# Size of the division of a segment in infinite intervals

1. Jun 8, 2006

### Iraides Belandria

Dear people of this forum,
Assume we have an interval of a line [ a, b].
Now, let us divide this interval in many ( infinite ) subintervals of finite size. ? What will be the size of these many intervals, equal or unequal, when the number of subintervals tends to infinite?. ¿there is any theorem to confirm possible answer?. Thank you.

2. Jun 8, 2006

### TD

If you divide it into a subintervals of finite size (let the smallest one be 'w' in width), then you can't have an infinite number of subintervals since you'll have at most (b-a)/w subintervals, which is finite.

3. Jun 8, 2006

### Hurkyl

Staff Emeritus
Intuitively, it could be just about anything.

Technically, this isn't a well-defined question... the sizes of the intervals form a finite sequence of numbers... and as we let the number of subintervals tend to infinity, then we have a sequence of finite sequences of numbers... and they are all different lengths. I how would you define a limit of such a thing?

You can't assume there will be a smallest! Their lengths might tend to zero. (Say... maybe the lengths form a geometric sequence)

4. Jun 8, 2006

### matt grime

You can't say anything about the sizes, other than none of them is larger than b-a.

5. Jun 8, 2006

### TD

Not even when the size of all subintervals are finite? When you let the number of subintervals tend to infinity (and hence the size to 0), I agree (and understand), but as long as you have a finite number of subintervals? Apparently I'm missing something :grumpy:

6. Jun 8, 2006

### matt grime

That the OP said there were an infinte number of subintervals, perhaps? They then wanted to let the number tend to infinity, which is odd. And in cany case, who is saying that the subintervals are equally sized?

7. Jun 8, 2006

### TD

Well that confused me, is it possible to have a infinite number of subintervals of a finite interval [a,b] while the sizes of all those subintervals are still finite? I'd say that you have a finite number of subintervals with possibly finite sizes and that the sizes will become infinitesimally small (tending to zero) when the number tends to infinity.

8. Jun 8, 2006

### matt grime

0 is a finite length, I don't understand why there is this emphasis of finite length at all. They are sub intervals of [a,b] which automatically makes them finite.

9. Jun 8, 2006

### TD

You're right, thanks.

10. Jun 8, 2006

### Hurkyl

Staff Emeritus
Yes.

You can even require that the sizes be positive, and that the intervals overlap only at their endpoints.

I even already gave a hint as to how this may be accomplished.

11. Jun 8, 2006

### Gokul43201

Staff Emeritus
The source of the confusion is that too many physicists (and possibly other non-mathematicians as well) incorrectly use the term 'finite' as almost synonymous with 'non-zero' (a "finite" quantity as something that is neither infinitely large nor infinitesimally small).

12. Jun 9, 2006

### TD

Guiltly, although I'm not a physicist
Usually not though, but apparently in this context I was thinking differently about 'finite'

13. Jun 9, 2006

### Iraides Belandria

Analizing your discusion, there is an additional restriction that I did not mention in my original thread.

The requirement is that to each nonzero small segment of the line we have to associate by a one to one transformation a number 1,2,3,4,5....n where n is a huge number near infinite.
In other words, let us divide a line of total lengh L equal to 1 meter in a huge number of non zero small segments L1,L2,L3,L4,L5............Ln and such way that
L1-------->1
L2-------->2
L3-------->3
L4-------->4
...............
...............

Ln-------->n

¿ If n tends to a number close to infinite in such a way that Ln is a nonzero small line segment, then

L1=L2=L3=L4=...........=Ln ?

14. Jun 9, 2006

### matt grime

that doesn't make mathematical sense. there is no concept of a number being 'near infinity'.

15. Jun 9, 2006

### Gokul43201

Staff Emeritus
Iraides, what's stopping me from making 1 segment of length 0.9, and then chopping the remaining interval of 0.1 into as many segments as I need?

16. Jun 9, 2006

### reilly

If all the sub-intervals are finite, then the partial sums of interval lengths, say starting from zero, will form a non-convergent series. That's why folks invented the definite integral.
Regards,
Reilly Atkinson

(Just to be clear, if the subintervals are all finite, then the limit L(n) ->. non-zero as n-> infinity, where L(n) is the length of the nth subinterval)

Regards,
Reilly Atkinson

17. Jun 10, 2006

### matt grime

Even if the n'th subinterval has length 2^n?

18. Jun 12, 2006

### HallsofIvy

Staff Emeritus
No, that's not clear at all- it's just wrong. As matt grime pointed out, if the length of the nth interval is 2-n the "the subintervals are all finite" but the limit is 0. In the case where we have in infinite number of intervals such that the length of the nth interval is 2-n then the length of the entire interval is 1.

19. Jun 13, 2006

### reilly

I'm right if I say "finite AND non-zero" which I did not, so I'm not right.
Regards,
Reilly

20. Jun 13, 2006

### matt grime

No, you are not right even with your correction. 2^{-n} gives an obvious subdivision and no subinterval has length 0 (or infinity).

Last edited by a moderator: Jun 13, 2006