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Story 1:

Let there be a bowl $A$ with countably infinite many of candies indexed by $\mathbb{N}$. Let bowl $B$ be empty.

  • After $1/2$ unit of time, we take candy number 1 and 2 from $A$ and put them in $B$. Then we eat candy 1 from bowl $B$.
  • After $1/4$ unit of time, we take candy number 3 and 4 from $A$ and put them in $B$. Then we eat candy 2 from bowl B.
  • After $(1/2)^n$ unit of time, we take candy number $2n-1$ and $2n$ from $A$ and put them in $B$. Then we eat candy $n$ from bowl $B$.

What happens after 1 unit of time? How many candies are there left in $A$, how many in $B$, how many have you eaten? Answer:

There are no candies in $A$ left. For any candy corresponding to a given natural number $k$, one can compute the time it was eaten. Similarly there are no candies in $B$. We ate as many candies as the cardinality of $\mathbb{N}$.

Story 2:

Let there be a bowl $A$ with countably infinite many of candies indexed by $\mathbb{N}$. Let bowl $B$ be empty.

  • After $1/2$ unit of time, we take candy number 1 and 2 from $A$ and put them in $B$. Then we eat candy 1 from bowl $B$.
  • After $1/4$ unit of time, we take candy number 3 and 4 from $A$ and put them in $B$. Then we eat candy 3 from bowl B.
  • After $(1/2)^n$ unit of time, we take candy number $2n-1$ and $2n$ from $A$ and put them in $B$. Then we eat candy $2n-1$ from bowl $B$.

What happens after 1 unit of time? How many candies are there left in $A$, how many in $B$, how many have you eaten? Answer:

As before, there are no candies in $A$ left. All the candies labelled by even numbers are in $B$, so there are countably many. We ate all the candies labelled by odd numbers, so we ate countably many.

Question:

The main question is, in both stories we do essentially the same thing: take two candies from $A$ to $B$, then eat one from $B$. The difference is that in the second case we have infinitely many candies left in $B$, but in the first case it was empty after 1 unit of time. So how did this happen?

Imagine this situation: let $X$ be conducting the eating according to story 1 with his labeling, let $Y$ be watching. But $Y$ secretly has a different labeling scheme in his mind, where the candy number $k$ in $X$'s labeling is candy number $2k-1$ in $Y$'s labeling. Then after unit time, according to $Y$, there should be infinitely many candies in $A$ but according to $X$ it should be empty. So the reality depends on the spectator?

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    $\begingroup$ We notice that $B$ cannot betreated as a black box. Also one needs to be careful what kind of limit process is to be used. We may assume that a certain candy is in a certain place at time 1 iff it is involved in only finitely many movements and its final place is the one assigned during that movement. The number of objects in a container is a secondary effect, the primary effect is the set of objects in the container. This is ok because the result of infinitely many additions/subtractions won't be defined in the first place. $\endgroup$ – Hagen von Eitzen Jul 29 '14 at 12:39
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    $\begingroup$ It is not a mere label. In the first case, you make sure to eventually come back and eat all the candies. In the second case, by skipping you ensured there are certain candies which you will never eat. $\endgroup$ – William Jul 29 '14 at 12:39
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    $\begingroup$ We must assume that the candies are "labelled" with the number written on them... otherwise at step $2$ we do not know what candy we have to eat. Is it so ? If so, we have "re-verified" the property of infinite sets that, e.g. the set of odd numbers is a proper subset of the set of natural numbes, and still the two sets have the same cardinality. $\endgroup$ – Mauro ALLEGRANZA Jul 29 '14 at 12:41
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    $\begingroup$ We can see it also as a clever variant of Achilles and the tortoise : after $n$ steps, for $n$ whatever, the player has moved $2n$ candies from A to B and eaten $n$ candies from B. Thus, after step $n$, exactly $n$ candies are left in B. $\endgroup$ – Mauro ALLEGRANZA Jul 29 '14 at 12:52
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    $\begingroup$ @did I have not found as many articles as I expected for supertasks. If I don't find at least ten posts for which supertasks is appropriate by this time tomorrow, I will delete the tag. I'll also delete it if almost all the articles I find are tagged with paradoxes. $\endgroup$ – MJD Jul 29 '14 at 14:31
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So as you said according to person X, there will be no candies left. Lets take a look, then, at what Y observes.

From person Y's point of view: First X moves candy 1 and 3 and from bowl A to B, then eats candy 1. Next, X moves candy 5 and 7 (3 and 4 by X's labeling) to B and eats 3. This continues for infinitely many steps. At this point person Y sees every odd candy eaten. Note that according to Y, the even candies never existed. So Y as well sees that there are no candies left in the bowls.

You may also be interested in these two links: A strange puzzle having two possible solutions Ross–Littlewood paradox

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You have essentially proven, with your stories, that $2\cdot \aleph_0 = \aleph_0$, where $\aleph_0=|\mathbb N|$. There is no sense in which either of your two stories is "impossible".

As for your last question with labeling: you described a scenario that cannot happen in real life, and something weird came out as a result. Why would that be surprising?

But OK, let's try to answer the question. So, what we have is an infinite amount of candy, labeled $1,2,3,4\dots$ by $X$, but they are labeled $1,3,5,7,\dots$ by $Y$. This means that there is no candy labeled $2$ in $Y$'s labeling, and the paradox does not exist.

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