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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.

ucim wrote: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,...}

Gwdion wrote: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.

xias wrote: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.

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.