# Trouble with the Cantor Set

1. Sep 20, 2010

### lugita15

I'm having trouble understanding the Cantor set. The idea of a nowhere dense uncountable set makes no sense to me, because of the following argument I thought of:

Let x be any element of the Cantor set. Since the Cantor set is nowhere dense, there exists some open interval I_x containing x but containing no other elements of Cantor set. Moreover, if x and y are distinct elements of the Cantor set then I_x and I_y are disjoint. Since the Cantor set is uncountable, the set {I_x: x is in the Cantor set} is also uncountable. But it is impossible to have an infinite number of pairwise disjoint open intervals, because each interval contains a rational number, and there are only countably many rational numbers.

Where is the error in my reasoning? It's probably something obvious, but I don't see any errors in reasoning.

Any help would be greatly appreciated.

2. Sep 21, 2010

### Hurkyl

Staff Emeritus
That does not follow! In fact, I believe every point of the Cantor set is a cluster point.

3. Sep 21, 2010

### D H

Staff Emeritus
In other words, any open interval in [0,1] that contains a member of the Cantor set C3 contains an infinitely many members of that set.

The error in your reasoning, lugita, is that you have a misconception of the meaning of the term "nowhere dense".

4. Sep 21, 2010

### lugita15

. My understanding is that there exists no open interval which is a subset of the Cantor set. Intuitively, that means that the Cantor set is "disconnected". If that is the case, how can every interval containing x also contain other elements of the Cantor set? This doesn't make sense to me.

5. Sep 21, 2010

### Office_Shredder

Staff Emeritus
Just because the whole interval is not contained in the Cantor set doesn't mean there can't be any elements of the Cantor set contained inside of it

6. Sep 21, 2010

### D H

Staff Emeritus
Correct. That is one way to interpret "nowhere dense".

Incorrect. Bottom line: Don't trust your intuition with things such as the Cantor set that are specifically constructed to be counter-intuitive.

7. Sep 21, 2010

### l'Hôpital

The "intuitive" way to think about the cantor set is that it contains infinitely many infinitely-small "intervals" and since intervals (no matter how small, so long as they're not a singleton) contain infinite amount of points, by grabbing one, you get all the others as well.

Of course, this is not rigorous or anything of the sort, but it is a way to sort of bring some sort of intuition into it.

8. Sep 21, 2010

### D H

Staff Emeritus
Emphasis mine:
That, I think, is the crux of lugita15's misunderstanding. He appears to be interpreting "nowhere dense" as meaning "comprised of singletons."

The points in the Cantor set are not singletons.

Let's look at the construction of the Cantor set. Start with the interval [0,1] and cast out the (open) middle third, (1/3,2/3), leaving the intervals [0,1/3] and [2/3,1]. Now cast out the (open) middle thirds of these two intervals, and so on. The endpoints of the surviving closed intervals will survive any subsequent removals. However, the Cantor set comprises more than these surviving endpoints. The number 1/4, for example, is not an endpoint of any of the finite subdivisions yet it is a member of the Cantor set: The trinary expansion of 1/4 is 0.20202...

This illustrates one way to look at the Cantor set: It comprises zero, one, and the numbers between 0 and 1 (exclusive) whose infinitely-long trinary expansion consists of only 0s and 2s. I can already hear the objection:
"Wait a second! 1/3 is a member of the Cantor set, and its trinary expansion is 0.1!"​
I said "infinite trinary expansion". In base 10, 0.999...=1 and 0.0999...=0.1. Similarly, in base 3, 0.222...3=1 and 0.0222...3=0.13. Any number that is regular in some base (has a finite expansion in that base) also has an infinitely-long repeating expansion in that base. It is these infinitely-long expansions rather than the finite expansions that are of interest here.

Consider some element x3 of the Cantor set exclusive of 0 and 1. Change all of the 2s in the infinitely-long trinary expansion of x3 to 1s and interpret the result as a binary number. Call this binary number x2. This will obviously be in the open interval (0,1). Now consider some open interval I2 in [0,1] that contains x2. Denote the endpoints of this interval as a2 and b2. These endpoints will have infinitely-long binary expansions. Change all of the 1s in their binary expansions to 2s and reinterpret these endpoints as infinitely-long trinary expansions. Call these numbers a3 and b3. The open interval I3=(a3, b3) will be in [0,1] and will contain our original member of the Cantor set x3. It will also contain infinitely many (uncountably many) other members of the Cantor set. Any element of I2 will have an infinitely-long binary expansion. Take some member of I2 that is not equal to x2. Change all of the 1s in this expansion to 2s and reinterpret the result as a number in base 3. The result will be a member of I3, it will not be equal to x3, and it will obviously be a member of the Cantor set. Finally, note that we can make I3 arbitrarily small by making I2 arbitrarily small.

9. Sep 21, 2010

### lugita15

That's interesting. I wonder if that intuition can be made rigorous using nonstandard analysis. In other words, if the Cantor set is constructed out of hyperreal numbers rather than real numbers, then does it contain a countably infinite number of intervals whose lengths are all infinitesimal? Of course, this wouldn't present a contradiction, because an interval whose length is an infinitesimal hyperreal number need not contain any rational numbers.

If that is the case, it would present a fascinating possibility. Let X_0 be the interval [0,1] in the hyperreal line, and let X_(k+1) be the set of hyperreal numbers obtained if the "middle third" is removed from each interval that makes up X_k. Then X_omega would be the Cantor set, where omega is the lowest infinite ordinal number. But X_omega still contains open intervals! So we can continue the process and obtain X_(omega + 1) and so on. I wonder what is the lowest ordinal a, if such an ordinal exists, such that X_a contains no intervals?

If hyperreal numbers don't suffice for these purposes, maybe we can try surreal numbers instead. Surreal numbers have the added advantage that all ordinal numbers are surreal numbers.

10. Sep 21, 2010

### lugita15

Then maybe that is an indication that the operation of removing middle thirds have not been applied a sufficient number of times! See my reply to l'Hôpital for my speculation along these lines.

11. Sep 22, 2010

### D H

Staff Emeritus
I suggest you stick to understanding Cantor set in the reals and understanding why 1/4 is a member of that set before diving off the deep end into a non-standard construction of the Cantor set.

12. Sep 22, 2010

### lavinia

Can you construct an open interval in which the Cantor set is dense?

13. Sep 22, 2010

### lavinia

Suppose instead of removing middle thirds one removes middle fifths or sevenths. The resulting set will have positive measure and is closed. Is is still totally disconnected?

Suppose we construct the analogue of the devil's staircase using one of these sets. Will the curve be non-differentiable on a set of positive measure?

Last edited: Sep 22, 2010