Suppose you have a bottle with infinite volume and infinite number of balls. Now it's 11 o'clock and an hour left till 12.

So you put 10 balls in and take one out 30 minutes later. You repeat this 15 minutes later. Then 7.5 and so on.

Basically repeat that every (Remaining time) / 2

How many balls are in the bottle at 12 o'clock?

  • My first thought was infinite but no.

  • Then i thought it will never be 12 o'clock. Like the Heracles and turtle example. Obviously no.

  • Now Another approach. If we consider the set of natural numbers and the set of even numbers. They have equal power or in other words equal "length" whatever that means. Taking the same approach here, infinite number of adding 10 and infinite number of subtracting 1 have equal powers so. There are no balls in the bottle at 12 o'clock ??

What is infinite - infinite ?

Despite the sets if they have equal powers then difference is always 0?

I would like to know everything about this. Every detail. Answer to the question is not the most important here.


  • $\begingroup$ Fun fact: suppose the removed ball is randomly* chosen each time. You can show that with probability one, the bottle will be empty at noon. See math.stackexchange.com/a/1817947/177399 for a proof. $\tag*{}$ *Uniformly at random, independent of previous choices. $\endgroup$ – Mike Earnest Sep 12 '18 at 15:18
  • $\begingroup$ @MikeEarnest I read the post but didn't really understand. Can you please post how you get the noon ? $\endgroup$ – R0xx0rZzz Sep 12 '18 at 17:55

There is no answer, I will show 2 ways to get 2 different outcomes:

Number your balls: $b_1,b_2,...$, now at set $n$ you will enter the balls $b_{10\times n-9}$ to $b_{10\times n}$ and take out the ball $b_n$, because every ball you enter at some point you will take out at some other point there are no balls in there.

But if instead of taking out the ball $b_n$ you will take out the ball $b_{10\times n}$ every step you will have 9 balls that will never be taken out, so there will be infinity many at the end.

In a similar way you can find a way such that there will be $54$ balls at the end or $3$ or $75291709$, or any natural number.

This is because $\infty-\infty$ is an Indeterminate form, it is not well defined(In terms of sets power, I suppose that you mean Cardinality, in that case, subtraction is not defined at all, or if you do define them you at least need to require them to have different cardinality, see here).

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  • $\begingroup$ Still confused, but I think I'll handle . Thanks $\endgroup$ – R0xx0rZzz Sep 12 '18 at 14:38

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