­dard in­fini­ties. In any case, the mea­sure the­ory solu­tion is de­pen­dent on an in­di­vi­d­ual be­ing a finite frac­tion of the agents in the uni­verse. While this is an ex­tremely plau­si­ble as­sump­tion, there doesn’t seem to be any prin­ci­pled rea­son why our solu­tion to in­finite paral­y­sis should de­pend on this as­sump­tion.

Para­dox of the Gods, Banach-Tarski: Your com­plaint is that I’m dis­al­low­ing the a se­quence con­sist­ing of all finite in­verses/​all finite in­te­gers. I ac­tu­ally be­lieve that ac­tual and po­ten­tial in­finity mod­els of these prob­lems need to be treated sep­a­rately, though I’ve only out­lined how I plan to han­dle ac­tual in­fini­ties. Hope­fully, you find my next post on this topic more per­sua­sive.

Two-en­velopes para­dox: “The prob­a­bil­ity that there is a finite amount of money in the en­velope is in­finites­i­mal”—Hmm, you’re right. That is a rather sig­nifi­cant is­sue.

Sphere of suffer­ing: “Sur­real num­bers are not the right tool for mea­sur­ing the vol­ume of Eu­clidean space or the du­ra­tion of for­ever”—why?

St Peters­berg Para­dox: Ah, but you have an in­finites­i­mal chance of win­ning a higher in­finity. So it be­comes an even more ex­treme ver­sion of Pas­cal’s Mug­ging, but again that’s its own dis­cus­sion.

Magic Dart­board: Yes, I’m aware that re­ject­ing the ex­is­tence of magic dart­boards could have far-reach­ing con­se­quences. It’s some­thing I hope to look into more.

Par­ity: See my re­sponse to gjm. Or­di­nal num­bers lack re­s­olu­tion and so can’t prop­erly de­scribe the length of se­quence.

• In­deed Pas­cal’s Mug­ging type is­sues are already pre­sent with the more stan­dard in­fini­ties.

Right, in­finity of any kind (sur­real or oth­er­wise) doesn’t be­long in de­ci­sion the­ory.

“Sur­real num­bers are not the right tool for mea­sur­ing the vol­ume of Eu­clidean space or the du­ra­tion of for­ever”—why?

How would you? If you do some­thing like tak­ing an in­creas­ing se­quence of bounded sub­sets that fill up the space you’re try­ing to mea­sure, find a for­mula f(n) for the vol­ume of the nth sub­set, and plug in , the re­sult will be highly de­pen­dent on which in­creas­ing se­quence of bounded sub­sets you use. Did you have a differ­ent pro­posal? It’s sort of hard to ex­plain why no method for mea­sur­ing vol­umes us­ing sur­real num­bers can pos­si­bly work well, though I am con­fi­dent it is true. At the very least, vol­ume-pre­serv­ing trans­for­ma­tions like shift­ing ev­ery­thing 1 me­ter to the left or ro­tat­ing ev­ery­thing around some axis cease to be vol­ume-pre­serv­ing, though I don’t know if you’d find this con­vinc­ing.

• “In­deed Pas­cal’s Mug­ging type is­sues are already pre­sent with the more stan­dard in­fini­ties.”
Right, in­finity of any kind (sur­real or oth­er­wise) doesn’t be­long in de­ci­sion the­ory.

But Pas­cal’s Mug­ging type is­sues are pre­sent with large finite num­bers, as well. Do you bite the bul­let in the finite case, or do you think that un­bounded util­ity func­tions don’t be­long in de­ci­sion the­ory, ei­ther?

• The lat­ter. It doesn’t even make sense to speak of max­i­miz­ing the ex­pec­ta­tion of an un­bounded util­ity func­tion, be­cause un­bounded func­tions don’t even have ex­pec­ta­tions with re­spect to all prob­a­bil­ity dis­tri­bu­tions.

There is a way out of this that you could take, which is to only in­sist that the util­ity func­tion has to have an ex­pec­ta­tion with re­spect to prob­a­bil­ity dis­tri­bu­tions in some re­stricted class, if you know your op­tions are all go­ing to be from that re­stricted class. I don’t find this very satis­fy­ing, but it works. And it offers its own solu­tion to Pas­cal’s mug­ging, by in­sist­ing that any out­come whose util­ity is on the scale of 3^^^3 has prior prob­a­bil­ity on the scale of 1/​(3^^^3) or lower.

• There’s definitely a part of me won­der­ing if in­fini­ties ex­ist, but be­fore I even con­sider tack­ling that ques­tion, I need to figure out the most con­sis­tent in­ter­pre­ta­tion of in­fini­ties as­sum­ing they ex­ist.

“At the very least, vol­ume-pre­serv­ing trans­for­ma­tions like shift­ing ev­ery­thing 1 me­ter to the left or ro­tat­ing ev­ery­thing around some axis cease to be vol­ume-pre­serv­ing, though I don’t know if you’d find this con­vinc­ing”—Well there are non-mea­surable sets that do this with­out sur­re­als, but do sur­re­als add more ex­am­ples?

I’ll have to read more about how sur­re­als ap­ply to vol­umes. It may be hard get­ting con­ver­gence to the ex­act in­finites­i­mal, but I don’t know if the prob­lems will ex­tent be­yond that.

(Also, the abil­ity to in­te­grate is mostly be­sides the point. In­stead of the sphere of suffer­ing, we could have defined the ex­pand­ing cube of suffer­ing. This will then let us solve some spe­cial cases of the sphere of suffer­ing)

• There are mea­surable sets whose vol­umes will not be pre­served if you try to mea­sure them with sur­real num­bers. For ex­am­ple, con­sider . Say its mea­sure is some in­finite sur­real num­ber . The vol­ume-pre­serv­ing left-shift op­er­a­tion sends to , which has mea­sure , since has mea­sure . You can do es­sen­tially the same thing in higher di­men­sions, and the shift op­er­a­tion in two di­men­sions () can be ex­pressed as the com­po­si­tion of two ro­ta­tions, so ro­ta­tions can’t be vol­ume-pre­serv­ing ei­ther. And since differ­ent ro­ta­tions will have to fail to pre­serve vol­umes in differ­ent ways, this will break sym­me­tries of the plane.

I wouldn’t say that vol­ume-pre­serv­ing trans­for­ma­tions fail to pre­serve vol­ume on non-mea­surable sets, just that non-mea­surable sets don’t even have mea­sures that could be pre­served or not pre­served. Failing to pre­serve mea­sures of sets that you have as­signed mea­sures to is en­tirely differ­ent. Non-mea­surable sets also don’t arise in math­e­mat­i­cal prac­tice; half-spaces do. I’m also skep­ti­cal of the ex­is­tence of non-mea­surable sets, but the non-ex­is­tence of non-mea­surable sets is a far bolder claim than any­thing else I’ve said here.

• Well shift­ing left pro­duces a su­per­set of the origi­nal, so of course we shouldn’t ex­pect that to pre­serve mea­sure.

• What about ro­ta­tions, and the fact that we’re talk­ing about de­stroy­ing a bunch of sym­me­try of the plane?

• I’m happy to bite that bul­let and de­stroy the sym­me­try. If we pick a ran­dom point and line in the uni­verse, are there more unit points to the left or right? Well, that de­pends on where the point is.

• It’s a bad bul­let to bite. Its sym­me­tries are es­sen­tial to what makes Eu­clidean space in­ter­est­ing.

And here’s an­other one: are you not both­ered by the lack of countable ad­di­tivity? Sup­pose you say that the vol­ume of Eu­clidean space is some sur­real num­ber . Eu­clidean space is the union of an in­creas­ing se­quence of balls. The vol­umes of these balls are all finite, in par­tic­u­lar, less than , so how can you jus­tify say­ing that their union has vol­ume greater than ?

• “Its sym­me­tries are es­sen­tial to what makes Eu­clidean space in­ter­est­ing”—Isn’t the in­ter­est­ing as­pect of Eu­clidean space its abil­ity to model our world ex­clud­ing rel­a­tivity?

Well, I just don’t think it’s that un­usual for func­tions to have prop­er­ties that break at their limits. Is this any differ­ent from 1/​x be­ing defin­able ev­ery­where ex­cept 0? Is there any­thing that makes the change at the limit par­tic­u­larly con­cern­ing.

• I don’t fol­low the anal­ogy to 1/​x be­ing a par­tial func­tion that you’re get­ting at.

Maybe a bet­ter way to ex­plain what I’m get­ting at is that it’s re­ally the same is­sue that I pointed out for the two-en­velopes prob­lem, where you know the amount of money in each en­velope is finite, but the uniform dis­tri­bu­tion up to an in­finite sur­real would sug­gest that the prob­a­bil­ity that the amount of money is finite is in­finites­i­mal. Sup­pose you say that the size of the ray is an in­finite sur­real num­ber . The size of the por­tion of this ray that is dis­tance at least from is when is a pos­i­tive real, so pre­sum­ably you would also want this to be so for sur­real . But us­ing, say, , ev­ery point in is within dis­tance of , but this rule would say that the mea­sure of the por­tion of the ray that is farther than from is ; that is, al­most all of the mea­sure of is con­cen­trated on the empty set.

• As I un­der­stand it, there is not yet a good the­ory of in­te­gra­tion on the sur­re­als. Par­tial progress has been made, but there are also some nega­tive re­sults es­tab­lish­ing limi­ta­tions on the pos­si­bil­ities. Here is a re­cent pa­per.

• In “Trumped”, it seems that if , the first in­finite or­di­nal, then on ev­ery sub­se­quent day, the re­main­ing num­ber of days will be for some nat­u­ral . This is never equal to .

Put differ­ently, just be­cause we count up to doesn’t mean we pass through . Of course, the to­tal or­der on days has has for each finite , but this isn’t a well-or­der any­more so I’m not sure what you mean when you say there’s a se­quence of de­ci­sions. Do you know what you mean?

• “Put differ­ently, just be­cause we count up to n doesn’t mean we pass through n/​3”—The first pos­si­ble ob­jec­tion I’ll deal with is not what I think you are com­plain­ing about, but I think it’s worth han­dling any­way. n/​3 mightn’t be an Om­nific Num­ber, but in this case we just take the in­te­ger part of n/​3.

I think the is­sue you are high­light­ing is that all finite num­bers are less than n/​3. And if you define a se­quence as con­sist­ing of finite num­bers then it’ll never in­clude n/​3. How­ever, if you define a se­quence as all num­bers be­tween 1 and x where x is a sur­real num­ber then you don’t en­counter this is­sue. Is this valid? I would ar­gue that this ques­tion boils down to whether there are an ac­tu­ally in­finite num­ber of days on which Trump ex­pe­riences or only a po­ten­tial in­finity. If it’s an ac­tual in­finity, then the sur­real solu­tion seems fine and we can say that Trump should stop say­ing yes on day n/​3. If it’s only a po­ten­tial in­finity, then this solu­tion doesn’t work, but I don’t en­dorse sur­re­als in this case (still read­ing about this)

• If the num­ber of days is, speci­fi­cally, , then the days are num­bered 0, 1, 2, …, with pre­cisely the (or­di­nary, finite) non-nega­tive in­te­gers oc­cur­ring. They are all smaller than . The num­ber isn’t the limit or the least up­per bound of those finite in­te­gers, merely the sim­plest thing big­ger than them all.

If you are tempted to say “No! What I mean by call­ing the num­ber of days is pre­cisely that the days are num­bered by all the om­nific in­te­gers be­low .” then you lose the abil­ity to rep­re­sent a situ­a­tion in which Trump suffers this in­dig­nity on a se­quence of days with ex­actly the or­der-type of the first in­finite or­di­nal , and that seems like a pretty se­ri­ous bul­let to bite. In par­tic­u­lar, I think you can’t call this a solu­tion to the “Trumped” para­dox, be­cause my read­ing of it—even as you tell it! -- is that it is all about a se­quence of days whose or­der-type is .

Rather a lot of these para­doxes are about situ­a­tions that in­volve limit­ing pro­cesses of a sort that doesn’t seem like a good fit for sur­real num­bers (at least so far as I un­der­stand the cur­rent state of the art when it comes to limit­ing pro­cesses in the sur­real num­bers, which may not be very far).

• I already pointed above to the dis­tinc­tion be­tween ab­solute and po­ten­tial in­fini­ties. I ad­mit that the sur­real solu­tion as­sumes that we are deal­ing with an ab­solute in­finity in­stead of a po­ten­tial one, so let’s just con­sider this case. You want to con­ceive of this prob­lem as “a se­quence whose or­der-type is ω“, but from the sur­real per­spec­tive this lacks re­s­olu­tion. Is the num­ber of el­e­ments (sur­real) ω, ω+1 or ω+1000? All of these are pos­si­ble given that in the or­di­nals 1+ω=ω so we can add ar­bi­trar­ily many num­bers to the start of a se­quence with­out chang­ing its or­der type.

So I don’t think the or­di­nary no­tion of se­quence makes sense. In par­tic­u­lar, it doesn’t ac­count for the fact that two se­quences which ap­pear to be the same in ev­ery place can ac­tu­ally be differ­ent if they have differ­ent lengths. Any­way, I’ll try to un­tan­gle some of these is­sues in fu­ture posts, in par­tic­u­lar I’m lean­ing to­wards hy­per­re­als as a bet­ter fit for mod­el­ling po­ten­tial in­fini­ties, but I’m still un­cer­tain about how things will de­velop once I man­age to look into this more.

• So far as any­one knows, no ac­tual pro­cesses in the ac­tual world are ac­cu­rately de­scribed by sur­real num­bers. If not, then I sug­gest the same goes for the “near­est pos­si­ble wor­lds” in which, say, it is pos­si­ble for Mr Trump to be faced with the sort of situ­a­tion de­scribed un­der the head­ing “Trumped”. But you can have, in a uni­verse very much like ours, an end­less suc­ces­sion of events of or­der-type . If the sur­real num­bers are not well suited to de­scribing such situ­a­tions, so much the worse for the sur­real num­bers.

And when you say “I don’t think the or­di­nary no­tion of se­quence makes sense”, what it looks like to me is that you have looked at the or­di­nary no­tion of se­quence, made the en­tirely ar­bi­trary choice that you are only pre­pared to un­der­stand it in terms of sur­real num­bers, and in­deed not only that but made the fur­ther ar­bi­trary choice that you are only pre­pared to un­der­stand it if there turns out to be a uniquely defined sur­real num­ber that is the length of such a se­quence, ob­served that there is not such a sur­real num­ber, and then said not “Oh, whoops, looks like I took a wrong turn in try­ing to model this situ­a­tion” but “Bah, the thing I’m try­ing to model doesn’t fit my pre­con­cep­tions of what the model should look like, there­fore the thing is wrong”. You can’t do that! Models ex­ist to serve the things they model, not the other way around.

It’s as if I’d just learned about the or­di­nals, de­cided that all in­finite things needed to be de­scribed in terms of the or­di­nals, was asked some­thing about a countably in­finite set, ob­served that such a set is the same size as but also the same size as and , and said “I don’t think the no­tion of countably in­finite set makes sense”. It makes perfectly good sense, I just (hy­po­thet­i­cally) picked a bad way to think about it: or­di­nals are not the right tool for mea­sur­ing the size of a (not-nec­es­sar­ily-well-or­dered) set. And like­wise, sur­real num­bers are not the right tool for mea­sur­ing the length of a se­quence.

Don’t get me wrong; I love the sur­real num­bers, as an ob­ject of math­e­mat­i­cal study. The the­ory is gor­geous. But you can’t claim that the sur­real num­bers let you re­solve all these para­doxes, when what they ac­tu­ally al­low you to do is to re­place the para­dox­i­cal situ­a­tions with other en­tirely differ­ent situ­a­tions and then deal with those, while re­ject­ing the origi­nal situ­a­tions merely be­cause your way of try­ing to model them doesn’t work out neatly.

• Maybe I should re-em­pha­sise the caveat at the top of the post: “I will provide in­for­mal hints on how sur­real num­bers could help us solve some of these para­doxes, al­though the fo­cus on this post is pri­mar­ily cat­e­gori­sa­tion, so please don’t mis­take these for for­mal proofs. I’m also aware that sim­ply not­ing that a for­mal­i­sa­tion pro­vides a satis­fac­tory solu­tion doesn’t philo­soph­i­cally jus­tify its use, but this is also not the fo­cus of this post.”

You wrote that I “made the en­tirely ar­bi­trary choice that you are only pre­pared to un­der­stand it in terms of sur­real num­bers”. This choice isn’t ar­bi­trary. I’ve given some hints as to why I am tak­ing this ap­proach, but a full jus­tifi­ca­tion won’t oc­cur un­til fu­ture posts.

• OK! I’ll look for­ward to those fu­ture posts.

(I’m a big sur­real num­ber fan, de­spite the skep­ti­cal tone of my com­ments here, and I will be ex­tremely in­ter­ested to see what you’re propos­ing.)

• You want to con­ceive of this prob­lem as “a se­quence whose or­der-type is ω“, but from the sur­real per­spec­tive this lacks re­s­olu­tion. Is the num­ber of el­e­ments (sur­real) ω, ω+1 or ω+1000? All of these are pos­si­ble given that in the or­di­nals 1+ω=ω so we can add ar­bi­trar­ily many num­bers to the start of a se­quence with­out chang­ing its or­der type.

It seems to me that mea­sur­ing the lengths of se­quences with sur­re­als rather than or­di­nals is in­tro­duc­ing fake re­s­olu­tion that shouldn’t be there. If you start with an in­finite con­stant se­quence 1,1,1,1,1,1,..., and tell me the se­quence has size , and then you add an­other 1 to the be­gin­ning to get 1,1,1,1,1,1,1,..., and you tell me the new se­quence has size , I’ll be like “uh, but those are the same se­quence, though. How can they have differ­ent sizes?”

• Be­cause we should be work­ing with la­bel­led se­quences rather than just se­quences (that is se­quences with a length at­tached). That solves the most ob­vi­ous is­sues, though there are some sub­tleties there

• Why? Plain se­quences are a perfectly nat­u­ral ob­ject of study. I’ll echo gjm’s crit­i­cism that you seem to be try­ing to “re­solve” para­doxes by chang­ing the defi­ni­tions of the words peo­ple use so that they re­fer to un­nat­u­ral con­cepts that have been ger­ry­man­dered to fit your solu­tion, while re­fus­ing to talk about the nat­u­ral con­cepts that peo­ple ac­tu­ally care about.

I don’t think think your pro­posal is a good one for in­dexed se­quences ei­ther. It is pretty weird that shift­ing the in­dices of your se­quence over by 1 could change the size of the se­quence.

• the or­di­nary no­tion of sequence

I as­sume here you mean some­thing like “a se­quence of el­e­ments from a set is a func­tion where is an or­di­nal”. Do you know about nets? Nets are a no­tion of se­quence preferred by peo­ple study­ing point-set topol­ogy.

• Thanks for the sug­ges­tion. I took a look at nets, but their pur­pose seems mainly di­rected to­wards gen­er­al­is­ing limits to topolog­i­cal spaces, rather than adding ex­tra nu­ance to what it means for a se­quence to have in­finite length. But per­haps you could clar­ify why you think that they are rele­vant?

• A net is just a func­tion where is an or­dered in­dex set. For limits in gen­eral topolog­i­cal spaces, might be pretty nasty, but in your case, you would want to be some to­tally-or­dered sub­set of the sur­re­als. For ex­am­ple, in the trump para­dox, you prob­a­bly want to:

in­clude and for some in­finite
have a least el­e­ment (the first day)

It sounds like you also want some co­her­ent no­tion of “to­mor­row” at each day, so that you can get through all the days by pass­ing from to­day to to­mor­row in­finitely many times. But this is equiv­a­lent to hav­ing your set be well-or­dered, which is in­com­pat­i­ble with the prop­erty “closed un­der di­vi­sion and sub­trac­tion by finite in­te­gers”. So you should clar­ify which of these prop­er­ties you want.

• “But this is equiv­a­lent to hav­ing your set be well-or­dered, which is in­com­pat­i­ble with the prop­erty “closed un­der di­vi­sion and sub­trac­tion by finite in­te­gers”″ - Why is this in­com­pat­i­ble?

• An or­dered set is well-or­dered iff ev­ery sub­set has a unique least el­e­ment. If your set is closed un­der sub­trac­tion, you get in­finite de­scend­ing se­quences such as . If your se­quence is closed un­der di­vi­sion, you get in­finite de­scend­ing se­quences that are fur­ther­more bounded such as . It should be clear that the two lin­ear or­ders I de­scribed are not well-or­ders.

A small or­der the­ory fact that is not to­tally on-topic but may help you gather in­tu­ition:

Every countable or­di­nal em­beds into the re­als but no un­countable or­di­nal does.

• Okay, I now un­der­stand why clo­sure un­der those op­er­a­tions is in­com­pat­i­ble with be­ing well-or­dered. And I’m guess­ing you be­lieve that well-or­der­ing is nec­es­sary for a co­her­ent no­tion of pass­ing through to­mor­row in­finitely many times be­cause it’s a re­quire­ment for trans­finite in­duc­tion?

I’m not so sure that this is im­por­tant. After all, we can imag­ine get­ting from 1 to 2 via pass­ing through an in­finite num­ber of in­finites­i­mally small steps even though [1,2] isn’t well-or­dered on <. In­deed, this is the cen­tral point of Zeno’s para­dox.

• Yes, there are good ways to in­dex sets other than well or­ders. A net where the in­dex set is the real line and the func­tion is con­tin­u­ous is usu­ally called a path, and these are ubiquitous e.g. in the foun­da­tions of alge­braic topol­ogy.

I guess you could say that I think well-or­ders are im­por­tant to the pic­ture at hand “be­cause of trans­finite in­duc­tion” but a sim­pler way to state the same ob­jec­tion is that “to­mor­row” = “the unique least el­e­ment of the set of days not yet vis­ited”. If to­mor­row always ex­ists /​ is uniquely defined, then we’ve got a well-or­der. So some­thing about the story has to change if we’re not fit­ting into the or­di­nal box.

• Your sec­ond ex­am­ple, 1 > 12 > 14 > … > 0, is a well-or­der. To make it non-well-or­dered, leave out the 0.

• A well-or­der has a least el­e­ment in all non-empty sub­sets, and 1 > 12 > 14 > … > 0 has a non-empty sub­set with­out a least el­e­ment, so it’s not a well-or­der.

• Yes, you’re right.

In gen­eral, ev­ery sub­or­der of a well-or­der is well-or­dered. In a word, the prop­erty of “be­ing a well-or­der” is hered­i­tary. (com­pare: ev­ery sub­set of a finite set is finite)

• I think gjm’s re­sponse is ap­prox­i­mately the clar­ifi­ca­tion I would have made about my ques­tion if I had spent 30 min­utes think­ing about it.

• For Eve and her ap­ple pieces. She may eat one piece per sec­ond and stay in Par­adise for­ever be­cause at any given mo­ment only a finite num­ber of pieces has been eaten by her.

If her eat­ing pace dou­bles ev­ery minute, she is still okay for­ever.

Only if she, for ex­am­ple, dou­bles her eat­ing pace af­ter ev­ery say 100 pieces eaten, then she is in trou­ble. If she su­per­tasks.

• Great post!

Satan’s Ap­ple: Satan has cut a deli­cious ap­ple into in­finitely many pieces. Eve can take as many pieces as she likes, but if she takes in­finitely many pieces she will be kicked out of par­adise and this will out­weigh the ap­ple. For any finite num­ber i, it seems like she should take that ap­ple piece, but then she will end up tak­ing in­finitely many pieces.

Pro­posed solu­tion for finite Eves (also a solu­tion to Trumped, for finite Trumps who can’t count to sur­real num­bers):

After hav­ing eaten n pieces, Eve’s de­ci­sion isn’t be­tween eat­ing n pieces and eat­ing n+1 pieces, it’s be­tween eat­ing n pieces and what­ever will hap­pen if she eats the n+1st piece. If Eve knows that the fu­ture Eve will be fol­low­ing the strat­egy “always eat the next ap­ple piece”, then it’s a bad de­ci­sion to eat the n+1st piece (since it will lead to get­ting kicked out of par­adise).

So what strat­egy should Eve fol­low? Con­sider the prob­lem of pro­gram­ming a strat­egy that an Eve-bot will fol­low. In this case, the best strat­egy is the strat­egy that will lead to the largest amount of finite pieces be­ing eaten. What this strat­egy is de­pends on the hard­ware, but if the hard­ware is finite, then there ex­ists such a strat­egy (per­haps count the num­ber of pieces and stop when you reach N, for the largest N you can store and com­pare with). Gen­er­al­is­ing to (finite) hu­mans, the best strat­egy is the strat­egy that re­sults in the largest amount of finite pieces eaten, among all strate­gies that a hu­man can pre­com­mit to.

Of course, if we al­low in­finite hard­ware, then the prob­lem is back again. But that’s at least not a prob­lem that I’ll ever en­counter, since I’m run­ning on finite hard­ware.

• We can definitely solve this prob­lem for real agents, but the rea­son why I find this prob­lem so per­plex­ing is be­cause of the bound­ary is­sue that it high­lights. Imag­ine that we have an ac­tual in­finite num­ber of peo­ple. Color all the finite placed peo­ple red and the non-finite placed peo­ple blue. Every­one one t the right of a red per­son should be red and ev­ery­one one to the left of blue per­son should be blue. So what does the bound­ary look like? Sure we can’t finitely trans­verse from the start to the in­finite num­bers, but that doesn’t af­fect the in­tu­ition that the bound­ary should still be there some­where. And this makes me ques­tion whether the no­tion of an ac­tual in­finity is co­her­ent (I re­ally don’t know).

• My best guess about how to clear up con­fu­sion about “what the bound­ary looks like” is via math­e­mat­ics rather than philos­o­phy. For ex­am­ple, have you un­der­stood the prop­er­ties of the long line?

• Thanks for that sug­ges­tion. The long line looks very in­ter­est­ing. Are you sug­gest­ing that the bound­ary doesn’t ex­ist?

• Yeah, I’d agree with the “bound­ary doesn’t ex­ist” in­ter­pre­ta­tion.