Generalized Cantor Set

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"Given (rn), rn E (0,1), define a generalized Cantor set E by removing the middle r1 fraction of an interval, then remove the middle r2 fraction of the remaining 2 intervals, etc.

Start with [0,1]. Take rn=1/5n. Then the material removed at the n-th stage has length < 1/5n, so the total length removed is < 1/5 + 1/52 + 1/53 +... = 1/4
Thus the length of E is >3/4. "
=========================

I don't understand why the material removed at the n-th stage has length < 1/5n. How can we derive this? At the n-th stage, we are removing 2n-1 pieces, so don't we have to multiply that by 2n-1?
I sat down and thought about this for half an hour, but I still can't figure it out.

I hope someone can explain this! Thank you!
 

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  • #2
jbunniii
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"Given (rn), rn E (0,1), define a generalized Cantor set E by removing the middle r1 fraction of an interval, then remove the middle r2 fraction of the remaining 2 intervals, etc.

Start with [0,1]. Take rn=1/5n. Then the material removed at the n-th stage has length < 1/5n, so the total length removed is < 1/5 + 1/52 + 1/53 +... = 1/4
Thus the length of E is >3/4. "
=========================

I don't understand why the material removed at the n-th stage has length < 1/5n. How can we derive this? At the n-th stage, we are removing 2n-1 pieces, so don't we have to multiply that by 2n-1?
I sat down and thought about this for half an hour, but I still can't figure it out.

I hope someone can explain this! Thank you!
It's true that you are removing [itex]2^{n-1}[/itex] intervals at stage [itex]n[/itex], but what is the LENGTH of each interval? It's not simply [itex]1/5^n[/itex], right? That is the FRACTION that you are removing from the preceding stage's intervals, so you have to multiply by the length of the preceding stage's intervals.

Since there are [itex]2^{n}[/itex] intervals remaining at the end of stage [itex]n[/itex], each one has length at MOST equal to [itex]1/2^{n}[/itex]. In reality the lengths are smaller because this assumes no material has been removed.

Therefore at stage [itex]n+1[/itex], you remove at most a length of [itex]2^n * (1/5^{n+1}) * (1/2^n) = 1/5^{n+1}[/itex]. (In reality you remove less.)
 
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Since there are [itex]2^{n}[/itex] intervals remaining at the end of stage [itex]n[/itex], each one has length at MOST equal to [itex]1/2^{n}[/itex]. In reality the lengths are smaller because this assumes no material has been removed.
I see what you're saying now: it is the FRACTION that you are removing from the preceding stage's intervals.

But I don't follow the reasoning here. Why are there 2n intervals remaining at the end of stage n?

And why each one has length at MOST equal to 1/2n? Can't it be the case that some of them has length less than 1/2n and some GREATER than 1/2n?

Thanks for explaining!
 
  • #4
jbunniii
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I see what you're saying now: it is the FRACTION that you are removing from the preceding stage's intervals.

But I don't follow the reasoning here. Why are there 2n intervals remaining at the end of stage n?
You start with one interval. In stage 1, you remove the middle part of that interval so at the end of stage 1 there are [itex]2 = 2^1[/itex] intervals left.

In stage 2, you start with 2 intervals, and you remove the middle parts of each, so you are left with [itex]4 = 2^2[/itex] intervals at the end of stage 2. And so on.

And why each one has length at MOST equal to 1/2n? Can't it be the case that some of them has length less than 1/2n and some GREATER than 1/2n?
No, because the way the process works is symmetric - in a given stage, you are removing the MIDDLE part of each interval, and you are removing the same FRACTION of each interval. So all of the intervals at the end of a given stage have the same length. And that length cannot be more than [itex]1/2^n[/itex] because there are [itex]2^n[/itex] intervals and the original interval had length 1.
 
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