Infinite Balls and Jugs [solution]

[Gwydion]

This was from a little while ago but I'm just now catching up.

Everything is set up so that the limits work that way - the discards are {1,2,3...n} followed by the keepers {n+1,n+2,...10n} which converges to N followed by another set isomorphic to N but higher than any number in N. How would you go about constructing such a set in a supertask, if you were asked to?

OK, so imagine an ordering of the naturals such that all numbers equal to 1 mod 10 are strictly less than all other numbers, but preserving order otherwise - {1,11,21...2,3,4...}. If we were to play a version of the original game (place 1-10, remove the lowest number, then place 11-20...) with this ordering, all the balls with label equal to 1 mod 10 would be removed at the end of the task, but the rest would remain. An alternative without such a reordering would be to leave the naturals ordered as they are normally, but modify the supertask such that after placing balls in the jug, the lowest ball equal to 1 mod 10 gets removed.

In these cases, the remaining naturals k =/= 1 mod 10 are all left in the jug at the end. In that sense, they are higher than the set of all numbers (10n-9), which are isomorphic to the naturals. Is this what you were looking for? It's worth mentioning, however, that this isn't equivalent to the initial supertask - here, we have a collection of balls that are explicitly never removed, and proving their existence is trivial.

[Xias]

Consider the following scenario:

What we are going to do is take two parallel lines and connect them with two line segments. Then, we will rotate those two line segments at different rates, such that the angles between the line segments and one of the parallel lines becomes zero. We will find that, if the angles are defined appropriately for each step, we get triangles with areas that are analogous to the sets in the original puzzle.

-

Graph the line y = 1 on the xy-plane for x>=0. Let (θ_n) be a sequence of angles, such that θ_n = (π/2)(1/2^n). Then for each θ_n, there is a line that can be drawn at angle θ_n from the origin, which intersects y = 1 at point (a_n, 1).

Now let (φ_n) be a second sequence of angles, such that for all n, when a line is drawn from the origin at angle φ_n, it intersects y=1 at the point (10*a_n, 1). That is, φ_n = arctan(tan(θ_n)/10) for all n.

Now, at each step n, consider the set of points bounded by the y axis, the line y = 1, and the line from the origin to (a_n, 1). These points form a triangle with area D_n=a_n/2. Also consider the set of points bounded by the line y = 1, the line from the origin to (a_n, 1), and the line from the origin to (10*a_n, 1). These points form a triangle with area K_n = 9*a_n/2. Also consider the entire area under the line y = 1 in the first quadrant. Call the defined area A. Now we can show that for any n, the area in K is 9 times the area in D.



All of the following are true for all n:
a) All points in the triangle K_n are to the right of the rightmost boundary of the triangle D_n;
b) The area of K_n is always increasing, and always 9 times larger than the area of D_n;
c) At no step n do the lines formed by theta and phi stop intersecting y=1, nor do the angles theta and phi ever stop being greater than zero.

Now, given the same supertask timeline as we've been working with:

1) What are the angles θ and φ at midnight?

Spoiler:

Like how the sequence of {1/2, 1/4, 1/8, 1/16, ...} converges to zero, the angles also converge to zero.

2) What are the intersection points of the line y=1 with the lines from the origin at angles θ and φ?

Spoiler:

There are no intersection points; the line that makes an angle of zero with the origin is y = 0, which never intersects y = 1.

3) What are the areas defined by D and K at midnight?

Spoiler:

The line that bounds D no longer intersects y=1, so the area of D at midnight is infinitely large: all of the area below y=1 in the first quadrant of the xy-plane. The area of K bounded above by the line y=0 and below by the line y=0, and so the area is zero.

Bonus:

4) What if instead of φ_n being defined in such a way as to intersect y=1 at the point (10*a_n, 1), φ is always zero, making its line y=0? What changes about K_n and K?

Spoiler:

K_n has an infinitely large area for all n, and K has an area of zero.

5) What if φ_n is defined in such a way as to create a line that intersects y=1 at the point (a_n+2, 1)? What changes about K_n and K?

Spoiler:

K_n has an area of 1 for all n, and K has an area of zero.

[kryptonaut]

@Demki - thanks for the clarification. When I say 'arbitrarily big' I am trying to describe a number that in comparison with any natural number is always greater. I don't know if 'infinite' is the right word either because I've been criticised for using that word. Maths is not my native language as you can tell, so I find some of the symbolic arguments hard to follow.

The concept I am trying to express is a number that is sorted after all natural numbers when counting them off, so you would never reach it - but if you start with this quantity you can still find a successor to it. It's an element in an ordered set, the element with ordinal omega, that's the best way to define it. If this has a name, please let me know.

It is a function of the way the set are ordered, so in the ordered set {odds then evens} the first even number is greater than any of the odds and has ordinal omega, you could not step through the set one at a time and reach an even number, but if you start with an even number you can keep going up. So omega is a positional quality, not a fixed property of a particular number. If you partition N relative to the start {1,2,...10} followed by {11,12,...} then every element's ordinal is the same as its actual number. But if you contrive to partition N somewhere 'partway along' or 'so many steps from the end' then omega is the ordinal of the first element of the second partition - this is because the first partition is countably infinite even if it's a 'fraction' of the original whole. It's the act of partitioning it that creates the omega.

Good. Before addressing the rest of your post, let me propose a minor elaboration. Partition the natural numbers into three sets:
L={1,2,3,4...n-1}
C={n}
R={n+1, n+2, n+3...}
with the supertask of moving the lowest element of R into C, the lowest element of C into L, and incrementing n.

Consider now the cardinalities of the sets. Specifically, I claim that at the end,
the cardinality of L is infinite (equal to aleph null),
the cardinality of R is finite (equal to zero), and
the cardinality of C is also finite (equal to zero). Anything else leads to a contradiction, for the same reasons.

Do you agree with this?

No, I would say that C contains a single element with a number I have previously described as 'arbitrarily bug' but if that is technically the wrong label, let's call it infinite in the sense that whatever natural number you compare it with, this infinite number is always greater. But that's not to say that there can't be other 'infinite' numbers greater than this - as there would be if C had started as {n,n+1} and R {n+2,n+3,...}

In these cases, the remaining naturals k =/= 1 mod 10 are all left in the jug at the end. In that sense, they are higher than the set of all numbers (10n-9), which are isomorphic to the naturals. Is this what you were looking for? It's worth mentioning, however, that this isn't equivalent to the initial supertask - here, we have a collection of balls that are explicitly never removed, and proving their existence is trivial.

That's along the lines I was thinking, creating a set that sorts higher than N. I'm not sure why the fact that 'balls are never removed' makes one supertask more trivial than the other though. The set theory stuff is a model for the thought experiment, and at any step we are specifying a sequence of numbers that are present in one place or another. Set theory doesn't care when or how a number was added to a sequence, just whether and where it is. What's important is the state of the sets at any stage of the supertask, surely.

What we are going to do is take two parallel lines and connect them with two line segments. Then, we will rotate those two line segments at different rates, such that the angles between the line segments and one of the parallel lines becomes zero. We will find that, if the angles are defined appropriately for each step, we get triangles with areas that are analogous to the sets in the original puzzle.

This is a very useful visual analogy, thanks for taking the time to describe and diagram it.
I would answer your questions as follows:
1) I agree, θ and φ converge to zero
2) The intersection points are at infinity if that's how you like to describe it, or further away than any distance you can compare them with (although I would argue that the φ intersection is still in some sense 'even further' than the θ)
3) Here I think things get interesting. The area above the blue line is 'infinite' as is the area above the red line, but I don't think that necessarily means the area in between the lines as defined in the problem is zero. Infinity minus infinity can be a variety of things as the Hilbert hotel teaches us.
I don't think it's valid to switch from taking the difference of areas for finite n, to saying that "the lines are coincident so the areas are the same" at infinite n. You need to consistently follow the area of the triangle (0,0)(a_n,1)(10a_n,1) and see how it behaves as n->infinity.
In fact if we calculate the area of K_n for any n it is proportional to 9n, so I'd argue for the area becoming infinite. The triangle is infinitely long and infinitesimally thin, so it's only decidable which one 'wins' if you look at how they are behaving before they get to infinity.

Spoiler:

This is like the behaviour of sinc(x)=sin(x)/x which converges to 1 at x=0 even though 0/0 is not defined. (But if you take the Taylor expansion of sin(x) and divide each term by x it is very clear that the result should be 1 at x=0)
Yet sqrt(x)/x diverges to infinity at x=0 even though it can also be seen as 0/0

4) Following the same argument as for Q3, I'd say that K_n has infinite area for all n, but K's area cannot be determined if it is to be described as the difference between the red-bounded and blue-bounded 'triangles'.
5) The area of K_n is 1 for all finite and infinite n for the same reasons given above - infinity minus infinity can make a variety of things, depending on how you get to the infinities. In this case it makes 1.

[ucim]

Good. Before addressing the rest of your post, let me propose a minor elaboration. Partition the natural numbers into three sets:
L={1,2,3,4...n-1}
C={n}
R={n+1, n+2, n+3...}
with the supertask of moving the lowest element of R into C, the lowest element of C into L, and incrementing n.

Consider now the cardinalities of the sets. Specifically, I claim that at the end,
the cardinality of L is infinite (equal to aleph null),
the cardinality of R is finite (equal to zero), and
the cardinality of C is also finite (equal to zero). Anything else leads to a contradiction, for the same reasons.

Do you agree with this?

No, I would say that C contains a single element with a number I have previously described as 'arbitrarily bug' but if that is technically the wrong label, let's call it infinite in the sense that whatever natural number you compare it with, this infinite number is always greater. But that's not to say that there can't be other 'infinite' numbers greater than this - as there would be if C had started as {n,n+1} and R {n+2,n+3,...}

Ok, let's look closer. With the substitution k=n-1, we can rewrite the example above as:
L={1,2,3,4...k}
C={k+1}
R={k+2, k+3, k+4...}
with the supertask of moving the lowest element of R into C, the lowest element of C into L, and incrementing k.

This is exactly the same puzzle, is it not?

Now, consider the union of C and R at the end of the supertask. This is the same as the set R of the two-partition puzzle, which you agreed was empty at the end of the supertask. It is empty, but it contains the element k+1, which implies that the element k+1 must not exist at the end of the supertask. Now, this is the only element of C at the end of the three-partition(k) supertask, so if that element doesn't exist, C must be empty.

Jose

[kryptonaut]

Ok, let's look closer. With the substitution k=n-1, we can rewrite the example above as:
L={1,2,3,4...k}
C={k+1}
R={k+2, k+3, k+4...}
with the supertask of moving the lowest element of R into C, the lowest element of C into L, and incrementing k.

This is exactly the same puzzle, is it not?

Now, consider the union of C and R at the end of the supertask. This is the same as the set R of the two-partition puzzle, which you agreed was empty at the end of the supertask. It is empty, but it contains the element k+1, which implies that the element k+1 must not exist at the end of the supertask. Now, this is the only element of C at the end of the three-partition(k) supertask, so if that element doesn't exist, C must be empty.

It's not really the same because we are 'keeping hold' of an element as we move up the numbers. This is really the nub of our different views I think. By keeping track of a particular element we partition the set into 'all those before' and then 'this' and then 'all that come after'. If we don't keep the 'this' part then at the end we have nothing. But if we keep 'this' then we end up holding on to it, and it has an infinite/higher-than-any-natural/whatever-we-call-it value. We'd end up dealing with something like {1,2,3,4,... k+1} in the ordered set of (L followed by C), where the element k+1 now has ordinal omega. (Having got this far we couldn't generate k from k+1, the element would have no predecessor)

Now people keep saying that 'an infinite value isn't in N', which is why I resist calling it infinite - but it's what you would get to if the endless counting ended - which also can't happen, but it does in this puzzle. (The jugs 'remove-the-lowest' puzzle is further complicated by growing an infinite ordered set for C)

The jugs puzzle doesn't specify whether the balls are allowed to end up with infinite/transfinite/whatever numbers on them, and I think this is the source of all our problems. If infinite numbers are allowed then I believe that my analysis is correct. If they are forbidden then I think there is no logical solution to the puzzle, like asking someone how they'd count their fingers if numbers higher than 9 didn't exist. "The last finger isn't there" would be as good an answer as any I guess, but it's not a very satisfying puzzle.

I think the physicist in me is happier to accept [edit: the appearance of] infinite numbers than the disappearance of (hypothetical) matter, but perhaps the more pure-mathematically inclined take a different view? But really I think the original question needs to make it clear what's allowed in the answer.

[Xias]

1) I agree, θ and φ converge to zero
2) The intersection points are at infinity if that's how you like to describe it, or further away than any distance you can compare them with (although I would argue that the φ intersection is still in some sense 'even further' than the θ)

Let's focus on these answers, since I think all of your further answers were informed by them.

So, you contend that the line formed by angle θ intersects y=1 at some infinitely large a, and the line formed by angle φ intersects y=1 at some infinitely large b, with b>a (perhaps even b = 10a).

If we follow the line formed by φ through the real values of x, we find that the points at any x are very close to y=0. So close that we can't tell the difference. At x = 1, or x = 10, or x = a google, or x = graham's number, the line is indistinguishable from y=0. But in order to intersect y=1 at b, the difference must be there, if only infinitesimally small.

Do we agree so far, that this infinitesimally small distance between y=0 and the φ-line follows from your answer?

Furthermore, the line formed by angle θ would be very close to both y=0 and the line formed by φ; only, at any value of x (let's take graham's number G), the infinitesimal distance between y=0 and the θ-line is still greater than the distance between y=0 and the φ-line.

Do we agree so far?

So if we zoom in incredibly close to the point (G, 0), infinitely close, we'll see that the φ-line takes a value (G, ε) where ε is an infinitesimally small number greater than 0 and smaller than any real number. Also, the θ-line will cross the point (G, δ), where δ is an infinitesimally small number greater than ε. In other words, 0 < ε < δ < y (for any real number y).

Do we agree so far?

If we draw a line from (0,0) through (G,ε), we get the φ-line. Since that line makes an angle φ with the x-axis, we can calculate φ as arctan(ε/G). ε/G is another infinitesimally small number, but still greater than 0, and so arctan(ε/G) is also infinitesimally small, but greater than 0. We can do the same with θ = arctan(δ/G), an infinitesimally small number greater than 0.

So then we have 0 < φ < θ. But then φ and θ are not zero; they are infinitesimal, greater than 0. But this contradicts your answer to the first question.

If the θ-line intersects y=1 at any point (even at a point where x is infinitely large), then θ cannot be zero. If θ is zero, the the point of intersection does not exist - not even for infinitely large x. The same goes for φ. The guest is out in the car park, so to speak.

How would you like to resolve this?

[ucim]

It's not really the same because...

Not the same as which? The three-partition(n) puzzle is exactly the same as the three partition(k) puzzle, except that we are "looking at" an element one earlier or one after (depending on whether we use n or k). If the one earlier exists, the one after does too. If the one earlier doesn't exist, then neither does the one after.

If you agree with this but were merely claiming that the three-partition puzzle is not the same as the two-partition puzzle, then rethink the two-partition puzzle, adding another step where after we increment n, we then (edit: temporarily) partition R into {the lowest number in R} and {the rest of R}. If R is empty, then both partitions are empty. If R is not empty, then {the lowest number in R} has cardinality 1 (and a finite element in it), and {the rest of R} has cardinality aleph null. Anything else leads to contradiction.

...we are 'keeping hold' of an element as we move up the numbers.

We do so in both (two-partition and three-partition) scenarios, the difference being that in the two-partition scenario we "keep hold" of it by leaving it in R. We never discard any element in R; in fact we never discard any element at all. We are just moving the partitions. Every element is always somewhere, either in L, C (for the three partition), or R.

If we don't keep the 'this' part then at the end we have nothing.

As far as C is concerned, we don't keep it. In the next step, it "moves" into L.

Now people keep saying that 'an infinite value isn't in N', which is why I resist calling it infinite - but it's what you would get to if the endless counting ended - which also can't happen, but it does in this puzzle.

No, the task doesn't "end". It "becomes finished". It's an important distinction, and is the one I alluded to and you seemed to agree with, here: "This happens, but there is no step at which it happens, because there are an infinite number of steps."

The jugs puzzle doesn't specify whether the balls are allowed to end up with infinite/transfinite/whatever numbers on them...

Yes it does. The numbers on the balls in the jug are all natural numbers. They can't "end up as" anything else, neither transfinites, nor emoji, nor ham sandwiches, so long as you don't alter the labels (which is the set of games I'm considering at this point). Even if a ball labeled omega appeared "somehow", did it come just after the ball labeled "omega minus one"? It's supposed to. And what about "omega minus two" all the way to "omega minus... a number bigger than any number you can think of", which has to come before any of the transifinites. (Note - you'd have this problem even if the supernatural numbers started out in the jug).

I think the physicist in me is happier to accept [edit: the appearance of] infinite numbers than the disappearance of (hypothetical) matter,

Is the physicist in you happier to accept the appearance of infinite (hypothetical) matter than the disappearance of natural numbers from a set?

Jose

[kryptonaut]

How would you like to resolve this?

The value 'driving' the puzzle is a_n, which is analogous to n in the jugs game. The angle subtended by a_n is zero when n is infinite. But that doesn't mean that the area of the triangle is zero. Zero times infinity can be any variety of things.

A point source at infinity subtends an angle of 0. A star (assume it's infinitely bright) at infinity will also subtend an angle of 0, but the star itself hasn't suddenly become a point. The galaxy that the star is in will subtend an angle of 0 but it hasn't become a point either, nor has it become the same size as the star. The area of the triangle (or volume of the cone if we're in 3D) subtended by the star is infinite, as is that of the galaxy, and that of the galaxy minus that of the star. The area/volume subtended by the point source is zero.

If we take your diagram and use a projection that maps the x axis to -2^-x, so we can fit all of infinity between -1 and 0, then the red and blue lines will map to curves which 'sag' more and more steeply as n increases. At the limit both curves follow along the x axis to (0,0) and then climb up to (0,1). The projections of the curves map to the same thing, a vertical line at x=0, but the 'density' at that location is infinite. What is zero times infinity? It all depends how it's set up.

If we don't keep the 'this' part then at the end we have nothing.

As far as C is concerned, we don't keep it. In the next step, it "moves" into L.

We keep a set which contains ball number (k+1). We are tracking the last number to be removed from R. At the end of the supertask we have a pointer to the last number removed. This is what makes all the difference. It is not a nameable, write-downable number. It is described by its property of being bigger than any natural number you compare it with. Call it infinity, call it whatever you like, but this is how it behaves.

The jugs puzzle doesn't specify whether the balls are allowed to end up with infinite/transfinite/whatever numbers on them...

Yes it does. The numbers on the balls in the jug are all natural numbers. They can't "end up as" anything else, neither transfinites, nor emoji, nor ham sandwiches, so long as you don't alter the labels (which is the set of games I'm considering at this point). Even if a ball labeled omega appeared "somehow", did it come just after the ball labeled "omega minus one"? It's supposed to. And what about "omega minus two" all the way to "omega minus... a number bigger than any number you can think of", which has to come before any of the transifinites. (Note - you'd have this problem even if the supernatural numbers started out in the jug).

I've explained several times that omega is not a fixed property of a particular number, it refers to the position of an element in an ordered set, like "third". In particular it represents the position of an element that has a countable infinite number of elements before it in the set. There is no such thing as omega minus one. You should read more about it if this isn't clear to you.

If a number is incremented for each element of N, or if we keep a pointer to the n'th element, such that by the end of the supertask we have a pointer to the final number X we got to even though we can't write it down, then in the ordered set of all naturals followed by X, X has ordinal omega. The set is {1,2,3,... X}. If we try to map N to this set we can match all of {1,2,3,...} but X will remain unmatched because it is greater than all N - even if it was picked out of N in the first place. It's weird but so is infinity. X is the number on the ball in your set {C} at the end of your supertask. X is the room number of the last room in the Hilbert Hotel. Every room number is less than it, if you check each room in turn there is always another one before you get to X, so you'll never get there even with a supertask.

I think the physicist in me is happier to accept [edit: the appearance of] infinite numbers than the disappearance of (hypothetical) matter,

Is the physicist in you happier to accept the appearance of infinite (hypothetical) matter than the disappearance of natural numbers from a set?

What do you mean? The puzzle stipulates a never-ending supply of balls, which are added faster than they are removed, so it seems very reasonable to imagine we'll end up with an infinite number of them. It also stipulates that the numbers on the remaining balls are constantly increasing so it also seems perfectly reasonable to imagine they'll end up infinitely big.


One more try at a visual analogy to the 'remove lowest' puzzle.
Let's use numbered blocks, 1cm high, and build a tower with 1 at the bottom, then 2, etc. Each step we add 10 new blocks, and a gremlin starts at the bottom of the tower painting one block black each step. Now let's view this tower from the ground, 10 cm away.

After step 1 the tower subtends an angle of 45 degrees, with the bottom 10% painted black. After step 2 we'll move back another 10cm, the tower will be 20cm high with 2 black blocks at the bottom. It will subtend an angle of 45 degrees, and the bottom 10% will be black.

In fact if we keep moving 10cm per step the tower will always look the same, subtending an angle of 45 degrees with the bottom 10% black. The first visible block will be numbered n+1.

Perform this as a supertask, so we can get infinitely far away - the tower still subtends an angle of 45 degrees with the bottom 10% black. For the tower to subtend a finite angle at infinity the tower must be infinitely tall, and the first non-black block must be infinitely high off the ground since any finite height will subtend an angle of zero with the horizon.

The 'remove highest' variant (if the gremlin actually removes the blocks) would produce a tower that looked the same height as the unpainted section in the previous example, but standing at horizon-level.

[Xias]

How would you like to resolve this?

The value 'driving' the puzzle is a_n, which is analogous to n in the jugs game.

θ is the thing that you change at each step, from which all other values in the puzzle are defined. So why is a_n the value that drives the puzzle?

If we think of the puzzle as a dial that we turn, which points a laser toward the line y=1, then the point of contact of the laser is equivalent to a_n. Forget the supertask aspect of it. You can just turn the dial from pointing straight upward, to pointing straight rightward. You can even point it downward at an angle of -π/2, then it will point to the line y=-1, and touch it at the point (1,1). It's clear that at any angle less than 0, the laser won't ever touch y=1. At any angle greater than 0, it won't ever touch y=-1. At the angle exactly zero, which line does it touch?

The angle subtended by a_n is zero when n is infinite.

If the angle is zero, then a does not exist. I went into a lot of detail in my reply to you as to why this is. I was very generous with you. You have accused me before of refusing to "explore the possibility" of infinite stuff and I spent a lot of energy exploring the possibility of the lines intersecting y=1 at points infinitely far away. So please forgive my frustration when you don't even acknowledge those arguments, refusing to "explore the possibility" yourself.

If the two lines intersect y=1 at different points, even infinitely large ones, then the angle between them must be greater than zero. Since the distance is infinitely large, the angle must be infinitesimally small, smaller than any real number. Are you arguing that all parallel lines intersect at infinity? I'd be happy to talk about this with consideration for the ideal line, but it doesn't support what you are saying either. Which brings me to this:

A point source at infinity subtends an angle of 0. A star (assume it's infinitely bright) at infinity will also subtend an angle of 0, but the star itself hasn't suddenly become a point.

Are we talking about infinity? Or are we talking about the numbers-larger-than-any-natural-number? If you want to talk about the latter, you can't just borrow the rules about infinity; you have to develop the rules around them so that they are consistent. A source at a distance larger-than-any-natural-number subtends a nonzero angle smaller-than-any-real-number. Do you think this is wrong? Why? What argument do you have against using numbers-smaller-than-any-real-number that you haven't completely dismissed when speaking about numbers-greater-than-any-natural-number?

I contend that if we are accepting numbers-greater-than-any-natural-number, then the angle subtended by the star is some ε, and the angle subtended by the galaxy is some δ, and that 0 < ε < δ < y (for any real number y).

What is zero times infinity? It all depends how it's set up.

I agree completely.

[ucim]

At the end of the supertask we have a pointer to the last number removed.

There is no last number removed. The "last number removed" doesn't exist. So, we have a pointer to null. The set whose only element is the "last number removed" has to be empty. This happens, but there is no step at which it happens, because there are an infinite number of steps.

There is no such thing as omega minus one. You should read more about it if this isn't clear to you.

First, let's drop the ad hominum if you want the discussion to continue.

Second, omega is an ordinal. It refers (as you say) to the position in an ordered set. Omega minus one refers to the position of the element before it.

then in the ordered set of all naturals followed by X, X has ordinal omega. The set is {1,2,3,... X}

Fine but irrelevant. X doesn't exist. It never existed. The supertask does not create it.

...such that by the end of the supertask we have a pointer to the final number X we got to even though we can't write it down

We never get to this number. It's not that we can't write it down, it's that it doesn't exist. Similarly, the smallest positive number doesn't exist even though you go through all the positive numbers on the way through zero. There simply is no such animal.

One more try at a visual analogy to the 'remove lowest' puzzle.

No. This particular one (ten remove one) adds a complication to an issue that is still not settled in its plainest form (one remove one).

The issue is: If you count to infinity, what's the last number you reach?

Your answer is infinity.

My answer is there is no such number.

It can be posed in the opposite way: If you count down to zero saying "1/n" at time t=1-(1/n), as n increases without bound, what number is the last number you said at time t=1 when you "reach zero"?

Your answer is "a number so small there's nothing smaller, but not zero".

My answer is "There is no such number".

Now go back to the two-partition puzzle. You're ok with R being empty. Are you ok with L being N? Because N doesn't contain "a number that's bigger than any number you can imagine". Such a number was never in R, and is never in L. How could it possibly get into (in the three-partition puzzle) C?

Jose

[Xias]

Second, omega is an ordinal. It refers (as you say) to the position in an ordered set. Omega minus one refers to the position of the element before it.

In fairness to kryptonaut, there is no element before an omega-th element for "omega minus one" to refer to.

[kryptonaut]

θ is the thing that you change at each step, from which all other values in the puzzle are defined. So why is an the value that drives the puzzle?

If we think of the puzzle as a dial that we turn, which points a laser toward the line y=1, then the point of contact of the laser is equivalent to an. Forget the supertask aspect of it. You can just turn the dial from pointing straight upward, to pointing straight rightward. You can even point it downward at an angle of -π/2, then it will point to the line y=-1, and touch it at the point (1,1). It's clear that at any angle less than 0, the laser won't ever touch y=1. At any angle greater than 0, it won't ever touch y=-1. At the angle exactly zero, which line does it touch?

I'd say that a_n is analogous to the number of a particular ball in the original puzzle. The area of the triangle subtended by line segments of y=1 between different values of a is analogous to the size of a set. Shining a laser pointer back towards the origin from a_n will also produce an angle of 0 at n=infinity, and I'd say that's more descriptive of the puzzle setup. It can be argued both ways. Which one is more appropriate depends on how you get there, like asking whether 1/0 should be + or -infinity.

If the angle is zero, then a does not exist. I went into a lot of detail in my reply to you as to why this is. I was very generous with you. You have accused me before of refusing to "explore the possibility" of infinite stuff and I spent a lot of energy exploring the possibility of the lines intersecting y=1 at points infinitely far away. So please forgive my frustration when you don't even acknowledge those arguments, refusing to "explore the possibility" yourself.

If a is infinite then the angle is zero. I'm sorry if I didn't acknowledge your discussion about Grahams numbers and so on - there are a lot of arguments being made on all sides and many of mine don't get a direct response either. But your argument appeared to be affirming that there is no upper limit to the sequence of numbers, real or natural - and I agree with this. But that doesn't make the concept of infinite numbers invalid.

When I am talking about a number X with ordinal omega in a set {1,2,3,... X} I mean that X is infinite in the sense that no natural number is greater (or equal) to it. But it is possible to talk about X+1 or X+n in a meaningful way. But there is no number whose successor is X.

Are we talking about infinity? Or are we talking about the numbers-larger-than-any-natural-number? If you want to talk about the latter, you can't just borrow the rules about infinity; you have to develop the rules around them so that they are consistent. A source at a distance larger-than-any-natural-number subtends a nonzero angle smaller-than-any-real-number. Do you think this is wrong? Why? What argument do you have against using numbers-smaller-than-any-real-number that you haven't completely dismissed when speaking about numbers-greater-than-any-natural-number?

In the jug and ball game we are talking about numbers such as the X I described above - actually an ordered set {X,X+1,X+2,...} isomorphic with N

Second, omega is an ordinal. It refers (as you say) to the position in an ordered set. Omega minus one refers to the position of the element before it.

omega is a limit ordinal, there is no element before it just as there is no -1'th element.

The issue is: If you count to infinity, what's the last number you reach?

Your answer is infinity.

My answer is there is no such number.

My answer is an infinite number. We take an infinite number of steps to reach it, so I see no contradiction there.

Now go back to the two-partition puzzle. You're ok with R being empty. Are you ok with L being N? Because N doesn't contain "a number that's bigger than any number you can imagine". Such a number was never in R, and is never in L. How could it possibly get into (in the three-partition puzzle) C?

The size of L or N is bigger than you can imagine. If you tack a number onto the end of it in an ordered set, {1,2,3,... X}, then by definition that number has to be bigger than any finite natural number. By introducing a partition into N with an unbounded number of elements before it, we create two sets, at least one of which is infinite. The second set by definition then has to start at an infinite number.

If the Hilbert hotel is full of people with red keyfobs, and an infinite number of guests with blue keyfobs are introduced using the wrong method, by shunting all the red keys up one place at every step, then we end up with a set {blue1, blue2, blue3,... red1, red2, red3...} In this scenario I suggest the reds are analogous to the 'keepers' and the blues to the 'discards' in the 'remove the lowest' jug game.

If your analysis is that the reds just cease to exist then that is the fundamental basis of our disagreement.
I assert that the reds are there but in rooms which are inaccessible. Effectively their room numbers are infinite.
What is your view of what happens?

[Xias]

If a is infinite then the angle is zero. I'm sorry if I didn't acknowledge your discussion

With that it becomes clear that you aren't interested in "exploring" anything. Take care.

[ucim]

omega is a limit ordinal, there is no element before it just as there is no -1'th element.

In fairness to kryptonaut, there is no element before an omega-th element for "omega minus one" to refer to.

Ok. I stand corrected. But in any case, we are not talking about ordinals, we are talking about cardinals. Members of the set N. There is no omega-th cardinal number in N. The set containing the omega-th cardinal number of N is the empty set. This remains true even after you "count to infinity".

If you tack a number onto the end of it in an ordered set, {1,2,3,... X}, then by definition that number has to be bigger than any finite natural number.

Fine, but at no point do we "tack a number onto the end of {1,2,3...}".

The second set by definition then has to start at an infinite number.

Or be empty. Since there are no infinite numbers in N, the set must be empty. This is true for R (after the supertask) and also for C (after the supertask) for the same reason. In the two-partition game you agreed that R ends up empty. In the three-partition game, C ends up empty for the same reason.

"How did it get that way?" There's no step at which this happens, but it does end up having happened, because there are an infinite number of steps.

Look, I'm not disputing the legitimacy of transfinite numbers. I'm just saying that there are none in N, and we are dealing solely with this set. It has cardinality aleph null, and does not contain any form of infinity as an element. No subset or partition of N can therefore contain any form of infinity as an element.

My answer is an infinite number. We take an infinite number of steps to reach it, so I see no contradiction there.

The contradiction is that there is no "infinite number" in N. Therefore this putative infinite number that is your answer, doesn't exist. Thus, your answer must reduce to the null set, or "there is no such". Which is my answer.

If the Hilbert hotel is full of people with red keyfobs, and an infinite number of guests with blue keyfobs are introduced using the wrong method, by shunting all the red keys up one place at every step [...]
What is your view of what happens?

They are "in" rooms that don't exist. It's the black hole of hotels... the Hilbert Hotel California. (It reminds me of the set of sets that don't contain themselves as members, or somesuch).

Jose

[kryptonaut]

If you tack a number onto the end of it in an ordered set, {1,2,3,... X}, then by definition that number has to be bigger than any finite natural number.

Fine, but at no point do we "tack a number onto the end of {1,2,3...}".

We have {C} specified as following L

My answer is an infinite number. We take an infinite number of steps to reach it, so I see no contradiction there.

The contradiction is that there is no "infinite number" in N. Therefore this putative infinite number that is your answer, doesn't exist. Thus, your answer must reduce to the null set, or "there is no such". Which is my answer.

Let me try to show you how the infinite numbers arise.

Let's build a tower using numbered blocks, putting the next highest number on top of the tower at each step. But let's make the blocks squishy so that when we add a block we press down on it until the tower is exactly 1 metre high with each block occupying the same proportion of the height. At every step n the tower is 1m high and contains n blocks. We make this a supertask and put all of N into the tower.

Now, if I want to point to any finite number x in the tower, I just point to the base of the tower. But if I point half way up the tower, what can I say about the number y I'm pointing to? Well, y is infinite. There are an infinite number of blocks below y. If I started counting from the bottom I would never get to y. There are also an infinite number of blocks above y, which must also have infinite numbers on them because numbers increase with height. If I started enumerating the blocks upwards from y I would never reach the top of the tower.

This is what I mean when I say that partitioning the set N can create an element with an infinite number. It is a number you can't count up to, because its specification is a function of the size of N.

The element y has ordinal omega in the tower, simply by virtue of it having been singled out in this way, because it can never be counted up to.

Does that help clarify things?

[ucim]

We have {C} specified as following L

Well, C follows L (note notation) inasmuch as all elements of C are strictly greater than any element of L. This doesn't force any elements of C to be infinite; It just ensures that at the end of the supertask, C does not contain any finite elements. C could be empty. Elements do get removed from C at every step; nothing requires there to be an element in C.

At the end of the two-partition supertask, R contains all elements of N that are greater than all finite elements of N. There are none, so R ends up empty. At the end of the three-partition supertask, C contains the smallest element of N that are greater than all finite elements of N. There is none, so C ends up empty.

Since there are no infinite elements in N (which is partitioned into L, C, and R), a set whose elements would all have to be infinite must therefore have no elements. It's the same reason R is empty at the end of both supertasks, even though there is no specific step that empties it, because there are an infinite number of steps.

Let's build a tower using numbered blocks, putting the next highest number on top of the tower at each step. But let's make the blocks squishy so that when we add a block we press down on it until the tower is exactly 1 metre high with each block occupying the same proportion of the height.

An interesting supertask, but this is not the same game. Remember, if you change the game, you change the outcome. In any case, my answer for the tower is that at the end of the supertask, it is infinitessimally tall. If you point "half way up" you're pointing to empty space. An infinite tower collapses under its own weight.

1: If you are pointing at a number, it must be an infinite number.
2: There are no infinite numbers in N.
3: Therefore, you are not pointing at a number, even if at every finite step there was one there.

Jose

[Demki]

The only way for that tower to have nonzero size is if each piece decreases in size(say the size of the section halves with each step), so wherever you point, it is part of some finite section corresponding to a natural number, if you point from above straight down at the top, there is no number you are pointing at, because there is no last element in the natural numbers.

There is no uniform distribution over the natural numbers.

[kryptonaut]

Well, C follows L (note notation) inasmuch as all elements of C are strictly greater than any element of L. This doesn't force any elements of C to be infinite; It just ensures that at the end of the supertask, C does not contain any finite elements. C could be empty. Elements do get removed from C at every step; nothing requires there to be an element in C.

...except for the fact that there is no mechanism for leaving C empty, since every step leaves something in it.

The way to show two infinite sets A and B are the same size is to show that for every n'th element in A there is an element in B, and vice versa. That's how N can be split into an odd and even partition each with the same size as N, by mapping n to 2n-1 in one case and to 2n in the other. In your 'R-draining-into-L' setup it is possible to map R and L in a 1:1 mapping, since they are both N. If one of them is N-followed-by-C then it is no longer possible to set up this mapping so the sets can no longer be considered equivalent.

An interesting supertask, but this is not the same game. Remember, if you change the game, you change the outcome. In any case, my answer for the tower is that at the end of the supertask, it is infinitessimally tall. If you point "half way up" you're pointing to empty space. An infinite tower collapses under its own weight.

1: If you are pointing at a number, it must be an infinite number.
2: There are no infinite numbers in N.
3: Therefore, you are not pointing at a number, even if at every finite step there was one there.

It was not intended to be the same game, I was trying to show how dividing a countably infinite ordered set leads to infinite numbers. Saying the tower collapses under its own weight is as helpful as claiming the jug explodes in the original game. Set theory isn't the original game either, it's a mathematical model. I am offering a visual model for dividing an infinite set.

But if you want a more concrete example, consider a 1m ruler. There are a countable infinite number of rational numbers between the 0m and 1m marks. If you start at 0 and try to step up the ruler one rational number at a time then you won't get anywhere. If you point at the 50cm mark and ask how many rational numbers are between there and zero the answer is 'a countably infinite number'. How many do you have to count to to get past the 50cm mark? How many do you have to count to get from 50cm to 1m? Well that section has another countable infinity of rationals. As does any section you care to point to. Starting from any division you have to count an infinite number of steps to get to the next one, even though the whole 1m ruler contains a countable infinite number of steps itself.

If you object to the fact that you can't enumerate the rationals in numerically ascending order, let's subdivide the ruler with n markings and let n increase to the size of N. Point to the 10th mark, the millionth mark, the googolplexth mark, they're all at the 0 position. You have to count an infinite number to get anywhere. If you start anywhere, you have to count another infinite number to get anywhere else. Every non-zero length of the ruler has an infinite number of markings on it.

The only way for that tower to have nonzero size is if each piece decreases in size(say the size of the section halves with each step), so wherever you point, it is part of some finite section corresponding to a natural number, if you point from above straight down at the top, there is no number you are pointing at, because there is no last element in the natural numbers.

Well I specified that the blocks are squishy so that the tower compresses with each block added. Zero times infinity can be pretty much anything - in this case, 1.

There is no uniform distribution over the natural numbers.

That is the gist of what I am demonstrating, by showing that anything above the ground floor of the block is infinite, even though it was made with naturally-numbered blocks.

This whole thing seems to be becoming a bit of a philosophical debate. Maybe we should all just go and try the experiment, and report back next week with our results.

[Poker]

The only way for that tower to have nonzero size is if each piece decreases in size(say the size of the section halves with each step), so wherever you point, it is part of some finite section corresponding to a natural number, if you point from above straight down at the top, there is no number you are pointing at, because there is no last element in the natural numbers.

Well I specified that the blocks are squishy so that the tower compresses with each block added. Zero times infinity can be pretty much anything - in this case, 1.

But is it 1?

Let me offer another take: suppose such a tower did have height 1 (or, indeed, any height greater than 0). Then, since the height of each "block" is 0, every two different vertical points on the tower correspond to two different numbers.

Now, since the only dimension that matters in this tower is the height, would you agree that the tower is effectively isomorphic to a line segment? (0 to 1, naturally) If not, you'll need to show me how it's not, so I'll proceed assuming you agree. In that case, since any such line segment contains an uncountably infinite number of points, the tower itself must contain an uncountably infinite number of blocks. But this is a contradiction, since the only blocks that we put on are natural numbers, and the natural numbers are merely countably infinite. Since we've run into a contradiction, our only assumption (that the power has height greater than 0) must be false.

I was trying to show how dividing a countably infinite ordered set leads to infinite numbers.

The trouble is, what we're dealing with in the original problem is the natural numbers in their usual ascending order. In order for your "infinite numbers" to exist, there needs to be an infinite amount of numbers followed by something. This doesn't happen with the natural numbers in their usual ascending order.

[Demki]

When we say '0×∞ can be anything', we don't mean the number zero times some notion of ∞, because that is undefined under most notions of infinity(and where it is defined we lose some properties, like multiplicative inverses).
What we mean is that if we have 2 real-valued functions f(x) and g(x), such that lim_x->c f(x) = 0 and lim_x->c g(x) = ∞ (remember this only means that we can make g(x) as large as we want by getting closer to c, not that g(c)=∞), then lim_x->c (f(x)×g(x)) can evaluate to anything, depending on the choice of f(x) and g(x).
Note that for lim_x->c (f(x)×g(x)) = (lim_x->c f(x)) × (lim_x->c g(x)) to hold, we must have that both limits in the rhs exist, which is not the case when lim_x->c g(x) = ∞.


Also, 0×aleph_0, when looking at multiplication for cardinalities is identically 0. This is because |A|×|B| is defined as the cardinality of the cartesian product of A and B, and the cartesian product of the empty set, which has cardinality 0, with any other set, is the empty set.
In fact, for ordinalities, 0•omega and omega•0 are both also 0.

So you can't have a uniform distribution over the naturals, thus your supertask either ends up with a tower of 0 height, or with the natural numbers distributed unevenly across it.
If you at some point added transfinite/superreal/not-natural-numbers, that should have been explicit in the description.

Edit: doing this on a phone and I forgot that it's a forward slash (/) and not a backslash (\) *facepalm*

[kryptonaut]

The trouble is, what we're dealing with in the original problem is the natural numbers in their usual ascending order. In order for your "infinite numbers" to exist, there needs to be an infinite amount of numbers followed by something. This doesn't happen with the natural numbers in their usual ascending order.

In the 'remove the lowest' problem we have exactly this scenario - the set of removed balls is {1,2,3... n} and the set of remaining balls is {n+1,n+2... 10n}
As n->infinity this becomes an infinite number of something followed by an infinite number of something. All in ascending order. My contention is that this means the upper set takes on infinite values, and is countably infinite in size. Others assert that since no infinite numbers were present to start with then this cannot be, so the set must be empty.

[kryptonaut]

@Poker, @Demki - Ok, I agree you can't have a uniform distribution of N. I was just trying to make a visual metaphor that would show how dividing N at some point relative to the size of N would make numbers with ordinal omega appear.

[Poker]

The trouble is, what we're dealing with in the original problem is the natural numbers in their usual ascending order. In order for your "infinite numbers" to exist, there needs to be an infinite amount of numbers followed by something. This doesn't happen with the natural numbers in their usual ascending order.

In the 'remove the lowest' problem we have exactly this scenario - the set of removed balls is {1,2,3... n} and the set of remaining balls is {n+1,n+2... 10n}
As n->infinity this becomes an infinite number of something followed by an infinite number of something. All in ascending order. My contention is that this means the upper set takes on infinite values, and is countably infinite in size. Others assert that since no infinite numbers were present to start with then this cannot be, so the set must be empty.

And it's possible that I've missed your explanation on where the infinite numbered balls came from since they did not exist at the start. Would you mind pointing it out to me?

[kryptonaut]

And it's possible that I've missed your explanation on where the infinite numbered balls came from since they did not exist at the start. Would you mind pointing it out to me?

They arise because the set N is divided in the way I just described, at a point relative to the size of N. The numbers in N go on for ever. If you divide N this way, you have a sequence that goes on for ever, then another sequence (which also goes on for ever in this case). The first number of the second sequence has ordinal omega, an infinite ordinal. These numbers are bigger than anything in N. The actual values are impossible to specify, just as it's impossible to specify the number that comes 10% of the way along the number line. Nevertheless the puzzle forces this division.

You could just as validly specify the set of remaining balls as {1,2,3...9Nn _(edit)} accompanied by an increasing counter that displayed n at every step. I guess then the question is "After an infinite number of steps, does the counter display infinity, or does it evaporate?"

[Poker]

And it's possible that I've missed your explanation on where the infinite numbered balls came from since they did not exist at the start. Would you mind pointing it out to me?

They arise because the set N is divided in the way I just described, at a point relative to the size of N. The numbers in N go on for ever. If you divide N this way, you have a sequence that goes on for ever, then another sequence (which also goes on for ever in this case). The first number of the second sequence has ordinal omega, an infinite ordinal. These numbers are bigger than anything in N. The actual values are impossible to specify, just as it's impossible to specify the number that comes 10% of the way along the number line. Nevertheless the puzzle forces this division.

You could just as validly specify the set of remaining balls as {1,2,3...9Nn _(edit)} accompanied by an increasing counter that displayed n at every step. I guess then the question is "After an infinite number of steps, does the counter display infinity, or does it evaporate?"

Can you prove that it's even possible to divide N this way? And can you prove that the puzzle does force this division?

[ucim]

...except for the fact that there is no mechanism for leaving C empty, since every step leaves something in it.

Every step also takes something away, and in the end, there's nothing left.

C is the set containing the next natural number. When you have used up all the natural numbers, there is no next number. This makes the set empty.

Why doesn't C contain an fraction? Because there are no fractions in N.
Why doesn't C contain an imaginary number? Because there are no imaginary numbers in N.
Why doesn't C contain an emoji? Because there are no emoji in N.
Why doesn't C contain a ham sandwich? Because there are no ham sandwiches in N.

Why doesn't C contain a transifinte? Because there are no transifintes in N.

C contains the natural number that is bigger than all natural numbers. There is no such number. Therefore C is {}.

Anything else leads to contradiction. To mangle Holmes - "When the impossible has been eliminated, what is left, no matter how weird, must be the truth."

Jose

[kryptonaut]

Can you prove that it's even possible to divide N this way? And can you prove that the puzzle does force this division?

It's not possible to take set N and pick a specific number partway up the set. But it's possible to construct two sets simultaneously, using every member from N, but keeping the two sets partitioned so that one is always higher than the other. That's what the puzzle does. When you do that, the first member of the second set has ordinal omega. That means it has an infinite value. Note that 'infinite' is not a property of the number itself, it's a property of where that element occurs in an ordered set.

The number 1 can have the ordinal omega if you decide to count through N in the order 'even numbers first, then odd numbers'. There is no defined element whose successor is '1' in this ordering. Every element in the combined set has a successor. There are a countably infinite number of elements in the first part (even numbers) and in the second part (odd numbers).

C is the set containing the next natural number. When you have used up all the natural numbers, there is no next number. This makes the set empty.

You never 'use up all the natural numbers'. There are always more.

Why doesn't C contain an fraction? Because there are no fractions in N.
Why doesn't C contain an imaginary number? Because there are no imaginary numbers in N.
Why doesn't C contain an emoji? Because there are no emoji in N.
Why doesn't C contain a ham sandwich? Because there are no ham sandwiches in N.

Why doesn't C contain a transifinte? Because there are no transifintes in N.

C contains the natural number that is bigger than all natural numbers. There is no such number. Therefore C is {}.

The property of being infinite is an ordinal property. It depends on how the set of numbers is ordered. Re-ordering the set N in the way specified in the puzzle gives some elements J_n this property, because there is a countably infinite set of elements with values lower than them.

Divide N into even and odd numbers. Count through them in order 'evens first then odds' You never get to the first odd number. Does that mean the odd numbers no longer exist? No, they just have an infinite ordinal in this particular ordering {2,4,6,... 1,3,5}

Mix an infinite set of numbered blue balls and and an infinite set of numbered red balls together. Count them like red1,blue1,red2,blue2,... giving each one a label with its number, and you'll get through them all in a supertask, every ball will have a corresponding finite numbered label. Count them like red1,red2,red3,.... blue1,blue2,blue3... giving each one a label and you won't get to the first blue, the blue labels would all have to be infinite, higher than any of the finite red labels. How did the blue balls suddenly get infinite values? Because of the ordering imposed on the set.

C has an infinite ordinal in the ordering {1,2,3,... C}

J has an infinite ordinal in the ordering {1,2,3,... J,J+1,J+2,...}

The fact that these numbers came from N tells us that they are integers, but not where they come in any particular ordering.


The claim that
lim_n->inf {n+1,n+2,...10n} = {}
because n+1 and 10n both tend to infinity is as spurious as claiming that
lim_n->inf (10n - n) = 0

[Poker]

Can you prove that it's even possible to divide N this way? And can you prove that the puzzle does force this division?

It's not possible to take set N and pick a specific number partway up the set. But it's possible to construct two sets simultaneously, using every member from N, but keeping the two sets partitioned so that one is always higher than the other. That's what the puzzle does. When you do that, the first member of the second set has ordinal omega. That means it has an infinite value. Note that 'infinite' is not a property of the number itself, it's a property of where that element occurs in an ordered set.

You haven't proved that the second set has a "first member" in the first place.

The number 1 can have the ordinal omega if you decide to count through N in the order 'even numbers first, then odd numbers'. There is no defined element whose successor is '1' in this ordering. Every element in the combined set has a successor. There are a countably infinite number of elements in the first part (even numbers) and in the second part (odd numbers).

We're all in agreement that this is the case. What exactly is your point? Are you trying to claim that because we have a bijection between the finite natural numbers and the natural numbers plus some ordinals, that the ordinals must have been in the natural numbers in the first place? I doubt that's your claim, but when you think about it, isn't that roughly the same as what you're claiming with the natural numbers in their usual order? You're taking what you claim is "10% of the naturals" and then making claims about the other 90%. But you can put the 10% of the naturals you took away in a bijection with the entire set of naturals, in their usual order. (For example, if you took out 1, 11, 21, etc. you could use 1 -> 1, 11 -> 2, 21 -> 3, etc.) 10% of the naturals is the exact same size as the naturals. And when you try to take 10% of the naturals in their usual order, 10% of the naturals is 100% of the naturals.

C is the set containing the next natural number. When you have used up all the natural numbers, there is no next number. This makes the set empty.

You never 'use up all the natural numbers'. There are always more.

Not on any finite step, no.

Why doesn't C contain an fraction? Because there are no fractions in N.
Why doesn't C contain an imaginary number? Because there are no imaginary numbers in N.
Why doesn't C contain an emoji? Because there are no emoji in N.
Why doesn't C contain a ham sandwich? Because there are no ham sandwiches in N.

Why doesn't C contain a transifinte? Because there are no transifintes in N.

C contains the natural number that is bigger than all natural numbers. There is no such number. Therefore C is {}.

The property of being infinite is an ordinal property. It depends on how the set of numbers is ordered. Re-ordering the set N in the way specified in the puzzle gives some elements J_n this property, because there is a countably infinite set of elements with values lower than them.

Divide N into even and odd numbers. Count through them in order 'evens first then odds' You never get to the first odd number. Does that mean the odd numbers no longer exist? No, they just have an infinite ordinal in this particular ordering {2,4,6,... 1,3,5}

Mix an infinite set of numbered blue balls and and an infinite set of numbered red balls together. Count them like red1,blue1,red2,blue2,... giving each one a label with its number, and you'll get through them all in a supertask, every ball will have a corresponding finite numbered label. Count them like red1,red2,red3,.... blue1,blue2,blue3... giving each one a label and you won't get to the first blue, the blue labels would all have to be infinite, higher than any of the finite red labels. How did the blue balls suddenly get infinite values? Because of the ordering imposed on the set.

C has an infinite ordinal in the ordering {1,2,3,... C}

J has an infinite ordinal in the ordering {1,2,3,... J,J+1,J+2,...}

The fact that these numbers came from N tells us that they are integers, but not where they come in any particular ordering.

It occurs to me where our disagreement lies. We're both in agreement on the proposition "If there is a ball in the urn at midnight, it must not have a finite value on it." Your claim is "There must be a ball in the urn at midnight, therefore it must have an infinite value." Our claim is "The ball must have a finite value on it, therefore there are no balls in the urn." Can you prove the assumption "There must be a ball in the urn at midnight"?

The claim that
lim_n->inf {n+1,n+2,...10n} = {}
because n+1 and 10n both tend to infinity is as spurious as claiming that
lim_n->inf (10n - n) = 0

The set is defined over the natural numbers, yes? Then for any natural number x, there is a value of n for which x stops being in the set (namely, n >= x). Therefore, the set cannot contain any natural number.

[ucim]

You never 'use up all the natural numbers'. There are always more.

Supertasks are designed to "use up all the natural numbers" in some form or other. Your squishy tower "uses up all the natural numbers" when it collapses. You can "use up all the natural numbers" by putting them in a set; when you are done, there are no natural numbers outside the set.

The property of being infinite is an ordinal property. It depends on how the set of numbers is ordered.

Fine but we are not dealing with a "cleverly reordered set". We are dealing with the natural numbers in natural order: {1,2,3...} There is no infinite cardinal and there is no infinite ordinal in this set. Partitioning the set into L, (sometimes C), and R do not change this.

No infinites. No fractions. No emoji. No ham sandwiches.

It occurs to me where our disagreement lies. We're both in agreement on the proposition "If there is a ball in the urn at midnight, it must not have a finite value on it." Your claim is "There must be a ball in the urn at midnight, therefore it must have an infinite value." Our claim is "The ball must have a finite value on it, therefore there are no balls in the urn." Can you prove the assumption "There must be a ball in the urn at midnight"?

This.

Jose

[kryptonaut]

But you can put the 10% of the naturals you took away in a bijection with the entire set of naturals, in their usual order. (For example, if you took out 1, 11, 21, etc. you could use 1 -> 1, 11 -> 2, 21 -> 3, etc.) 10% of the naturals is the exact same size as the naturals. And when you try to take 10% of the naturals in their usual order, 10% of the naturals is 100% of the naturals.

And then if you try to count the 10% followed by the 90%, you'll never get to start on the 90% because they are infinitely far away in this ordering.
Labelling discarded balls as D and balls in the jug as J, then at step n of the remove-the-lowest jug puzzle we have {D₁,D₂,D₃,...D_n,J₁,J₂,... J_9n}
At the end of the supertask, we have {D₁,D₂,D₃,... (forever),J₁,J₂,... (forever)} - all the D's can be mapped 1:1 to N, so where does that leave the J's? They have higher values than any natural number.

The set is defined over the natural numbers, yes? Then for any natural number x, there is a value of n for which x stops being in the set (namely, n >= x). Therefore, the set cannot contain any natural number.

And for any x removed from the set, there is an even bigger set starting at x+1. Rinse, repeat.

Fine but we are not dealing with a "cleverly reordered set". We are dealing with the natural numbers in natural order: {1,2,3...} There is no infinite cardinal and there is no infinite ordinal in this set. Partitioning the set into L, (sometimes C), and R do not change this.

The input is the natural numbers in order. The output is two or three sets of natural numbers. The first set can be mapped to N. That doesn't mean the other sets are empty, as demonstrated by the even-odd sorting.

It occurs to me where our disagreement lies. We're both in agreement on the proposition "If there is a ball in the urn at midnight, it must not have a finite value on it." Your claim is "There must be a ball in the urn at midnight, therefore it must have an infinite value." Our claim is "The ball must have a finite value on it, therefore there are no balls in the urn." Can you prove the assumption "There must be a ball in the urn at midnight"?

Yes, clearly that's the disagreement.
As proof I would say:
The set of balls in the jug is of size 9n, starting at n+1. As n grows infinitely, the limit is an infinite number of balls starting at an infinite number.

As an illustration I would say to lay the balls in a line, 1cm apart, starting at the 1cm mark. Each step add 9 balls and step 1cm away, so the distance of any ball from us represents its number in the set of 'keepers'. At step 1 we have balls at 2,3,..10cm and at step 2 we have balls at 3,4,...20cm, and so on. At the end of the task there is an infinite line of balls infinitely far away. They got there because although there are no infinite numbers in N, it is an infinitely big set. If you do things an infinite number of times, infinite quantities can arise.

As another illustration I would say that balls don't care what's written on them. If we just add 10 unmarked balls and remove 1 at each step then we'll clearly end up with an infinite number of them. If we allocate numbers to them then the only difference is that they will have numbers allocated to them at the end. The particular numbers will depend on which ones we chose to remove.

As another illustration I would repeat the scenario mentioned many posts earlier, where instead of adding balls 10n-9...10n and removing ball n, we add balls 10n-9...10n-1 and multiply the value of ball n by 10. We can tack a zero onto its label if you like to think of it that way. So no ball leaves the jug. Nine balls are added an infinite number of times. The numbers have ever-increasing numbers of zeros appended, resulting in infinitely long strings of digits. You end up with infinite numbers on infinite balls. There is no difference between the sets in this game and the sets in the puzzle. They map exactly 1:1 with each other, number to number. There is no mysterious essence-of-original-n that gets propagated through the game - once a ball is renumbered it becomes exactly equivalent to that new number. So how do they vanish? I say they don't.

I'm getting pretty tired of going round in circles - it's a neat paradox but I feel I've resolved it to my satisfaction, and learned some interesting stuff in the process. I might check in here from time to time but unless something novel appears I think I'll not be able to put my points any better than I have already.

See you in the next puzzle...

[ucim]

The input is the natural numbers in order. The output is two or three sets of natural numbers. The first set can be mapped to N. That doesn't mean the other sets are empty, as demonstrated by the even-odd sorting.

You are absolutely correct. The reason the other sets are empty is not that the first set can be mapped to N. Nonetheless, the other sets are empty. Therefore, there must be another reason.

The reason is that anything else leads to a contradiction. However, having the sets be empty does not lead to a contradiction.

Just like R in the two-partition game, the set ends up empty, even though there isn't a step that empties it, because there are an infinite number of steps.

Jose

[Poker]

But you can put the 10% of the naturals you took away in a bijection with the entire set of naturals, in their usual order. (For example, if you took out 1, 11, 21, etc. you could use 1 -> 1, 11 -> 2, 21 -> 3, etc.) 10% of the naturals is the exact same size as the naturals. And when you try to take 10% of the naturals in their usual order, 10% of the naturals is 100% of the naturals.

And then if you try to count the 10% followed by the 90%, you'll never get to start on the 90% because they are infinitely far away in this ordering.
Labelling discarded balls as D and balls in the jug as J, then at step n of the remove-the-lowest jug puzzle we have {D₁,D₂,D₃,...D_n,J₁,J₂,... J_9n}
At the end of the supertask, we have {D₁,D₂,D₃,... (forever),J₁,J₂,... (forever)} - all the D's can be mapped 1:1 to N, so where does that leave the J's? They have higher values than any natural number.

You have not proved there are any J's at the end. The 10% is the 100%, which means the "90%" must be the empty set.

The set is defined over the natural numbers, yes? Then for any natural number x, there is a value of n for which x stops being in the set (namely, n >= x). Therefore, the set cannot contain any natural number.

And for any x removed from the set, there is an even bigger set starting at x+1. Rinse, repeat.

And that x+1 will also get removed, as will x+2, x+3, and so on. There is no infinite number in N, and the balls in the initial problem are numbered with the numbers in N and the numbers in N only.

It occurs to me where our disagreement lies. We're both in agreement on the proposition "If there is a ball in the urn at midnight, it must not have a finite value on it." Your claim is "There must be a ball in the urn at midnight, therefore it must have an infinite value." Our claim is "The ball must have a finite value on it, therefore there are no balls in the urn." Can you prove the assumption "There must be a ball in the urn at midnight"?

Yes, clearly that's the disagreement.
As proof I would say:
The set of balls in the jug is of size 9n, starting at n+1. As n grows infinitely, the limit is an infinite number of balls starting at an infinite number.

An infinite number which you have yet to prove exists in the set of N, and which in fact does not exist in N, so there cannot be any numbers present, so there cannot be any balls present. Balls are NOT interchangeable.

As an illustration I would say to lay the balls in a line, 1cm apart, starting at the 1cm mark. Each step add 9 balls and step 1cm away, so the distance of any ball from us represents its number in the set of 'keepers'. At step 1 we have balls at 2,3,..10cm and at step 2 we have balls at 3,4,...20cm, and so on. At the end of the task there is an infinite line of balls infinitely far away. They got there because although there are no infinite numbers in N, it is an infinitely big set. If you do things an infinite number of times, infinite quantities can arise.

As another illustration I would say that balls don't care what's written on them. If we just add 10 unmarked balls and remove 1 at each step then we'll clearly end up with an infinite number of them. If we allocate numbers to them then the only difference is that they will have numbers allocated to them at the end. The particular numbers will depend on which ones we chose to remove.

As another illustration I would repeat the scenario mentioned many posts earlier, where instead of adding balls 10n-9...10n and removing ball n, we add balls 10n-9...10n-1 and multiply the value of ball n by 10. We can tack a zero onto its label if you like to think of it that way. So no ball leaves the jug. Nine balls are added an infinite number of times. The numbers have ever-increasing numbers of zeros appended, resulting in infinitely long strings of digits. You end up with infinite numbers on infinite balls. There is no difference between the sets in this game and the sets in the puzzle. They map exactly 1:1 with each other, number to number. There is no mysterious essence-of-original-n that gets propagated through the game - once a ball is renumbered it becomes exactly equivalent to that new number. So how do they vanish? I say they don't.

Your illustrations all differ from the original puzzle. Balls are NOT interchangeable.

Your second illustration is ill-defined, because it doesn't specify exactly how the balls are removed. You could leave any number of balls in there you wanted. Your first and third illustrations renumber the balls at every step:

Balls in a line - This is equivalent to adding balls 10n+1 through 10n+9, then adding 1 to each numbered ball.
Multiply by 10 - This is self-evident.

In each case, the finite sets of the original and the finite sets of the modifications are equivalent. However, they differ completely at the limit. Consider the following two limits, just as an example of differing at the limit:

lim_x->0 (x-1) and lim_x->0 (x^2-x)/x

Now, of course, the equation on the right can be reduced to the equation on the left, and so they both have the same limit (-1), but at x=0 itself, the equation on the left does equal -1 but the equation on the right has no defined value (division by 0).

You see? Just because two things appear exactly the same each step along the way, does not mean they have to be the same at the limit. If they happen to be the same at the limit, it is for other reasons. I have yet to see those other reasons.

[Xias]

I have yet to see those other reasons.

Several people spent the better part of five pages asking for them. I think if there were any, they would have been provided.

I gave up trying to teach kryptonaut mathematics, but I think that for posterity the following deserves comment (emphasis mine)

As another illustration I would repeat the scenario mentioned many posts earlier, where instead of adding balls 10n-9...10n and removing ball n, we add balls 10n-9...10n-1 and multiply the value of ball n by 10. We can tack a zero onto its label if you like to think of it that way. So no ball leaves the jug. Nine balls are added an infinite number of times. The numbers have ever-increasing numbers of zeros appended, resulting in infinitely long strings of digits. You end up with infinite numbers on infinite balls. There is no difference between the sets in this game and the sets in the puzzle. They map exactly 1:1 with each other, number to number. There is no mysterious essence-of-original-n that gets propagated through the game - once a ball is renumbered it becomes exactly equivalent to that new number. So how do they vanish? I say they don't.

The emphasized part leads to a contradiction. If no ball leaves the jug, then after step 10, where is the ball that was originally labeled '1'? It is still in the jug, only now it has a label 100 on it. During which step was it added? During step 1. Compare this to the original puzzle: When was the ball with a label 100 on it added to the jug? During step 10. The answers are different, because the balls can be identified by their original ordering. Since the balls labeled the same in both games have different well-defined properties, they are different balls (despite having the same label.) Then the statement "there is no difference between the sets in this game and the sets in the puzzle" is false.

On the other hand, let's assume that the identity does not remain constant, so that the sets remaining at each step are the same for each game. Then renumbering a ball changes its identity. Then after step 10, the ball originally labeled '1' is no longer in the jug. Then sometime between being added in step 1 and being searched for in step 10, the ball left the jug. Then the statement "no ball leaves the jug" is false.

-

It doesn't matter that we are adding a zero. What matters is that we can track the identity of each ball b_n in order to perform some action on each ball in a well-defined way. If instead of adding a zero to the end, we put a dot on b_n, then we are no longer doing anything based on the "lowest numbered ball" at each step (during step 2, we put a dot on ball 2, not ball 1, for example). Instead, we are doing actions based on the lowest indexed ball. So when step 10 comes around and there's no ball labeled or indexed 10, what do we do? In order to be consistent with the append-0 game, we must add a dot to the ball indexed '1,' which is now labeled with a 1 and a dot, ending the step with b₁ having a 1 and two dots.

It's for this reason that the append-zero game is closely identical to a game where we merely add balls {b_10n-9, ... , b_10n-1} and do nothing. It's clear that changing labels (by adding a zero, or a dot, or a squiggle, or anything) merely changes the labels we end up with, but not which balls remain.

If you accept that the original jug ends up empty, then this should resolve the paradox of the append-0 game ending in a different result: they are different games. If instead you want to insist that the original jug ends up full, then you have to justify that on its own. You can't use the result of the append-0 game as "proof," since they are different games.

[kryptonaut]

Several people spent the better part of five pages asking for them. I think if there were any, they would have been provided.

I'll say just this, then I'm gone.

If you want to model the game with sets of numbers then you have to accept that a number is a number is a number, whether it's represented as a string of decimal digits, a knotted piece of string, Egyptian hieroglyphics, Unicode, whatever - if it represents 100 then it is 100. You then have to reconcile your belief that the numbers vanish with the puzzle setup in which no ball is ever removed.

If you want to model the game using sets of entities that have a mysterious knowledge of their past then you can do that too, but then they are no longer numbers and you can't treat them as such.

Finally, if you want to understand where the infinite numbers came from in my explanation, consider the ever-expanding labels in this game. At any stage they have a finite length. At the end of the game they have infinite length. They are the infinite numbers you are looking for.

[Xias]

I'll say just this, then I'm gone.

I thought you were gone already. But in any case, you once again did not acknowledge anything I said in my post. We're not on a live debate stage. You have an opportunity to sit back and take your time reflecting on and responding to something. Yet you don't. If you are wondering why you are unable to convince anyone that what you are saying is true, it is because all you have done is repeat yourself; and when people very patiently explain why they believe what you have said is incorrect, you respond by repeating yourself again with no regard for what they actually said. You very well could be right, kryptonaut; we could all be completely wrong about set theory, or perhaps you could have the next great theorem in mathematics. And none of us would ever know because you never actually address anything that anyone has said.

I went into great detail about how the balls can be identified as different, and that ball b₁ does not "become" ball b₁₀ when you put an arbitrary mark on it. You responded by continuing to act as though I'm referring to some "mysterious knowledge of their past."

I went into great detail about how, if a line from the origin intersects y=1 at a value of x greater than any real number, the angle must be a nonzero number smaller than any real number. You responded by simply stating that the angle is zero, without regard for my arguments that it cannot be.

I spent a great deal of energy constructing and formatting a proof that a particular sequence of sets converges to the null set. Everyone else has also provided various explanations for why it is the null set. You repeat it back to us as "because n+1 and 10n both tend to infinity," which has not been an argument ever presented to you.

When we ask you for proof that in this game, we end up with infinite balls, your response is to point at a different game, repeat yourself without proof, and then be completely bewildered when we disagree with your conclusions there (since you haven't proven them yet) or disagree that the games have identical outcomes (and show you why they do not).

I could continue.

I thoroughly enjoy talking about this paradox. If you would like to come back and actually honestly engage with the arguments being presented, then perhaps you'll convince me that I'm wrong, and I welcome that. If you would prefer to come back and continue merely repeating yourself while failing to address the majority of the points made against you, then I doubt you will have any luck.

[ucim]

Finally, if you want to understand where the infinite numbers came from in my explanation, consider the ever-expanding labels in this game.

Let's skip the labels and the balls and all the distractions. Consider two sets: A and E.

Set A contains all the natural numbers.
Set E contains all the natural numbers that are not elements of set A.

Is set E empty?

Now consider the two-partition L-R scenario. As you have already agreed, at the end of the supertask, L contains all the natural numbers. R contains all the natural numbers that are not in L. (It so happens that they are also "greater than" any element in L, but that does not matter. The point is that they are natural numbers that are not in L, which contains all the natural numbers).

Is set R empty? (You have already agreed that it is; do you still agree?)

Now consider the three-partition L-C-R scenario. At the end of the supertask, all the natural numbers are in L. It doesn't matter how it gets there, it just matters that it gets there. Now, we'll call set U the union of C and R. Set U contains all the natural numbers that are not in L.

Do you agree that set U is empty?
If so, this implies that set C must also be empty. Do you agree here?

Jose

[phlip]

↶
If you want to model the game using sets of entities that have a mysterious knowledge of their past then you can do that too, but then they are no longer numbers and you can't treat them as such.

I think it's worth reiterating that the limit we're looking at here is a pointwise limit. That is, to find the limit at the end of the supertask of the sets (ie what balls are in each jug) we look at the individual limits over the motions of each ball. If the limit of a given ball's position is a particular jug, then the limit of that jug's contents will contain that ball. There are other definitions of limits you can have different implications depending on the situation, but this is the most common.

What this means is the individual balls themselves don't innately have a "mysterious knowledge of their past". But we do when we analyse their paths to find out where they are in the limit. The balls are interchangeable in the broad sense, in that you could swap two over entirely through the whole operation and be fine. But they're not interchangeable in individual states, as the definition of the limit involves tracking each ball from one state to the next, and if you shuffle the balls you break that tracking.

Compare it to a pointwise limit of a function. Look at, for example, a sequence of functions f_i each N->N
f_i(n) = {1 if n<i, otherwise 0}
so f₀ is 0 everywhere, then f₁ is 0 everywhere except at 0. Then f₂ is 0 everywhere except at 0 and 1. And so on.
Then define another, different sequence of functions:
g_i = {1 if i≤n<2i, otherwise 0}
so g₀ is 0 everywhere, then g₁ is 0 everywhere except at 1. Then g₂ is 0 everywhere except at 2 and 3. And so on.

Now, we define a pointwise limit of these functions... calling f_∞ = lim(i->∞) f_i, such that f_∞(n) = lim(i->∞) f_i(n). That is, each point in the limit-function is the limit of that point in each of the sequence functions.
It should be pretty clear that f_∞ is 1 everywhere, while g_∞ is 0 everywhere.

However, if you just look at the individual sequence functions... f₀ and g₀ are both 0 everywhere. f₁ and g₁ are both 0 everywhere except for 1 point. f₂ and g₂ are both 0 everywhere except for 2 points. And this equivalence continues for every element of the sequence.

Because it matters which of those points are 1s and which ones aren't, and that's information that's being lost when you boil down the functions to just "has so many points that are 1". Just like you lose information when you boil down the jugs to "contains so many balls". It matters which balls... not because the balls are different, but because we need to track each ball's trajectory.

There was a comment a few pages back that I didn't pick up properly at the time, where it was said "but if the sequences are the same at every point, then how can the limit be different"? And here's the answer to that... f_i and g_i are different functions... the sequences of functions is different at every point. The sequences of counts of 1s of those functions is the same at every point, but that isn't what we're taking the limit of.

[Cauchy]

There was a comment a few pages back that I didn't pick up properly at the time, where it was said "but if the sequences are the same at every point, then how can the limit be different"? And here's the answer to that... f_i and g_i are different functions... the sequences of functions is different at every point. The sequences of counts of 1s of those functions is the same at every point, but that isn't what we're taking the limit of.

I know I mentioned this, when we were talking about the append-0 game vs. the remove balls game. I said that since a_i and r_i (for append step i and remove step i) are the exact same function (not just have the same counts of 1's and 0's), they must have the same limit. It was argued that since balls have special memories of their past digits and Hilbert's Hotel has an incompetent assistant manager, the limits should actually be different.

[Xias]

It was argued that since balls have special memories of their past digits and Hilbert's Hotel has an incompetent assistant manager, the limits should actually be different.

You sound unconvinced.

1. In the original, remove-the-lowest-ball game, there exists at some point a ball with the label "100" on it. When was it put in the jug?
2. In the append-0 game, there exists at some point a ball with the label "100" on it. When was it put in the jug?

[phlip]

↶
I know I mentioned this, when we were talking about the append-0 game vs. the remove balls game. I said that since a_i and r_i (for append step i and remove step i) are the exact same function (not just have the same counts of 1's and 0's), they must have the same limit. It was argued that since balls have special memories of their past digits and Hilbert's Hotel has an incompetent assistant manager, the limits should actually be different.

They're not the same function, though. In order to view them as the same, you have to view the "append a 0" action as changing the ball's identity. But it's not, it's the same ball, it's just got a different label.

That is... we have a countably-infinite set of balls B, all identical except as parameters to the functions we're about to describe. At each step, each ball has a position p_i:B->{in, out} and a label l_i:B->N.

For the remove-balls game, the pointwise sequence of p_i(x) for a constant x goes from out to in to out again, then stays as out forever, so p_∞(x) = out. Meanwhile, l_i(x) is a constant sequence, so l_∞(x) is the same value.

For the append-0 game, p_i(x) goes from out to in, then stays as in forever, so p_∞(x) = in. Meanwhile, l_i(x) periodically grows by a factor of 10, so l_∞(x) does not converge. To a certain degree of notation abuse, you could say l_∞(x) = ∞ (in the sense that lim[i->∞] l_∞(x) = ∞, not in the sense that this l_∞ actually has a value here, and that value equals ∞).

The function you're considering is the composition: f_i(n) = p_i(l_i^-1(n)) being a mapping of labels to positions.
This is, indeed, the same for both games. But composition doesn't preserve limits, so that doesn't tell us much about p_∞ and l_∞.