$\mathcal{C}$ denote the cantor middle third set.

$$\mathcal{C}_t=\{(1-t)x+ty : x, y\in \mathcal{C} \}$$

$\mathcal{C}_0=\mathcal{C}_1=\mathcal{C}$ and we can prove that that $\mathcal{C}$ contains no non empty open interval.

What can be said for other $t\in [0, 1]$? Does it contains a non empty open interval ?

Can you list some resources where I can find such type of problems?

  • $\begingroup$ Do you mean $\mathcal{C}_t=\{(1-t)x+ty : x, y\in \mathcal{C} \}$, with $t\in[0,1]$? $\endgroup$
    – Martin R
    May 20 at 17:40
  • 1
    $\begingroup$ As written, your definition for $C_t$ has no dependence on the parameter $t$. I think you want to remove the $t \in [0,1]$ from the set-builder notation. $\endgroup$
    – Joe
    May 20 at 18:08

2 Answers 2


While not a complete characterization of all the $C_t$, we may easily see that $C_t$ can contain a non-empty open interval for some values of $t$. Set $t := \frac{1}{2}$. Then we may compute:

$$\begin{align} C_{1/2} & = \{\frac{1}{2}x + \frac{1}{2}y : x,y \in C\} \\ & = \frac{1}{2} \cdot \{x + y : x,y \in C\}\\ & = \frac{1}{2} (C + C) \end{align}$$ It’s easy to see from the “points in $[0,1]$ with ternary expansions consisting of only $0$s and $2$s” definition $C$ that $C + C = [0,2]$.

Therefore $C_{1/2} = [0,1]$.

EDIT: I gave it a little more thought, and we can say quite a bit. Let $C^n$ denote the $n$’th stage of the middle thirds construction of $C$, so that $C = \bigcap_n C^n$. I know this is non-standard notation, but I don’t want it to be confusing with $C_t$.

For $\alpha \in [0,1]$, we may easy see that: $$C_{\alpha} = \bigcap_{n} [\alpha C^n + \beta C^n]$$ Where $\beta = (1 - \alpha)$. Set $X^n := \alpha C^n + \beta C^n$. What does $X^n$ look like as we vary $\alpha$?

When $\alpha \in \{0,1\}$, we get that $X^n = C^n$, and we recover that $C_0 = C_1 = C$.

When $\alpha = \frac{1}{2}$, we get that $X^n = [0,1]$, and we recover that $C_{1/2} = [0,1]$.

What happens for $\alpha \in (0, \frac{1}{2})$? Well, we’ll have that $C^n \subsetneq X^n$. But we’ll also have that $X^{n+1}$ splits every interval in $X^n$. Hence we’ll end up with $C_\alpha$ being totally disconnected. Further, I believe that the measure of $C_t$ will monotonically increase as $t$ moves from $0$ to $\frac{1}{2}$, and then start monotonically decreasing again.

EDIT EDIT: I no longer believe this last part because it contradicts the paper in the other answer.


Can you list some resources where I can find such type of problems?

Maybe this is of interest:

Pawłowicz, Marta. Linear combinations of the classic Cantor set. Tatra Mt. Math. Publ. 56 (2013), 47–60.

From Math Review:

In this paper, linear combinations of classic Cantor sets are studied. The problem goes back to a result by Hugo Steinhaus [in Selected papers, 205–207, PWN, Warsaw, 1985], who proved in 1917 that $C+C=[0,2]$, where $C$ is the classic Cantor set and $C+C=\{c_1+c_2; c_1,c_2∈C\}$. This result was extended and generalized by several authors during the last hundred years. The main result of the present paper is the topological classification of linear combinations of $C$, i.e., sets of the form $aC+bC=\{ac_1+bc_2; c_1,c_2∈C\}$ where $a,b∈R$ are fixed. It is shown that this problem can be reduced to characterization of $C+mC$, where $m∈(0,1)$. This is given by the following theorem.

Theorem 1. $$C+mC=\bigcup_{n=1}^{2^k}[l_k^{(n)} ,r_k^{(n)}+m],$$for all $m∈(0,1)$, where $k$ is such that $ m∈[\frac{1}{3^{k+1}},\frac{1}{3^k})$, $k∈N_0$, where $l_k^{(n)}$ and $r_k^{(n)}$ are the left and right endpoints of the $n$-th component of the $k$-th iteration of the Cantor set.

Reviewed by Ladislav Mišík.


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