1. Jul 13, 2003

### sage

you have heard this before perhaps. its about a runner trying to run d metres. he covers d/2 in t1 second, then half of the distance that is left in t2 seconds, then half of the rest in t3 seconds and so on.as there is always a finite distance left, according to the paradox he can never cover d metres. so how does he do it?

2. Jul 13, 2003

### Hurkyl

Staff Emeritus
This infinite sequence of actions can be accomplished in finite time, so he does them all and then keeps going.

3. Jul 13, 2003

### AndersHermansson

We had this over at sciforums recently. The short answer is that a sum of infinite series can be finite, which is where it might seem confusing. So that if you add an infinite amount of lengths the total length can still be finite. So the original question simply assumes it is not so.

4. Jul 13, 2003

### quartodeciman

"there is always a finite distance left"

really means

"there is for any time before d/v (, with v being the speed of the runner) a finite distance left".

5. Jul 14, 2003

### sage

yes this infinite sequence converges. but the point is if we go on adding the successive elements of the sequence one by one (as must be done here) we never reach the end of the sequence precisely because it is infinite. as we cannot reach the end of the sequence we cannot cover this finite distance in the calculated finite time. consider the finite time interval between n-th second and n+1-th second. first half a second passes by, then another 1/4-th, then another 1/8-th and so on. another infinite sequence converging at the limit, but that limit can never be attained. that is the problem.

6. Jul 14, 2003

### Hurkyl

Staff Emeritus
But why should one think that sequence of events cover the entire range of motion? Try this transfinite sequence:

Cover half the distance.
Cover half of what's left.
Cover half of what's left.
...
(countably finite repetitions)
...
Arrive at the destination.

Each step in the sequence picks up right there you left off if you perform all previous steps, includes the "Zeno sequence", and continues on afterwards to arrive at the destination.

7. Jul 14, 2003

### drnihili

You have to be quite clear on what the question is. If you take Zeno to merely be asking how an infinite sequence can occupy a finite space, then calculus indeed answers the question. However, if you taking him to be asking the question of how one can complete and infinite sequence one member at a time, then calculus not only doesn't answer the puzzle but is entirely irrelevant to it. I think the latter question is the better way to understand the point of the paradox.

There are a host of related paradoxes which highlight the central issues. SOmetimes it helps to look at them instead of just the runner paradox.

8. Jul 14, 2003

### drnihili

Ah, but this sequence can't be right. It presumes that after you've completed all the half distances you still have to do something further to arrive. If your sequence were correct, it would be possible to travel all the distances and yet still fail to arrive. But arriving cannot amount to traversing a distance or you give up the continuity of the reals. So on your account two runners could travers precisely the same distance and yet one of them would run d meters and the other wouldn't.

9. Jul 14, 2003

### Hurkyl

Staff Emeritus
Covering all of the half distances means covering the interval [0, d). If I run 1 meter per second, I cover all the half distances over the time interval [0, d).

You actually have to get to time d to have arrived at distance d. Zeno's paradox is a paradox because it presumes that you can't continue beyond the infinite sequence of covering half distances.

By continuity, any possible continuation of motion would have to include being at distance d at time d.

Last edited: Jul 14, 2003
10. Jul 17, 2003

### drnihili

The problem is that the open and closed intervals have the same distance. Closing the interval does not add any distance. Continuity comes in because the LUB of the two intervals is the same. If the runner really has completed all of the open intervals, he must have arrived at d.

Suppose otherwise, i.e that the runner has completed [0, d) but has not yet arrived at d. Call the runner's position r. r must be between the open interval and d. But this contradicts the fact that d is the least upper bound of the interval. So if r<d, then r must be in the open interval. But if r is in the open interval, then the runner has not yet completed the interval. This is because for every point in the interval there are infinitely many other points beyond it that are still in the interval. So r cannot be in the interval. thus the earliest point which can be r is d.

And the paradox isn't that you can't continue beyond the open interval, it's that you can't complete the interval at all.

Last edited: Jul 17, 2003
11. Jul 17, 2003

### Hurkyl

Staff Emeritus
I'm aware the lengths of [0, d) and [0, d] are the same.

Anyways, a paradox is typically a contradiction that arises from an unfounded assumption. They usually get cleared up once you try to do everything rigorously.

So tell me, as precisely as possible, what you think the problem is.

12. Jul 17, 2003

### drnihili

Well, I don't think I agree with your view of what a paradox is, but we'll leave the general theory of paradox for another thread.

The paradox in this case is that the runner, Achilles, must accomplish an infinite sequence of tasks. We know that he can complete them, we can even calculate precisely by when he will have completed them. The problem is in explaining how he completes them.

Achilles starts out with an infinite number of tasks to do. By the description of the problem, he must complete them one at a time. After he has accomplished his first task, there are an infinite number of tasks left. After he completes his second taks, there are an infinite number of tasks left. In fact after each task that he completes, there's always an infinite number left. As he moves down his list of tasks, he never gets any closer to the end of it. He always has just as many left to do as he started out with. As long as he is still working on the list, he has infinitely many left. The first point at which he has fewer than infinitely many tasks left is when he is all done, and at that point he has zero. He never decreases his list, he just suddenly finds that it is already done. So how is it that he manages to get to the end?

Geometry can predict the point at which Achilles will be done. Calculus can explain how it is that all the decreasing segments have a finite sum. But neither of them explains how it is that Achilles counts through the list, one task at a time - how he manages to complete an endless sequence.

13. Jul 17, 2003

### Hurkyl

Staff Emeritus
You still haven't answered the big question; why should an infinite sequence of tasks be impossible?

In particular (if I'm predicting your response correctly), why should every task in a sequence of tasks have a previous and a next task? (except, of course, for the first and last task, should they exist)

14. Jul 17, 2003

### drnihili

Because there's a function that given any task in the sequence returns the next task, and another function that returns the previous. If you take an ordering that lack that property it gets even more difficult. But Zeno's ordering does have the property.

15. Jul 18, 2003

### Hurkyl

Staff Emeritus
But why should an infinite series of tasks be impossible?

The resposne I was anticipating was something equivalent to saying that in my sequence of tasks, there is no task previous to "arrive at d". (it is eqiuvalent to say that there is no last task in Zeno's sequence)

16. Jul 22, 2003

### drnihili

That response doesn't quite get it right. I've tried to explain it a couple times, but I'll have another go at it.

If Achilles accomplishes an infinite series of tasks, there must be some action of his which counts as completing all the tasks. But none of the tasks can be that action as each of the tasks leaves an infinite number remaining. So, if Achilles accomplishes all the tasks, then there must be something he does beyond the tasks themselves in virtue of which he can be said to have completed them all. By the description of the problem, there is no such action.

If there were such an action, then it would be theoretically possible for Achilles to accomplish each of the tasks and yet still fail to complete all of them. This is absurd. Hence there can be no such action.

17. Jul 22, 2003

### Hurkyl

Staff Emeritus
For the problem at hand, there must be some task which counts as the completion of all (previous) tasks, though this isn't always the case. But the question is why must that task be one of the infinite series of tasks?

Continuity (and completeness) guarantees that there must be a unique limiting event, but it does not guarantee that the unique limiting event must be one of the members of the infintie sequence.

In particular, the limiting task is the "arrive at destination" step I listed.

18. Jul 22, 2003

### drnihili

Obviously it can't be one of the listed tasks. But your proposal is no solution. What exactly does one do to arrive at the destination and when does one do it? Do you really mean to imply that one might complete each of the tasks and still not arrive at the destination?

19. Jul 22, 2003

### Hurkyl

Staff Emeritus
One traverses the position interval [0,d) over the time interval [0, d). That is sufficient to be at position d at time d. (I'm assuming the traversal is in the manner being discussed)

I mean to imply that one does not reach the destination during the time interval in which one is performing Zeno's tasks. In this case, the time interval [0, d). One arrives at the destination at time d, after all of Zeno's tasks have been completed.

20. Jul 22, 2003

### drnihili

Here you've essentially said that completing all the tasks is sufficient for arrival. But you haven't said how that is accomplished. I agree that it's sufficient, that's not the issue. The issue is saying how it is done.

This can't be right. One doesn't first complete the tasks and then arrive. If that were the case then there would have to be a moment in between finishing the tasks and arriving. (given infinite divisibility.) But that would contradict what you said above about completing the tasks being sufficient for arriving. Arriving can't be separate from completing all the tasks. It can't occur after completing them, nor can it occur before completing them. It has to occur simultaneously with completing them. But this still leaves the problem of saying what it means to complete and endless sequence.