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.

Re­s­olu­tion Paradoxes

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.

Last-Half Paradoxes

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.

Par­ity Paradoxes

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.

Boundary Paradoxes

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.

Triv­ial Paradoxes

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.

This post was writ­ten with the sup­port of the EA Hotel