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September 7th, 2012, 08:33 PM   #1
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Dichotomy paradox reverse engineering

t = 0: move pen from 0 to 0.5 (taking 1/2 second), naming it segment #1
t = 0.5: move pen from 0.5 to 0.75 (taking 1/4 second), naming it segment #1
t = 0.75: move pen from 0.75 to 0.875 (taking 1/8 second), naming it segment #1
...

At which step should we stop using #1 as the segment's index in order to have enough segments of non-zero size left to assign the rest of the natural numbers?

If we don't stop using #1 all segments in [0, 1) will be named #1 at t = 1, right? What about the size of all other segments (#2, #3, ...) at t = 1 then? I mean it like this:

t = 0: move pen from 0 to 0.5, naming it segment #1, [0.5, 0.75] is #2, [0.75, 0.875] is #3, ...
t = 0.5: move pen from 0.5 to 0.75, naming it segment #1, [0.75, 0.875] is #2, [0.875, 0.9375] is #3, ...
t = 0.75: move pen from 0.75 to 0.875, naming it segment #1, [0.875, 0.9375] is #2, [0.9375, 0.96875] is #3, ...

and so on.
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September 8th, 2012, 12:49 AM   #2
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Re: Dichotomy paradox reverse engineering

Quote:
Originally Posted by netzweltler
t = 0: move pen from 0 to 0.5 (taking 1/2 second), naming it segment #1
t = 0.5: move pen from 0.5 to 0.75 (taking 1/4 second), naming it segment #1
t = 0.75: move pen from 0.75 to 0.875 (taking 1/8 second), naming it segment #1
...
You get your t values by defining a function f from the natural numbers 0, 1, 2, ... to the reals defined as

f(n) = 1 - (1/2^n)

Clearly t = 1 is not in the range of f. So the behavior of your system at t = 1 is not defined. Most of these supertask problems can be resolved by noting that the task is not well-defined at the limit.
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September 8th, 2012, 01:53 PM   #3
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Re: Dichotomy paradox reverse engineering

Quote:
Originally Posted by Maschke
Quote:
Originally Posted by netzweltler
t = 0: move pen from 0 to 0.5 (taking 1/2 second), naming it segment #1
t = 0.5: move pen from 0.5 to 0.75 (taking 1/4 second), naming it segment #1
t = 0.75: move pen from 0.75 to 0.875 (taking 1/8 second), naming it segment #1
...
You get your t values by defining a function f from the natural numbers 0, 1, 2, ... to the reals defined as

f(n) = 1 - (1/2^n)

Clearly t = 1 is not in the range of f. So the behavior of your system at t = 1 is not defined. Most of these supertask problems can be resolved by noting that the task is not well-defined at the limit.
I guess that's true. So, what I am trying here is some kind of reverse engineering, i.e. I am defining what I think is needed and look for a process that fits the needs.

The task is to design a process that enables us to draw a line from point 0 to point 1 in 1 second in infinitely many steps. Any suggestions how to accomplish that (i.e. which steps have to be defined)? The pencil is not allowed to change direction. Point 1 is the last point covered by the line. There is no last step in this process.
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September 8th, 2012, 05:07 PM   #4
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Re: Dichotomy paradox reverse engineering

Quote:
Originally Posted by netzweltler
The task is to design a process that enables us to draw a line from point 0 to point 1 in 1 second in infinitely many steps. Any suggestions how to accomplish that (i.e. which steps have to be defined)? The pencil is not allowed to change direction. Point 1 is the last point covered by the line. There is no last step in this process.
Why don't we just do it in one step. At t = 0 put your pen at x = 0. Let the pen move to the right according to the rule x = f(t) = t. Then at t = 1 you'll have been moving to the right for exactly one second, and you'll have hit every single one of the points in the closed unit interval.

Continuous motion. What a concept!
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September 9th, 2012, 12:03 AM   #5
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Re: Dichotomy paradox reverse engineering

Quote:
Originally Posted by Maschke
Quote:
Originally Posted by netzweltler
The task is to design a process that enables us to draw a line from point 0 to point 1 in 1 second in infinitely many steps. Any suggestions how to accomplish that (i.e. which steps have to be defined)? The pencil is not allowed to change direction. Point 1 is the last point covered by the line. There is no last step in this process.
Why don't we just do it in one step. At t = 0 put your pen at x = 0. Let the pen move to the right according to the rule x = f(t) = t. Then at t = 1 you'll have been moving to the right for exactly one second, and you'll have hit every single one of the points in the closed unit interval.

Continuous motion. What a concept!
I guess, the greeks did it that way before Zeno came up with his ideas. Now we have to reinvent continuity. So, try it again: Infinitely many steps - no last step, point 1 is the last point covered.
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September 10th, 2012, 12:50 PM   #6
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Re: Dichotomy paradox reverse engineering

Quote:
Originally Posted by netzweltler
Infinitely many steps - no last step, point 1 is the last point covered.
If there's no last step, how can 1 be the last point covered, unless you allow redundant steps? If you allow redundant steps you can just have each step cover [0, 1].
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September 10th, 2012, 07:23 PM   #7
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Re: Dichotomy paradox reverse engineering

Quote:
Originally Posted by CRGreathouse
If there's no last step, how can 1 be the last point covered, unless you allow redundant steps? If you allow redundant steps you can just have each step cover [0, 1].
A possible solution might be:

t = 0: move pen from 0 to 0.5 (taking 1/2 second)
t = 0.5: move pen from 0.5 to 0.75 (taking 1/4 second)
t = 0.75: move pen from 0.75 to 0.875 (taking 1/8 second)
t = 0.875: move pen from 0.875 to 0.9375 (taking 1/16 second)
...
t = 1: move pen from 1 to 1 (taking 0 second)
t = 1: move pen from 1 to 1 (taking 0 second)
t = 1: move pen from 1 to 1 (taking 0 second)
...

we are ending up in an endless loop which allows the last point 1 to be drawn without introducing a last step of the process. Even if we didnít define a last step we can lift the pencil, since there is no difference between executing this endless loop and not executing it (the endless loop does do nothing in time and space).

The problem is: This model is lacking a description of the process that enables us to adjoin the endless loop. That's why I think it might be useful to clarify this question:
Quote:
t = 0: move pen from 0 to 0.5 (taking 1/2 second), naming it segment #1
t = 0.5: move pen from 0.5 to 0.75 (taking 1/4 second), naming it segment #1
t = 0.75: move pen from 0.75 to 0.875 (taking 1/8 second), naming it segment #1
...

At which step should we stop using #1 as the segment's index in order to have enough segments of non-zero size left to assign the rest of the natural numbers?
First of all we might have to clarify, if there is a limit case for this task:
Quote:
t = 0: move pen from 0 to 0.5, naming it segment #1, [0.5, 0.75] is #2, [0.75, 0.875] is #3, ...
t = 0.5: move pen from 0.5 to 0.75, naming it segment #1, [0.75, 0.875] is #2, [0.875, 0.9375] is #3, ...
t = 0.75: move pen from 0.75 to 0.875, naming it segment #1, [0.875, 0.9375] is #2, [0.9375, 0.96875] is #3, ...

and so on.
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September 11th, 2012, 10:05 AM   #8
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Re: Dichotomy paradox reverse engineering

Quote:
Originally Posted by netzweltler
First of all we might have to clarify, if there is a limit case for this task:
I assume you understand that the sequence 1/2, 3/4, 7/8, 15/16, ... has the limit 1; but no term of the sequence is equal to 1.

You can take the millionth term of this sequence, or the billionth, or the googol-th or the googolplex-to-the-googolplex-th term, and none of them are 1.

1 is the limit of the sequence, but it's not attained by any element of the sequence.

That's why the limit concept is so clever. It packages up this counterintuitive mystery of getting "arbitrarily close without attaining" in a concise definition that we can teach to calculus students. [Not that they understand it, but that's what Real Analysis is for ]

But I don't understand what you're trying to do. The facts I outlined are well-understood and universally accepted in mathematics. No finite halving process will let you hit the endpoint.

Another way to put this is that in binary, the sequence .1, .11, .111, .1111, ... has 1 as its limit, but none of the members of the sequence is 1. So in fact the binary number .111... is equal to 1; but that expression really denotes an infinite sum 1/2 + 1/4 + 1/8 + ... No finite sum of powers of 1/2 can be equal to 1.
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September 11th, 2012, 11:37 AM   #9
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Re: Dichotomy paradox reverse engineering

I think I can understand the concept of limits, but I am not that sure about it. If I try to apply the concept of limits to the task
Quote:
t = 0: move pen from 0 to 0.5, naming it segment #1, [0.5, 0.75] is #2, [0.75, 0.875] is #3, ...
t = 0.5: move pen from 0.5 to 0.75, naming it segment #1, [0.75, 0.875] is #2, [0.875, 0.9375] is #3, ...
t = 0.75: move pen from 0.75 to 0.875, naming it segment #1, [0.875, 0.9375] is #2, [0.9375, 0.96875] is #3, ...

and so on.
I guess the (most) correct assumption is, that all segments in [0, 1) are named #1 in the limit case. But what happens to the other segments which are always present at the finite steps. There is always a segment #2. The size of the segment named #2 is converging towards 0. Is there a segment #2 present in the limit case? Is it's size 0 in the limit case? If so, does it mean it is no longer an element of the set of segments in [0, 1]?
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September 11th, 2012, 03:14 PM   #10
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Re: Dichotomy paradox reverse engineering

Quote:
Originally Posted by netzweltler
I think I can understand the concept of limits, but I am not that sure about it. If I try to apply the concept of limits to the task
Quote:
t = 0: move pen from 0 to 0.5, naming it segment #1, [0.5, 0.75] is #2, [0.75, 0.875] is #3, ...
t = 0.5: move pen from 0.5 to 0.75, naming it segment #1, [0.75, 0.875] is #2, [0.875, 0.9375] is #3, ...
t = 0.75: move pen from 0.75 to 0.875, naming it segment #1, [0.875, 0.9375] is #2, [0.9375, 0.96875] is #3, ...

and so on.
I guess the (most) correct assumption is, that all segments in [0, 1) are named #1 in the limit case. But what happens to the other segments which are always present at the finite steps. There is always a segment #2. The size of the segment named #2 is converging towards 0. Is there a segment #2 present in the limit case? Is it's size 0 in the limit case? If so, does it mean it is no longer an element of the set of segments in [0, 1]?
As I indicated in my first post, you would have to tell us. Because your process is not defined at t = 1.

Let me repeat that. Your process is not defined at t = 1.

So for all we know, at step n you have your pen moving to 1 - 1/2^n or whatever. And at t = 1 your pen turns upside down and whistles Dixie.

Nobody knows what your pen does at t = 1 because you haven't taken the trouble to defined the behavior of the pen at t = 1.
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