In the Ross-Littlewood paradox, there is a supertask that goes as follows: At step 1, you add 10 balls to a jar labeled 1 through 10 and remove ball #1. At step 2, you add 10 balls to a jar labeled 11 through 20 and remove ball #2. At step 3, you add 10 balls to a jar labeled 21 through 30 and remove ball #3. This process goes on inductively, with each step taking half the time of the previous step (so that all this takes place in finite time). The question is, how many balls are left in the jar after the supertask is done? Based on the literature, it seems there are two sensible conclusions: either you can say the scenario is not physically sensible and thus there is no answer, or you can say there are precisely zero balls left.

Now, my question is not about the paradox, so please leave the debate about it aside. Under some suitable interpretation, we can try to think about the process of adding and removing balls in terms of set theory with the final result being zero balls left.

When I try to think about this, I am thinking of defining the sets $$ S_{1} = (\varnothing\cup\{1, \ldots, 10\})\setminus\{1\}, \qquad S_{2} = (S_{1}\cup\{11, \ldots, 20\})\setminus\{2\}, \qquad S_{3} = (S_{2}\cup\{21, \ldots, 30\})\setminus\{3\}, \qquad \ldots. $$

In general,

$$ S_{n+1} = (S_{n}\cup\{10n+1, \ldots, 10(n+1)\})\setminus\{n+1\}. $$

We can either think of this as an infinite sequence of sets $S_{1}, S_{2}, \ldots$ or as an infinite sequence of operations (union, setminus, union, setminus, etc.) done to sets. Either way, it seems like there ought to be some sort of "set limit" which would be the empty set $\{\}$. Is there any way in which this can be formalized?

I am aware there are cases where no "set limit" could be assigned. For example, if we define $T_{1} = \{5\}$, $T_{2} = \varnothing$, $T_{3} = \{5\}$, $T_{4} = \varnothing, \ldots$, then I think there likely can't be any "set limit" assigned. However, I don't think this spoils the original question, because there are plenty of sums in real analysis that don't have any result such as $\sum_{n\ge 0} (-1)^{n}$. The fact that the result of this sum is undefined doesn't mean other sums can't have a well-defined result. Indeed, we can formalize infinite sums by limits (under the epsilon-delta definition of limits) perfectly well and we can even tell which series converge and which don't. Is there anything analogous to this for sets?


1 Answer 1


You can use $\liminf\limits_{n \to \infty} A_n = \bigcup\limits_{n \geq 1} \bigcap\limits_{j \geq n} A_j$ and $\limsup\limits_{n \to \infty} A_n = \bigcap\limits_{n \geq 1} \bigcup\limits_{j \geq n} A_j$ or the equivalent with indicator functions, and then see if they are equal.

  • If so, as with $\liminf$ and $\limsup$ for numerical sequences, they are the limit $\lim\limits_{n \to \infty} A_n$.
  • If not, as with $\liminf$ and $\limsup$ for numerical sequences, then there is no limit.

Illustrating this with your Ross-Littlewood example, $A_j=\{j+1,j+2,\ldots,10j\}$, giving:

  • $\bigcap\limits_{j \geq n} A_j = \emptyset$ since $A_k \cap A_{10k}=\emptyset$ for any $k$,
    • so $\bigcup\limits_{n \geq 1} \bigcap\limits_{j \geq n} A_j = \emptyset$ since this is $\bigcup\limits_{n \geq 1} \emptyset$,
    • i.e. $\liminf\limits_{n \to \infty} A_n =\emptyset$;
  • $\bigcup\limits_{j \geq n} A_j = \{n+1,n+2,\ldots\}$ since all numbers higher than $n$ are in the union,
    • so $\bigcap\limits_{n \geq 1} \bigcup\limits_{j \geq n} A_j = \emptyset$ since $k\not \in \bigcup\limits_{j \geq k} A_k$ for any $k$,
    • i.e. $\limsup\limits_{n \to \infty} A_n =\emptyset$;
  • and since these are equal, $\lim\limits_{n \to \infty} A_n$ exists and is $\emptyset$.
  • 1
    $\begingroup$ +1. For the proposer: A nice description of the set-theoretic lim inf and lim sup is that $x\in \lim\inf A_n$ iff $\{n:x\not\in A_n\}$ is finite, i.e. $x\in A_n$ for "almost all" $n, $ and $x\in \lim\sup A_n$ iff $x\in A_n$ for infinitely many $n.$ $\endgroup$ Commented Jul 23, 2023 at 1:38
  • $\begingroup$ @DanielWainfleet so there is no limit when there is any $x$ with $\{n:x\not\in A_n\}$ and $\{m:x\in A_m\}$ both being infinite; if there is no such $x$ then there is a limit $\endgroup$
    – Henry
    Commented Jul 23, 2023 at 1:49
  • $\begingroup$ No. ,,,,,,,,,,,,,,,,, $\endgroup$ Commented Jul 23, 2023 at 2:26

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