# Does the set of convex combination of points in Cantor set contains a non empty open interval?

$$\mathcal{C}$$ 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?

• Do you mean $\mathcal{C}_t=\{(1-t)x+ty : x, y\in \mathcal{C} \}$, with $t\in[0,1]$? May 20 at 17:40
• 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.
– Joe
May 20 at 18:08

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.