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Positions in Infinite Space |
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| Jun26-12, 03:14 PM | #18 |
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Positions in Infinite Space |
| Jun27-12, 12:58 AM | #19 |
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The union of the segments [0, 0.5], [0.5, 0.75], [0.75, 0.875], ... is [0, 1) and not [0, 1]. If I am drawing these segments of the unit interval I am drawing infinitely many segments, but I am not drawing the last point 1.0 (which in some sense could be named the [itex]\omega[/itex]th segment). If the drawing of one segment takes a second, I am drawing these segments forever, and I am not reaching a point infinitely apart (in some sense point 1.0). That's what I meant, when I was asking: |
| Jun27-12, 03:21 PM | #20 |
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The fallacy in your analogy is that each step takes a fixed amount of time, no matter how small the interval.
I don't see the connection to the original question where (as I have already said) you need to give a precise definition of what you mean. |
| Jun27-12, 05:37 PM | #21 |
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To see the analogy it might be helpful to biject the previous example to the trip of the set of mathematicians who haven't reached their destination. Let's say
1. travelling the first meter → drawing the segment [0, 0.5] 2. travelling the second meter → drawing the segment [0.5, 0.75] 3. travelling the third meter → drawing the segment [0.75, 0.875] ... Do we have two different definitions of infinite travelling? 1. Travelling between two points infinitely apart 2. Passing every finite meter/segment of infinitely many meters/segments In the first case we arrive at the point infinitely apart. In the second case we don't. According to the first definition the set of mathematicians who haven't reached their destination is empty at arrival at the point infinitely apart. According to the second definition the set of mathematicians who haven't reached their destination is non-empty for the whole infinite trip, and so it might be valid to state, that the mathematicians of this set are on an infinite trip, thus questioning that there are only finite distances in infinite space. |
| Jun27-12, 06:01 PM | #22 |
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This whole conversation seems to have gone somewhere where I don't know where it is. First, the original question: The travel time of every mathematician is finite. We've settled that. It's done. As for this comment...
If that's not what you're talking about...then I have no idea. 1) There are no such points. The euclidean metric assigns to every two points in the plain a finite distance. 2) I really have no idea what this is supposed to mean. |
| Jun28-12, 01:29 AM | #23 |
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To get an idea of what I mean it might help to clarify this first:
I am shifting the line [0, 1] in infinitely many steps to the left step 1: I am shifting the line [0, 1] to position [-0.5, 0.5] step 2: I am shifting the line at [-0.5, 0.5] to position [-0.75, 0.25] step 3: I am shifting the line at [-0.75, 0.25] to position [-0.875, 0.125] ... At which position is the line after infinitely many steps? |
| Jun28-12, 03:15 AM | #24 |
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| Jun28-12, 11:39 AM | #25 |
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| Jun28-12, 12:08 PM | #26 |
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| Jun28-12, 12:31 PM | #27 |
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| Jun28-12, 03:33 PM | #28 |
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| Jun28-12, 06:53 PM | #29 |
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Honestly, I have no idea what it is that we're supposed to be talking about anymore. All of the initial questions about "infinite travelling" have been answered, but you seem to have spontaneously created some sort of brand new definition that you're not sharing with anyone. Please clearly explaining what it is that you're asking. |
| Jun29-12, 12:46 AM | #30 |
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step 1: at t = 0 I am shifting the line [0, 1] to position [-0.5, 0.5]
step 2: at t = 0.5 I am shifting the line at [-0.5, 0.5] to position [-0.75, 0.25] step 3: at t = 0.75 I am shifting the line at [-0.75, 0.25] to position [-0.875, 0.125] ... All shifting is done before t = 1. No action is done at t = 1. Countable infinity doesn't allow a last shift at t = 1, which clearly is needed for point -1 to be covered. If this is my private math solely, further clarification doesn't make sense. |
| Jun29-12, 03:12 AM | #31 |
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