Let $A$ and $B$ be two subsets of $\mathbb{R}^d$. We say that $A$ can be affinely embedded into $B$ if there is an affine map $f$ on $\mathbb{R}^d$ of the form $f(x)=Mx+t$ such that $f(E)\subset F$, where $M$ is an invertible $d\times d$ matrix and $t\in \mathbb{R}^d$.

Let $\Phi=\{\varphi_i\}_{1\leq i\leq k}$ be a family of contractive mappings on $\mathbb{R}^d$, we say $\Phi$ is an iterated function system (IFS in short) on $\mathbb{R}^d$. There is a unique non-empty compact set $E\subset \mathbb{R}^d$, such that

$E=\bigcup_{i=1}^k \varphi_i(E).$

We call $E$ the attractor of the IFS $\Phi$. See [2, 6].

It is seen that the attractor $E$ is a singleton if and only if the contractive mappings $\varphi_{i}(1\leq i\leq k)$ have the same fixed point. In this paper, we always suppose the attractor $E$ is not a singleton.

We say that $\Phi$ satisfies the strong separation condition (SSC in short) if $\varphi_i(E)(1\leq i\leq k)$ are pairwise disjoint subsets of $E$.

A contractive mapping $\varphi$ on $\mathbb{R}^d$ is called a similitude if $\varphi$ is of the form $\varphi(x)=\rho A(x)+a$, where $0<\rho<1$ is the contractive ratio of $\varphi$, $A$ is an $d\times d$ orthogonal matrix and $a\in \mathbb{R}^d$. If all maps of an IFS $\Phi$ are similitudes, the attractor $E$ of $\Phi$ is called a self-similar set.

Suppose that $\Phi=\{\varphi_i\}_{1\leq i\leq k}$ and $\Psi=\{\psi_j\}_{1\leq j\leq l}$ are two families of contractive similitudes of $\mathbb{R}^d$ with the form

$\begin{equation} \varphi_i(x)=\rho_{i}A_i(x)+a_i, \ \ \ \ \psi_j(x)=\lambda_j B_j (x)+b_j, \ \ \ 1\leq i\leq k, \ 1\leq j\leq l, \end{equation}$

(1.1)

where $0<\rho_i,\lambda_j<1, a_i,b_j\in \mathbb{R}^d$ and $A_i,B_j$ are orthogonal $d\times d$ matrixes. Let $E$ and $F$ be the attractors of $\Phi$ and $\Psi$, respectively.

In this paper, we discuss when the self-similar set $E$ can be affinely embedded into $F$. It was conjectured in Feng, Huang and Rao [4] that if $E$ and $F$ are totally disconnected and $E$ can be affinely embedded into $F$, then the contractive ratios $\rho_i,\lambda_j$ should satisfy the following arithmetic conditions.

Conjecture 1.1 Suppose $E$ and $F$ are totally disconnected and $E$ can be affinely embedded into $F$. Then for each $1\leq i\leq k$, there exists non-negative $t_{i,j}\in \mathbb{Q}$ such that $\rho_i=\prod_{j=1}^{l}\lambda_j^{t_{i,j}}$. In particular, if $\lambda_j=\lambda(1\leq j\leq l)$, then $\log\rho_i/\log \lambda \in \mathbb{Q}$ for $1\leq i\leq k$.

Falconer and Marsh [3] proved the arithmetic conditions when $E,F$ satisfy the SSC and are Lipschitz equivalent. Deng, Wen, Xiong and Xi [1] showed that if $E,F$ satisfy the SSC and $\dim_H E<\dim_H F$, then $E$ can be Lipschitz embedded into $F$.

Let $E$ and $F$ be central Cantor sets in $\mathbb{R}$. That is, $E=C_\rho$ and $F=C_\lambda$ are the attractors of the IFS $\{\rho x, \rho x+1-\rho\}$ and $\{\lambda x, \lambda x+1-\lambda\}$, respectively. Feng, Huang and Rao [4] proved the following result.

Theorem 1.2 If $0< \rho<\lambda< \sqrt{2}-1$, and $C_\rho$ can be affinely embedded into $C_\lambda$, then $\log\rho/\log \lambda$ is a rational number.

In this paper, we partially generalize the above Theorem 1.2 to the case of multi-branches when $d=1$. Let $E,F$ be attractors of the IFS defined by (1.1). It seen that $A_i,B_j$ equal to $1$ or $-1$ for each $1\leq i\leq k$ and $1\leq j\leq l$. Without loss of generality, we always assume that $B_j=1$ for $1\leq j\leq l$ and

$\begin{equation}\label{A-1} 0=b_1<b_2<\cdots<b_l\leq 1-\lambda. \end{equation}$

(1.2)

Theorem 1.3  Let $E,F$ be attractors of the IFS defined by (1.1) with $\lambda_j=\lambda$ for all $1\leq j\leq l$. If

$\begin{equation}\label{A-2} b_j-b_{j-1}\geq 2\lambda, \quad 2\leq j\leq l, \end{equation}$

(1.3)

and $E$ can be affinely embedded into $F$, then $\log\rho_i/\log \lambda$ is a rational number for $1\leq i\leq k$.

Let $\varphi_i=\varphi$ be a mapping in the IFS $\Phi$, denote its contraction ratio $\rho_i$ by $\rho$. We shall show $\log \rho/\log \lambda\in {\mathbb Q}$.

Without loss of generality, we assume that $A_i=1$ in (1.1). In the case of $A_i=-1$, we just replace $-\rho$ by $\rho$.

Let $a$ be the fixed point of $\varphi$, i.e., $\varphi(a)=a$. Then it is easy to see that $\varphi$ has the form $\varphi(x)=\rho(x-a)+a$. Since $E$ is not a singleton, there exists $b\in E$ with $b\neq a$. By induction, it is easy to show that

$ \varphi^n(b)=\rho^n(b-a)+a, \quad n\geq 1, $

where $\varphi^n$ stands for the $n$-th iteration of $\varphi$. Since $\varphi^n(b)\in \varphi^n(E)\subset E\subset F$, we infer that $\rho^n(b-a)+a \in F$ for all $n\geq 1$. Especially

$\begin{equation}\label{eq-5} \rho^{2n}(b-a)+a \in F, \quad \forall n\geq 1. \end{equation}$

(2.1)

Suppose on the contrary that $\log \rho/\log \lambda$ is irrational. Then by the classical Kronecker Theorem the set

$ \left \{m-n \frac{2\log \rho}{\log \lambda}: m,n\in {\mathbb N}\right \} $

is dense in ${\mathbb R}$ [5]. We choose $\varepsilon$ such that $0<\varepsilon<\rho^2$. Assume $b-a>0$ in what follows. The case $b-a<0$ can be treated exactly in the same way.

By the above dense property, there exist $m,n\in {\mathbb N}$ such that

$ \frac{\log (b-a)}{\log \lambda}<m-2n \frac{\log\rho}{\log \lambda}<\frac{\log (b-a)}{\log \lambda}-\frac{\log(1+\varepsilon)}{\log \lambda}, $

in other words, $1<\rho^{2n}(b-a)\lambda^{-m}<1+\varepsilon$, or

$\begin{equation}\label{eq-10} \lambda^m<\rho^{2n}(b-a)<\lambda^m(1+\varepsilon). \end{equation}$

(2.2)

It is seen that $F\subset [0,1]$ by (1.2). We call $\psi_{j_1}\cdots \psi_{j_m}([0,1])$ a basic interval of order $m$ w.r.t. $\Psi$. Clearly the length of a basic interval of order $m$ is $\lambda^m$. By our assumption (1.3), the distance between two basic intervals of order $m$ is $\geq \lambda^m$.

Let $c=\rho^{2n}(b-a)+a$. Then $a,c\in F$. $c-a=\rho^{2n}(b-a)>\lambda^m$ implies that $a$ and $c$ belong to different basic intervals of order $m$. Let us denote the basic intervals of order $m$ containing $a$ and $c$ by $I$ and $J$, respectively.

Let us denote by $x$ the distance between $a$ and the right end point of $I$. Then

$c-a\geq x+\lambda^m, $

and by (2.2), we have

$x\leq c-a-\lambda^m\leq \rho^{2n}(b-a)-\lambda^m<\varepsilon \lambda^m.$

Let $y$ be the distance between $a$ and the left end point of $J'$, where $J'$ is the basic interval of order $m$ on the right hand side of $I$ and next to $I$. (It is possible that $J=J'$.) Then $y\geq x+\lambda^m$ and so that

$ \frac{x}{y}\leq \frac{x}{x+\lambda^m}<\frac{\varepsilon \lambda^m}{\varepsilon \lambda^m+\lambda^m} <\rho^2. $

Hence there exists $s\geq 1$ such that

$x<\rho^{2n+2s}(b-a)<\lambda^m.$

It follows that $z=\rho^{2n+2s}(b-a)+a$ is between $a+x$ and $a+\lambda^m$, which is fallen into the gap between $I$ and $J'$. This contradicts (2.1) which asserts $z\in F$. This proves the theorem.