kryptonaut wrote:gmalivuk wrote:Your problem is your continued insistence that somehow balls like ω and ω+1 and 2ω just magically show up. All the balls we started with had natural numbers on them. So which natural number was originally on the now-lowest ball in the jug, which you claim is now numbered at least ω? When did it acquire its new label? How did this happen when all we've been doing is coloring some balls and removing balls one-by-one from the jug?

No, the problem is in treating ω as a cardinal number. N is an infinite set. If you set aside the numbers 1 to 10, what's left is still an infinite set, isomorphic to ω. If you set aside the number 1 million then what's left is still an infinite set isomorphic to ω. If you contrive to set aside the last number, by some procedure such as ucim's transfer jug, then what's left is an infinite set, isomorphic to ω. That last number, in this particular arrangement, can be given the label ω. If you contrive to set aside the last 10 numbers then they can be labelled ω,ω+1,ω+2,...ω+9. They have the property that there are an infinite number of numbers smaller than them. If you contrive to set aside a countably infinite set from the top end, as is done by the 'remove-lowest' variant of the original puzzle, then you still leave behind a set ω, and the infinite set that is set aside is also isomorphic to ω, it is {ω,ω+1,ω+2,...}. The actual numbers have the property of having an infinite number of numbers smaller than them.

First, as far as I can tell, it's not obvious that ucim's transfer jug contains a number at the end. Certainly, in the modify-number case it does, but there, as in so many examples before, the numbers are not fixed to the balls.

Second, you're saying there is a last number, yet there are numbers after it? If I'm understanding you correctly, anyway. If so, how could that number possibly be considered to be the last one?

kryptonaut wrote:Xias wrote:Let's examine what is in the set at the beginning of step ω, and call a ball in that set bx. If x is finite, then bx would have been removed at tx, which is prior to midnight. It's midnight right now, so bx is not in the jug. So x is not finite. If x is infinite, then it must have been added at or after midnight, but it's midnight right now and we haven't added any balls yet. So x is not infinite. If x is neither finite nor infinite, then it does not exist.

At midnight, because of the construction of the supertask, it's not possible to declare which (non-finite) steps have or haven't happened.

What about at the point before any non-finite step?

kryptonaut wrote:Xias wrote:To illustrate this, let's perform the supertask twice. The first time we do it, let's paint every ball removed blue, and every ball remaining in the jug red. So we have a blue ball for every natural number, and a red ball for every ball in the set of balls never removed (infinite numbers or unknown numbers or whatever). Dump out the red balls and start the supertask again.

You will find that there is never a time when there is both a red and a blue ball in the jug.

You will find that all blue balls are added before midnight, and no red ball is added before midnight.

You will find that all blue balls are removed before midnight, and no red ball is removed before midnight.

The only way, then, to end up with red balls in the jug is to have an empty jug at midnight, and then to add red balls at or after midnight. And since the puzzle does not prescribe adding anything at all at or after midnight, there is no reason for us to do anything to the empty jug that we have at midnight. Then it was also empty the first time we did it, and there were no balls to paint red in the first place.

I'm not 100% sure I follow your argument, or understand the procedure(s) you are describing. But dividing the infinite input set up into two infinite sets, and then repeating the process on one of those infinite sets can produce two more infinite sets.

I'm going to try to set up an equivalent (though not completely identical) example, and maybe that'll help. Xias, could you double-check and make sure my example really is equivalent, and that there's not something weird going on that I didn't expect?

We're going to run the supertask twice, with the same set of balls each time. The first time we run it, every time we put a ball in the jug we paint it red, and every time we take a ball out of the jug we paint it blue. So after the supertask has finished, every ball that got removed is blue, while if there are any balls left in the jug, they would be red. Now, we're going to empty the jug, so that any red balls are in the same place as the blue balls. Finally we take that whole set of balls - whether each one wound up blue or red - and run the whole process again, exactly the same as before, with exactly the same ordering as before, only now we're not going to paint any balls - we're just going to run the normal supertask, only the balls are now colored based on their final position.

Tell me, under what circumstances would there be a blue ball and a red ball in the jug at the same time?

kryptonaut wrote:Take two Hilbert hotels, A and B, where A is occupied and B is empty. Tell everyone in A to write their current room number on a piece of paper and put it in their pocket. Now fill each room B_{n}by taking the occupant of A_{1}and moving everyone in A down one room. A will still be full at the end of the supertask, as will B - but what number is written on the piece of paper held by the current resident of A_{1}?

We have divided the original infinite set into two, one of which is numerically higher than the other. We could repeat the process, filling a new hotel C with the occupants of B, still leaving B full. And so on.

Can you prove A will still be full? Sure, we only remove one person at a time, but by the end we have removed an infinite number of people, so to suddenly claim that it's still infinite is not immediately obvious. Infinity minus infinity does not have to be infinity.