How Many Rocks Will Be in the Hole at 12:00?

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The discussion centers on a thought experiment involving the addition and removal of rocks from a hole, exploring the implications of countably infinite sequences. At 11:00, 100 rocks are added while a rabbit removes one, and this pattern continues with halving time increments. The key question is whether the assumption that countably infinite rocks added minus countably infinite rocks removed equals zero is valid. Different removal strategies by the rabbit lead to contrasting outcomes: if the rabbit removes the most recent rock, infinite rocks remain; if it removes the oldest rock, none remain. The conclusion highlights the complexity of infinite sets and the importance of the removal method in determining the final count of rocks.
Jim Kata
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Say it's 11:00 and you throw 100 rocks in a hole and at the exact same second a rabbit throws one rock out of the hole. Then at 11:30 you throw a 100 rocks in and the rabbit throws one rock out. At 11:45 you throw 100 rocks in and the rabbit throws one rock out and so on, with the time increments halving each time. How many rocks will be in the hole at 12:00? Well one thought is that you've thrown a countably infinite number of rocks in and the rabbit has thrown countably infinite number out so the answer is 0, but say you threw 99 rocks in the hole each time and there was no rabbit then obviously the answer is infinity. Where is the contradiction?
 
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Jim Kata said:
Well one thought is that you've thrown a countably infinite number of rocks in and the rabbit has thrown countably infinite number out so the answer is 0

You assume countably-infinite-X minus countably-infinite-Y is equal to zero, if X=Y.

Is that a valid assumption?
 
I would have to assume that to get an answer of zero.
 
It depends on the rabbit's algorithm for throwing rocks out of the hole.

Suppose that at each step he throws out the rock most recently thrown into the hole. Then at each step you throw 99 rocks into the hole which are never removed by the rabbit, and you will have infinite rocks in the hole at the end.

On the other hand suppose that at each step he takes the rock which is on the bottom of the pile (the rock which was thrown in before any other rock currently in the hole) and removes that one. Then every rock which is thrown into the hole is removed by the rabbit, and you will end up with no rocks in the hole
 
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