# Two questions about the Cantor set construction

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## Solution 1

As Andre Nicolas commented, if we only remove the end points we are left with more or less the entire Cantor set in terms of cardinality. Topologically speaking, however, this is not the case.

A very interesting thing happens when removing these end points, though. I will elaborate on that in a short while. First we introduce a few concepts (and while at it take this question from into and )

First I want to introduce the Baire space, denoted often by $\omega^\omega$ and $\mathcal N$. This is the space of infinite sequences of natural numbers, but we can look at this space as though it were a tree.

In the root we have the empty sequence, on the first level we have sequences in length one, that is $\langle n\rangle$, then we have sequences in length of two, three, and so on. Each vertex in the tree has infinitely many immediate successors. The Baire space is the space formed when the tree has infinitely many levels, and we consider the branches (and not the vertices) as the elements.

To compare, the Cantor set has a similar construction when considering the underlying set $\{0,1\}$ as Dylan showed in his answer with $3$-adic expansions limited to digits $0,2$ in the expansion.

The Baire space (and every tree of this form as well) can be endowed with a topology - that is we say which sets are open and which are closed. In this case we let a basic open set (a ball of some positive radius, if you will) to be all branches that share the same initial segment of a prescribed length.

Examples would be all the sequences in which the first ten coordinates are $42$, or the sequences that in the first six coordinates are $\langle 2,574,87144781, 2^{1000}, 314159\rangle$.

We can also define a metric function on the Baire space, turning it into a metric space (similar to $\mathbb R$, yet very very different). If we have two distinct branches, they must have at least one $n$ in which they are different. Therefore they have a minimal $n$ with this property, so we say that they are in distance of $2^{-n}$. Formally: $$d(u,v) = \begin{cases} 2^{-n} & \text{ where } n = \min\{k\mid u(k)\neq v(k)\} \\ 0 &\text{ otherwise} \end{cases}$$

This is an ultrametric, which means that every element inside an open ball is its center (kind of weird, compare with $p$-adic numbers).

A very interesting fact about the Cantor set is that it is the only (up to homeomorphism) space which is totally disconnected, separable, has no isolated points and compact.

Two interesting facts about the Baire space:

1. It is homeomorphic to the irrationals, that is $\mathbb R\setminus\mathbb Q$, with the subspace topology. It is also a complete metric space. While this seems somewhat strange, since clearly the irrationals are incomplete (take a sequence going to $0$). However the two metrics are only topologically equivalent, this means that if we were to measure distance in a way similar to the way we do in the Baire space, a sequence of irrationals will converge to an irrational limit.

2. This is a totally disconnected space, without isolated points, it has a countable dense subset, and every compact subset of the Baire space has an empty interior. To compare the Cantor set has an empty interior in the reals, that is there is no open interval contained within the Cantor set. To some extent this means that compact subspaces of the Baire space "look" a bit like the Cantor set (to be fair, they look exactly like the Cantor space, but this is not the point here).

Like the Cantor set, the Baire space is unique (up to homeomorphism, of course) space with this (the second) property. Which makes both the Cantor set and the Baire space very interesting in the field of descriptive set theory.

Now to the main punch, suppose we take the Cantor set, and remove all the end points of the intervals (i.e. we remove closed intervals in the construction) the result is a space which zero dimensional, totally disconnected, without isolated points, in which every compact subspace has an empty interior!

That means that we start with the Cantor set, removed a relatively small collection of points and ended up with the irrational numbers, which is a completely different space.

As for the compact notation for $C_n$, note that every stage you remove more and more intervals. In fact at the $n$-th stage we remove $2^n$ intervals (as well all that you removed before). The Wikipedia page suggests the following notation:

$$C_n = [0,1]\setminus\bigcup_{k=0}^{2^m-1} (\frac{3k+1}{3^m},\frac{3k+2}{3^m})$$

Note that in this notation the $C_n$ are not a decreasing sequence of sets, if you want this property as well replace $[0,1]$ by $C_{n-1}$ (and let $C_0 = [0,1]$ of course).

## Solution 2

I don't think you'll produce the same set. At each stage of the usual construction, whatever endpoints you pick up are always in the Cantor set; for example, $1/3, 2/3 \in C$. If I understand you correctly, then your construction removes these points.

As for a sensible way of writing the $C_n$, that depends on your definition of "sensible". I think that cutting off the formula for $C$ given at Wikipedia at a finite stage is something you've considered. You could look at $C_n$ via $3$-adic expansions: to begin, prove that $C_1$ consists of points that can be written as $$\sum_{i = 1}^\infty \frac{a_i}{3^i}$$ where $a_i = 0, 1, 2$ and we restrict to $a_1 = 0$ or $2$. Such descriptions are pretty compact, but I always found it hard to use them to prove anything about $C$. On the other hand, they make constructing the Cantor function very simple.

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### Noteventhetutorknows

Updated on June 01, 2022

• Noteventhetutorknows 22 days

Let $C_0$ be the segment $[0,1]$ $C_2$ will be $[0,1]$, with the middle third, an open set removed, so $[0,1/3]\cup[2/3,1]$

First, if we removed closed sets would the cantor set, the limit of what remains from this process, be the same?

I was able to show that the limit of the sequence of partial sums of the pieces converges to $1$. And this happens if the removed bits are open or closed, it doesn't matter. Don't know if that helps or not.

Second, I've been trying for two hours and I still can't find a sensible way to write $C_n$, in some compact notation.

• André Nicolas almost 11 years
Not the same, but all the endpoints removed are rationals of the shape $k/3^n$ (however, not all such rationals are endpoints). Thus in your "remove closed intervals" version, we are removing only countably many more points than in the Cantor set construction. Since the Cantor set has the cardinality of the continuum, this means that in your version, "almost all" of the original Cantor set remains.
• Gerry Myerson almost 11 years
I want to edit the title - "acceding" is a typo, right? Is it supposed to be "ascending"? But then again, the question doesn't seem to be about any kind of order on the Cantor set "endpoints". Oh, and I see $C_0$ and $C_2$, but what happened to $C_1$?
• Pete L. Clark almost 11 years
"Acceding" was edited to "accending"? The world is a strange place sometimes.
• Gerry Myerson almost 11 years
OK, I got impatient and edited the title, and while I was there I did away with the descriptive-set-theory tag. OP can take care of $C_1$, or not.
• Asaf Karagila almost 11 years
@Gerry: Why did you remove the [descriptive-set-theory] tag?
• Gerry Myerson almost 11 years
@Asaf, descriptive-set-theory is a technical term - it doesn't just mean, "questions about describing sets." Having said that, I may have been too hasty. en.wikipedia.org/wiki/Descriptive_set_theory says "Descriptive set theory begins with the study of Polish spaces and their Borel sets," and gives the Cantor set as an example of a Polish space. If someone who, unlike me, actually knows something about descriptive set theory wants to put the tag back, I'll get out of the way.
• Asaf Karagila almost 11 years
@Gerry: Seeing how I just finished taking a course in descriptive set theory, I figured I had the basic judgment skills whether this tag relates to my answer or not :-) I will put it back now, if you don't mind.