Articles of cantor set

Intuitive way to understand the Smith–Volterra–Cantor set

The Smith–Volterra–Cantor set (or ε-Cantor set) is a set of points on $ℝ$ that is nowhere dense, yet has positive measure. As I understand it, being nowhere dense means containing no intervals. An interval is a set of real numbers where any number that lies between two numbers in the set is also included in […]

the elements of Cantor's discontinuum

Let $(A_n)_{n \in \mathbb{N}}$ the sequence of subsets of $\mathbb{R}$, given by $A_0 := \bigcup_{k \in \mathbb{Z}}[2k, 2k + 1]$ und $A_n := \frac{1}{3}A_{n-1}$ for $n ≥ 1$. Also, we define $$ A := \bigcap_{n=0}^\infty A_n, \,C:= A \cap [0, 1]$$ We call $C$ Cantor’s discontinuum. Given this definition, I now want to prove that […]

Cantor set: Lebesgue measure and uncountability

I have to prove two things. First is that the Cantor set has a lebesgue measure of 0. If we regard the supersets $C_n$, where $C_0 = [0,1]$, $C_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$ and so on. Each containig interals of length $3^{-n}$ and by construction there are $2^n$ such intervals. The lebesgue measure of each […]

Continuous functions and uncountable intersections with the x-axis

Let $f : \mathbb{R} \to \mathbb{R}$ such that the set $X = \{x \in \mathbb{R} : f(x) = 0\}$ does not contain any interval (i.e. there is no interval $I \subset X$) Of course the set $X$ can be uncountable (see Cantor Set). If we add that $f$ is continuous, is it true that X […]

Does every continuous map between compact metrizable spaces lift to the Cantor set?

I’m interested in the universal properties of the Cantor set. It is well-know that the Cantor set $2^\mathbb{N}$ is “universal” in the category of metrizable compact spaces, in the sense that every compact metrizable space is a quotient of $2^\mathbb{N}$ (although the quotient map is not unique). My main question is: Does every continuous map […]

Find a Continuous Function with Cantor Set Level Sets

This was a problem from a class that I thought was really interesting. It asked to find function $f\in C[0,1]$ such that the sets $\{x:f(x)=c\}$ form a Cantor Set for all $0\leq c\leq 1$. I found a non-constructive proof of the existence of such functions, but would be curious if anyone could give a constructive […]

Does every homeomorphism of a compact metric space lift to the Cantor set?

This is a follow-up to this question. It is well-known that any compact metrizable space can be expressed as a quotient of the Cantor set. But can every homeomorphism of such a space be lifted to a homeomorphism of the Cantor set? More precisely, let $K$ be a compact metric space, and let $h\colon K\to […]

Pairing function for ordered pairs

Is there a pairing function like Cantor’s (https://en.wikipedia.org/wiki/Pairing_function) that would map ordered pairs (of integers) to different integers? ie: (M, N) -> L1 (N, M) -> L2 Where L1 != L2 All input integers could be positive, but the output does not have to be positive..so perhaps something like: newpair(M, N) = if (M < […]

Dense subset of Cantor set homeomorphic to the Baire space

Does anyone know a proof that the Cantor set, $\{0,1\}^{\mathbb{N}}$, has a dense subset homeomorphic to the Baire space, $\mathbb{N}^{\mathbb{N}}$? Thank you.

Is the Cantor set made of interval endpoints?

The Cantor set is closed, so its complement is open. So the complement can be written as a countable union of disjoint open intervals. Why can we not just enumerate all endpoints of the countably many intervals, and conclude the Cantor set is countable?