1. May 12, 2012

### cragar

When we use the nested interval property, we have closed intervals that are nested inside of each other and eventually when we do this a countable number of times out to infinity we eventually enclose a singlet point. So my question is can we apply this to the cantor set?
I mean when we remove the middle third of our line segment we keep the first third and the last third and they are closed intervals. And if I look at part of the cantor set I will have nested closed intervals that are subsets of each other. But we know that the cantor set has no isolated points. So what is wrong with my reasoning?

2. May 12, 2012

### Stephen Tashi

It isn't clear to me what you mean by "the closed interval property". You can have a countable sequence of nested closed intervals whose intersection is not a single point.

3. May 12, 2012

### cragar

I mean $[a_1,b_1]\supset [a_2,b_2]\supset......[a_n,b_n]....$
here is also a video of a guy talking about what i mean.

Last edited by a moderator: Sep 25, 2014
4. May 12, 2012

### micromass

Staff Emeritus
You forgot something extremely important: that

$$\lim_n b_n-a_n = 0$$

Also, I don't see how you would write the cantor set as an intersection of intervals.

5. May 12, 2012

### HallsofIvy

Staff Emeritus
As Stephen Tashi said, this is not necessarily true. You also need that the lengths of the intervals go to 0. Without that, you can only say that the limit is a non-empty set.

Each step is a union of intervals, not single intervals.

6. May 12, 2012

### Stephen Tashi

Perhaps a correct way to phrase the question is to say that one may define a nested sequence of intervals by taking one interval from each stage of the construction of the Cantor set. The length of these intervals approaches zero. This sequence of intervals will converge to a single point.

This doesn't imply that the single point must be an isolated point. If you picked a different sequence of nested intervals from the stages of construction, you can get other points in the Cantor set.

7. May 12, 2012

### cragar

ok you get what im saying. Lets start at the left edge of the cantor set, zero. and that will be our $a_n$ Now lets start with the original line segment and remove the middle 1/3, now are first b_n will be 1/3, then our next b_n will be 1/9 and then 1/27. our b_n will always be the endpoint of the interval containing 0 at the nth stage of removal.
SO I think the limit of this will go to zero $lim|b_n-a_n|$
So our nested intervals will be $[0, \frac{1}{3^n}]$ ok I guess the only problem might be that it wont be isolated, but it will converge to a point. But we always have endpoints so why wouldn't it be isolated?

8. May 13, 2012

### Stephen Tashi

For any $\epsilon > 0$ there will be a point of the Cantor set within $\epsilon$ of the point 0. Following the construction you described until the length of the intervals involved becomes less than $\epsilon$. Then begin using the right third of the intervals (instead of the left third) to get the next interval. This constructs a sequence of nested closed intervals whose length approaches 0 and they converge to apoint in the Cantor set within $\epsilon$ of the point 0.

9. May 13, 2012

### cragar

The reason I am using endpoints is because eventually the width of the endpoints will go to zero. And we always pick an endpoint because we do not want to have some numbers to the right of our endpoint. It seem like this construction would eventually just enclose 0 and it seems like 0 would be an isolated point, That is why I picked the endpoints. But I am probably missing something.

10. May 13, 2012

### Stephen Tashi

Points already have zero width. You need to practice using language precisely. This is not merely because it is a communication skill; it will clarify ideas in your own mind.

You missing the fact that there are other constructions of nested intervals involved. Just because your construction excludes all points to the right of 0 doesn't mean that someone else's different choices of nested intervals must do that. To show 0 is an isolated point, you would have to find an interval E containing 0 and show that nobody can pick a nested sequence of close intervals that converges to a point inside E. Can you find such an interval?

11. May 14, 2012

### cragar

OK I understand that someone can pick some other nested intervals. But it seems like the cantor set itself on the left edge has nested closed interval that go to zero width. Not because I picked them that way, but on that part of the cantor set by construction at the infinite step it has closed intervals that go to zero width. I mean I did pick it in a sense, I am looking at that part of the set. Im not trying to fight you on this, Im just trying to understand what is going on because it is not clear to me. Thanks for your responses by the way.

12. May 14, 2012

### Stephen Tashi

The fact that there is a sequence of nested closed intervals on the leftmost edge of the Cantor set whose widths approach zero, does not imply that the Cantor set has an isolated point on its left edge. You seem to believe that when the intersection of one particular sequence of nested closed intervals excludes a point P from the Cantor set that P cannot be in the Cantor set. That isn't true.

(Apparently you haven't thought about trying to find an open interval containing the left endpoint, which contains no other points of the Cantor set. Try to find one. What would its width be?)

13. May 14, 2012

### cragar

why not can you give me a concrete example.

14. May 15, 2012

### Stephen Tashi

I've done the best I can. You'll have to read over the previous posts.

15. May 18, 2012

### cragar

ok I think I understand it now. When I remove the middle one third of a line segment and lets say I pick the left half, instead of the right half. Then we do this again and then i pick another interval below, the distance between the line segments is getting smaller and will eventually go to zero. So I could pick different segments and it will converge to a point that is epsilon close to zero but not zero. I originally thought this was not possible because I thought that If i picked another line segment that did not contain 0 at some finite step that it couldn't get close to zero because their would be some positive distance between it and zero but I now see why the distance between them goes to zero. So i could think of picking my line segments as picking the left or right half at each step. SO I have 2 choices at each step. so I have
$2^{\aleph_0}$ number of paths that I can take to construct my nested closed intervals. you probably were getting at stuff like this above I just didn't see the connection and I talked to someone else about it and I think I got it figured out now. Thanks for your help by the way.

16. May 18, 2012

### Stephen Tashi

Yes, that's a key observation.