I have infinity number of balls and a large enough vase. I define an action to be "put ten balls into the vase, and take one out". Now, I start from 11:59 and do one action, and after 30 seconds I do one action again, and 15 seconds later again, 7.5 seconds, 3.75 seconds...

What is the number of balls in the vase at 12:00?

My attempt

It seems like that it should be infinity (?), but if we consider the case:

Number each balls in an order of positive integers. During the first action, I put balls no. 1-10 in, and ball no.1 out, and during the $n^{\text{th}}$ action I take ball no. $n$ out.

In this way, suppose it is at noon, every ball must have been taken out of the vase. So (?) the number of balls in the vase is


My first question: if I take the ball randomly, what will be the result at noon? (I think it may need some probability method, which I'm not familiar enough with.)

Second one: is it actually a paradox?

Thanks in advance anyway.

  • $\begingroup$ It is not a paradox, it just does not make sense. In real life, this of course cannot be done. In mathematics, it is not a well-defined process since the series does not converge. $\endgroup$ – Luke May 23 '17 at 8:31
  • $\begingroup$ It, I think, will be zero in your way. $\endgroup$ – Paul May 23 '17 at 8:32
  • $\begingroup$ No, Paul, the point is that by different ways of looking at the problem, you can get different numbers of balls. You might also say that at each time you add $10-1=9$ balls, so there must be infinitely many balls in the vase at the end. By the other argument, there must be zero in it, and here you already note that it is just not a well-defined process and one should be careful with infinite sums. $\endgroup$ – Luke May 23 '17 at 8:35
  • $\begingroup$ Why should it be zero? You seem to be assuming that $\infty - \infty = 0$ $\endgroup$ – Slug Pue May 23 '17 at 8:35
  • $\begingroup$ This is the Ross-Littlewood Paradox ... Google it. $\endgroup$ – Bram28 May 24 '17 at 1:18

What you just discovered is that the cardinality of a set (the number of elements) is not a continuous function, that is, for a convergent sequence $S_n$ of sets you may have $$\lim_{n\to\infty}\left|S_n\right|\ne \left|\lim_{n\to\infty} S_n\right|$$ where $\left|S_n\right|$ is the cardinality of $S_n$ (e.g. $\left|\{2,3,4,5\}\right|=4$). In your case, the left hand side diverges (giving you the infinite number of balls), while the right hand side gives $0$ (the cardinality of the empty set).

This is not a paradox, but a warning that you have to be careful with such limits.


This is essentially a pointwise convergence versus uniform convergence situation. For each ball separately, we have convergence to the state where that ball is not in the vase: as you say, every ball gets taken out before $12{:}00$. But if you look at the number of balls in the urn, that tends to infinity. So whether the process converges to the empty vase depends on what metric we use to define convergence.

A similar thing happens in real analysis. Define a sequence of functions $f_n:[0,1]\to\mathbb R$ as follows. $f_n(0)=0$, $f_n(1/2n)=2n$, $f_n(1/n)=0$, with $f_n(x)$ increasing linearly between $0$ and $1/2n$, decreasing linearly between $1/2n$ and $1/n$, and $0$ between $1/n$ and $1$.

Now for any $x$, $f_n(x)\to 0$. This means that $f_n$ converges pointwise to the zero function. But $\int_0^1f_n(x)\mathrm dx=1$ for every $n$. So pointwise convergence doesn't tell us how the integral behaves, and if that's what we care about then we need to define convergence differently.


It is not a well defined problem. You talk about a result after infinite time (or at least an infinite amount of actions), which requires a notion of a limit.

We can compute limits of numbers because we have defined what it means to approach a number. So when you just look at the number of balls and not which balls, then the answer is indeed $\infty$. This is because the sequence of the number of balls $0, 9, 18, 27,...$ is divergent.

If we want to include the identity of the balls, we have to talk about sets instead of just numbers. We can compute the limit of sets (via intersection and union) if the sequence of sets is increasing or decreasing, i.e.

$$ A_0\subset A_1\subset A_2\subset\cdots \quad\text{or}\quad A_0\supset A_1\supset A_2\supset \cdots$$

For more general sequences of sets there might be a set-theoretic limit (Thanks to celtschk in the comments). Applied to this problem, it will give indeed the empty set $\varnothing$.

The paradox arises because the problem intentionally leaves open the kind of limit to consider. As soon as the definition is fixed, the ambiguity vanishes.

  • $\begingroup$ "But we have no definition for convergence of general sequences of sets" — Yes, we have. Moreover, as can be easily checked with the definitions, the limit of that specific sequence indeed is the empty set. $\endgroup$ – celtschk May 23 '17 at 8:59
  • $\begingroup$ @celtschk Very interesting. I might include this in my answer. $\endgroup$ – M. Winter May 23 '17 at 9:00

From a physics point of view, not more than 72 balls (maybe 81).

Otherwise, imagine somebody putting ten and taking out 1 balls in less than 1 second!!

  • $\begingroup$ I see why this answer sort of answers the question by explaining how the situation is not realistically possible. However, the question is about the hypothetical situation and people's intuition that if each individual ball is there at noon if and only if it was there at all times sufficiently close to noon before noon, there should be infinitely many balls at noon, not whether it could really be done. I downvoted this answer. This answer cannot be fixed up to actually answer the question without being fundamentally rewritten. That's when a question is meant to be downvoted. The Stack Exchange $\endgroup$ – Timothy Feb 27 '20 at 0:43
  • $\begingroup$ community has a lot of things on their plate and the down vote directs their attention to answers that are more worthy of their time. $\endgroup$ – Timothy Feb 27 '20 at 0:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.