 # Math paradox: putting infinite number of balls in a box

There’s a math paradox that goes like this:

You have an infinite supply of balls, each labeled with a natural number (1,2,3,4, etc). At step 1, you put the 10 balls numbered 1 through 10 in a (very large) box and then remove the smallest numbered ball in the box, number 1. At step 2, you put the next 10 balls, 11 through 20, in the box and then remove the smallest numbered ball, number 2. At step 3, you put in the next 10 balls (21 through 30) in the box and remove the smallest, number 3. And so on. In the limit, after you have performed step N for all natural numbers N, what is in the box?

By one argument, the number of balls in the box increases with each step (in fact after any step N there are 9*N balls in the box). So there should be an infinite number of balls in the box at the end. But by another argument, every ball has been removed at some step – ball N was removed at step N for every N. So the box will be empty at the end.

Ok, feel free to discuss this paradox if you wish (it seems to generate strong opinions, like the .999… issue), but what I really want to know is what is the NAME of this paradox? My google-fu has failed so far. Perhaps it has already been discussed on this board?

My calculus is rusty, but 9N approaches infinity faster than N approaches 0 so there are infinite balls in the box. Right?

Well, there is no end, so that makes the question moot.

Who knows how many balls you have in the box at the end? It’s never specified in the problem. Now, usually with problems of this sort, you can take a limit of the problem’s state as it approaches the unknown point, but this problem’s state is discontinuous, so the limit does not exist.

Ross-Littlewood Paradox it is! Thanks, The Other Waldo Pepper.

Chronos’ answer is one of several possible ones, and is the one that I personally think is most compelling. There are certainly similarly structured supertask problems where there is no well-defined answer, like “Start with a face up playing card. At each step, turn the card over. After an infinite number of steps, which side of the card is up?” Ross-Littlewood is interesting to me because both of the arguments I gave seem superficially plausible, while most supertask problems don’t have such plausible seeming alternate answers.

beowulff, note that Wikipedia’s formulation of the problem avoids the “there is no end” issue by making each step take half as long as the previous step. That’s the way I’ve always heard it, but I simplified it slightly in my post.

BTW, anyone who wants to argue for either the “infinite balls” or “no balls” solution should address these counter arguments:

1. If you say there are an infinite number of balls, name one. For any number N, if you claim that ball N is in the box, I say no, it was removed at step N.

2. If you say the box is empty, consider this variant. Instead of removing the smallest numbered ball at each step, remove the largest. Now the argument that every ball was removed at some step fails, and it seems that the box unequivocally contains an infinite number of balls. How can the result change so dramatically when we added and removed the same number of balls as in the original problem, simply choosing a different ball to remove?

I’d guess that my calc is far rustier than snfalknur’s, but the problem appears straightforward to me. As you stated, at any step, n, the amount of balls in the box is 9n. So as n approaches infinity, so does 9n. I don’t recall doing this in any calc class. In fact, it doesn’t even seem like there’s a question about this.
Looking at the Ross-Littlewood wiki page, they seem to add a spin to it, which is where the question, or doubt shows up. They’re showing, if I’m interpreting it correctly, that a random ball is removed, not the one with the smallest number. Since at any given time you could pick a number between 1 and 9n+1 and that ball may not still remain, you can argue that it’s empty.

Similarly, if you only take out the smallest numbered ball (like in the OP), an argument can be made that as the amount of balls in the box (9n) goes to infinity, so does the number of the ball being removed does as well. IOW, as long as balls are being added, they’re also being subtracted. Kinda like saying there’s no highest number because you can always add 1. They’re both going to infinity, one of them is just getting there faster.

Personally, I’d argue that there are an infinite number of balls in the box. But then, I really have no memory of what happens as these types of things approach infinity.

Good thing I got that B.S. in math. Seriously, 4 years of mostly math classes. My Jr and Sr year were made up almost entirely of this stuff. Gone now.

On preview, saw this:

At step 5, balls 6-45 are in the box. Sure, ball 5 has been removed, but there’s still 39 easy to pick numbers leftover. Said differently, at step N, I can guarantee balls numbered N+1 through 9N (inclusive).
(I didn’t spend too much time checking and double checking my formulas, but I think at least the theory is correct)

As I recall from calculus it is because you can name any removed ball that proves that the number of balls would approach infinity, as with any ball named, you are well past that one by a order of magnitude. So the limit of 10x - x as x approaches infinity would be 9x. It is not important for the actual name of the balls, just the quantity.

There are also other problems you can construct, which look very similar, but which do have well-defined answers. For instance, how are we writing these integers on the balls: I propose a simple method, by number of dots: The 1 ball has 1 dot on it, the 2 ball has 2 dots on it, and so on. And now, to keep things simple, I propose that, instead of taking out a ball, we instead leave it in the bucket, but put enough extra dots on it to make it one of the new balls we’re adding.

After any finite number of steps, this new problem has the same state as the original problem. But here, a solution seems clear: After an infinite number of steps, we have an infinite number of balls, but that none of the balls has any integer on it. Each ball instead has an infinite number of dots.

How can you claim that there are ANY balls in the box, let alone an infinite number, if every single ball that you can name has been removed? Yes, at any step N, you are removing a single ball and there are vast hordes of larger numbered balls in the box, but eventually you will remove every single one of them (at the step matching the ball’s number).

Seems like the problem is illustrating that our language can’t adequately describe the problem, resulting in what appear to be paradoxical results.

The reality is that the last ball is never placed in the box in the real world.

There’s a difference between “every single ball that you can name has been removed” and the actual case which would be “every single ball that you can name either has been or will be removed”. It the 'will be" that changes everything. You alluded to it when you said “there are vast hordes of larger numbered balls in the box”. What if I name one of the balls that are still int he vast hordes? Your statement only works if, instead of removing the lowest, you remove a random ball.

IOW, if you remove a random ball, as N approaches infinity, it’s been removed. If you only remove the smallest one, then as N approaches infinity, N+1 remains. ISTM, with a formula so easy to write, there’s no case to be made that there are zero balls remaining. In fact, perhaps someone that remembers this stuff better than I can actually write it…is there an online mathematica? Does wolfram still do these equations?

This is like saying how can a ship sink with a gaping hole in it’s side if I keep emptying the water coming in one cup at a time.

Or it doesn’t matter at all what name is on the ball, but the shear number of balls.

Put *X[SUB]n/SUB *= 1 if ball n is in the box at time t and 0 otherwise. The total number of balls in the box at a fixed time t is then just \sum X[SUB]n/SUB, the sum of the X[SUB]n/SUB. The paradox then becomes the statement that lim[SUB]t->∞[/SUB] \sum[SUB]n[/SUB] X[SUB]n/SUB and \sum lim[SUB]t->∞[/SUB] X[SUB]n/SUB differ, but that’s hardly surprising; infinite sums don’t commute with limits in general.

(Also, trying to typeset math here sucks.)

Just for kicks I looked at my old math books. They don’t list this one, but most of them speak of Zeno’s Paradox. One example of which is walking from a given starting point to a wall and suggesting that it can’t be done. The reason being that to get to the wall, you must walk from the starting point to the halfway point, then you have to walk from that point to the midway point between where you are and the wall and so on.

I think this may be the disconnect.
If you start from the beginning and at each step add, then subtract balls from the box, you have the ‘empty box’ situation. However, if you look at this as a sequence, each step giving us an independent number, you’ll find that as N approaches infinity, so do the numbers in the sequence.

My takeaway:
Numbering the balls and forcing the person to remove either a specific or random one is what trips people up and, IMO, is what changes this from an actual paradox to a riddle (as opposed to an actual paradox that helps with critical thinking).

For any given N>0, there are at least nine balls in the box. N+9 is never removed before N, therefore there will always be balls in the box for any N>0. The idea that it could be emptied is simply faulty inductive reasoning.

The rate of change in the number of balls in the box is constant with respect to the number of steps. 9, to be exact. With a constant, positive derivative, the number of balls never decreases. If you extend that to infinity, we can comfortably conclude there are infinite balls. I mean, what number do you expect to find if you just keep adding 9?

Lets go deeper, though.

For some reason, schools don’t teach people that there isn’t one infinity. The idea that all infinities are equal is completely wrong. Consider, for every whole number there is half that number. There are therefore two “half” numbers for every one whole number. There are an infinite number of whole numbers, ergo there is an infinity twice as large as the first infinity of “half” numbers. And then again for “quarter” numbers. And then again for, well, any subdivision of numbers.

The number of balls being placed is a larger infinity than the number of balls being removed.

If you say the box is empty, you’re wrong. Faulty premise, no consideration required.

I think there are multiple issues:
1 - that you can complete infinite tasks in a finite amount of time (if so, what was the number of the third to last ball placed in the box?)

2 - that you can describe a problem using a formula based on the state at discrete time steps, and then ask for the state at a nonsensical time step (there is no time step called infinity)

It seems to me (without reading up on it to see if this line of reasoning has been pursued) that what we have here is an infinite sequence of sets (i.e. A[sub]n[/sub] = the set of balls in the box after step n), and the question is, does this sequence converge? What is the limit of the sequence as n approaches infinity?

So, what does it mean for a sequence of sets to converge? Do we have to have a metric defined on the set of sets of balls?

The series is 9+9+9+…, or 9(1+1+1+…). Therefore there is 9*-1/2 = -9/2 of a ball at the end.

What?