You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: |
    Let $C$ be the middle-third Cantor set. Define $C*C=\{x*y:x,y\in C\}$, where $*=+,-,\cdot,\div$ (when $*=\div$, we assume $y\neq0$). Steinhaus [@HS] proved in 1917 that $$C-C=[-1,1], C+C=[0,2].$$ In 2019, Athreya, Reznick and Tyson [@Tyson] proved that $$C\div C=\bigcup_{n=-\infty}^{\infty}\left[ 3^{-n}\dfrac{2}{3},3^{-n}\dfrac
    {3}{2}\right] .$$ In this paper, we give a description of the topological structure and Lebesgue measure of $C\cdot C$. We indeed obtain corresponding results on the uniform $\lambda$-Cantor sets.
address:
- 'Department of Mathematics, Ningbo University, 315211, Ningbo, P. R. China'
- 'Department of Mathematics, Ningbo University, 315211, Ningbo, P. R. China'
- 'Department of Mathematics, Ningbo University, 315211, Ningbo, P. R. China; Key Laboratory of Computing and Stochastic Mathematics (Ministry of Education), School of Mathematics and Statistics, Hunan Normal University, Changsha, Hunan 410081, P. R. China'
- 'Department of Mathematics, Ningbo University, 315211, Ningbo, P. R. China'
author:
- jiangwen Gu
- kan Jiang
- lifeng Xi
- bing Zhao
title: 'Multiplication on uniform $\lambda$-Cantor sets'
---

[^1]

Introduction
============

The middle-third Cantor set, denoted by $C$, is a celebrated set in fractal geometry. Many aspects of this set were analyzed [@FG; @FW2000; @Hutchinson]. In this paper, we shall investigate the arithmetic on the middle-third Cantor set. Arithmetic on the fractal sets has some connections to the geometric measure theory, dynamical systems, and number theory, see [@Yoccoz; @BABA1; @KarmaCor1; @Bond; @KCor; @KM; @DajaniDeVrie; @KarmaOomen; @Hochman2012; @JM; @MO; @Kan2019; @Palis; @PS] and references therein. However, relatively few papers investigate the structure of the fractal sets under some arithmetic operations. The theme of this paper is to explore this problem.

Let $C*C=\{x*y:x,y\in C\}$, where $*=+,-,\cdot,\div$ (when $*=\div$, we assume $y\neq0$). Arithmetic on the middle-third Cantor set was pioneered by Steinhaus [@HS] who proved that $$C+C=[0,2],C-C=[-1,1].$$ It is natural to ask the multiplication and division on $C$. For this question, recently, Athreya, Reznick and Tyson [@Tyson] proved that $17/21\leq\mathcal{L}(C\cdot C)\leq8/9$, where $\mathcal{L}$ denotes the Lebesgue measure. Moreover, they also proved that $$C\div C=\bigcup_{n=-\infty}^{\infty}\left[ 3^{-n}\dfrac{2}{3},3^{-n}\dfrac
{3}{2}\right] .$$ Let $f$ be a continuous function defined on an open set $U\subset
\mathbb{R}^{2}$. Denote the continuous image by $$f_{U}(C,C)=\{f(x,y):(x,y)\in(C\times C)\cap U\}.$$ Jiang and Xi [@JX2018] proved that if $\partial_{x}f$, $\partial_{y}f$ are continuous on $U,$ and there is a point $(x_{0},y_{0})\in(C\times C)\cap U$ such that one of the following conditions is satisfied, $$1<\left\vert \frac{\partial_{y}f|_{(x_{0},y_{0})}}{\partial_{x}f|_{(x_{0},y_{0})}}\right\vert <3 \mbox{ or } 1<\left\vert \frac{\partial_{x}f|_{(x_{0},y_{0})}}{\partial_{y}f|_{(x_{0},y_{0})}}\right\vert <3,$$ then $f_{U}(C,C)$ has a non-empty interior. In particular, $C\cdot C$ contains infinitely many intervals. To date, to the best of our knowledge, there are few results for the structure of $C\cdot C$. This is one of the main motivations of this paper. We shall give an answer to this question. Let $K_{\lambda}$ be the uniform $\lambda$-Cantor set generated by the IFS $$\{ f_{i}(x)=\lambda x+i\delta\}_{i=0}^{m-1},$$ where $m\geq2, 0<\lambda<1/m,g=\dfrac{1-m\lambda}{m-1},$ and $\delta
=\lambda+g$. Suppose $\sigma=i_{1}i_{2}\cdots i_{n}\in\{0,1,\cdots m-1\}^{n} $ with its length $|\sigma|=n\geq0$. If $n=0$, then $\sigma=\emptyset$. In this case, $f_{\sigma}([0,1])=[0,1].$ For any $I=f_{\sigma}([0,1])$, let $I^{(i)}=f_{\sigma i}([0,1]), 0\leq i\leq m-1$. Denote $$I^{(i)}\cdot I^{(j)}=\{xy:x\in I^{(i)}, y\in I^{(j)}\}.$$ Define $$\widehat{I}=\left\lbrace
\begin{array}
[c]{cc}\cup_{0\leq p<q}\left( I^{(p)}\cdot I^{(q)}\right)  & \mbox{ if } |\sigma
|\geq1\\
\cup_{1\leq p<q}\left( I^{(p)}\cdot I^{(q)}\right)  & \mbox{ if }
\sigma=\emptyset.
\end{array}
\right.$$ We denote $$\mathcal{A}=\{f_{\sigma}([0,1]) : \sigma=i_{1}i_{2}\cdots i_{n}, i_{1}\geq1\}\cup\{[0,1]\}.$$ Let $$K_{R}=\cup_{i=1}^{m-1}f_{i}(K_{\lambda}).$$ Now we state the first result of this paper.

\[main1\] Let $K_{\lambda}$ be the uniform $\lambda$-Cantor set. If $1/(m+1)\leq\lambda<1/m$, then $$K_{\lambda}\cdot K_{\lambda}= \left( \bigcup_{n=0}^{\infty}\lambda^{n}
(K_{R}\cdot K_{R})\right) \bigcup\left\{ 0\right\} ,$$ where $$K_{R}\cdot K_{R}=\left( \bigcup_{I\in\mathcal{A}}\widehat{I}\right)
\bigcup\{x^{2}:x\in K_{R}\}.$$

When $\lambda=1/3, m=2$, we give the structure of $C\cdot C$. With the help of computer program, we are able to calculate the value of the Lebesgue measure of $C\cdot C$, which is about $0.80955.$

The main ingredient of $K_{R}\cdot K_{R}$ is $\bigcup_{I\in\mathcal{A}}\widehat{I}$ as the Lebesgue measure of $\{x^{2}:x\in K_{R}\}$ is zero. We shall use this fact in the proof of next result. It is natural to consider that whether some points in $\{x^{2}:x\in K_{R}\}$ can be absorbed by $\bigcup_{I\in\mathcal{A}}\widehat{I}$. We may prove, for instance, that for $m=2$, if $0.44<\lambda<1/2$, then $$K_{R}\cdot K_{R}\setminus\bigcup_{I\in\mathcal{A}}\widehat{I}=\{(1-\lambda
)^{2},1\}.$$ Equivalently, in this case $$\bigcup_{I\in\mathcal{A}}\widehat{I}=((1-\lambda)^{2},1).$$

The next result is to consider the function $\Phi(\lambda)=\mathcal{L}(K_{\lambda}\cdot K_{\lambda})$. In terms of some basic results in analysis, we prove the following result.

\[main2\] If $\lambda\in[1/(m+1),1/m),$ then the function $\mathcal{L}(K_{\lambda}\cdot K_{\lambda})$ is continuous.

This paper is arranged as follows. In section 2, we give the proofs of main theorems. In section 3, we give some remarks and problems.

Proof of Main results
=====================

In this section, we shall give the proofs of the main theorems.

Proof of Theorem \[main1\]
--------------------------

Let $E=[0,1]$. For any $(i_{1}\cdots i_{n})\in\{0,1,\cdots m-1\}^{n}$, we call $f_{i_{1}\cdots i_{n}}(E)$ a basic interval with length $\lambda^{n}$. Denote by $E_{n}$ ($n\geq1$) the collection of all the basic intervals with length $\lambda^{n}$. Let $J=f_{\sigma}([0,1])\in E_{n}$ be a basic interval for some $\sigma\in\{0,1,\cdots, m-1\}^n$. Denote $\widetilde{J}=\cup_{i=0}^{m-1}f_{\sigma i}([0,1])$. Let $[A,B]\subset[0,1]$, where $A$ and $B$ are the left and right endpoints of some basic intervals in $E_{k}$ for some $k\geq1$, respectively. $A$ and $B$ may not be in the same basic interval. Let $F_{k}$ be the collection of all the basic intervals in $[A,B]$ with length $\lambda^{k}, k\geq k_{0}$ for some $k_{0}\in
\mathbb{N}^{+}$, i.e. the union of all the elements of $F_{k}$ is denoted by $G_{k}=\cup_{i=1}^{t_{k}}I_{k,i}$, where $t_{k}\in\mathbb{N}^{+}$, $I_{k,i}\in
E_{k}$ and $I_{k,i}\subset[A,B]$. Clearly, by the definition of $G_{n}$, it follows that $G_{n+1}\subset G_{n}$ for any $n\geq k_{0}.$ Similarly, let $M$ and $N$ be the left and right endpoints of some basic intervals in $E_{k}$ for some $k\geq k_0$. Denote by $G_{k}^{\prime}$ the union of all the basic intervals with length $\lambda^k$ such that all of these basic intervals are subsets of $[M,N]$, i.e. $G_{k}^{\prime}=\cup_{i=1}^{t_{k}^{\prime}}I_{k,i}^{\prime}$, where $t_{k}^{\prime}\in\mathbb{N}^{+}$, $I_{k,i}^{\prime}\in
E_{k}$ and $I_{k,i}^{\prime}\subset[M,N]$. Now, we state a key lemma of our paper.

\[key1\] Assume $F:\mathbb{R}^{2}\to\mathbb{R}$ is a continuous function. Suppose $A$ and $B$ ($M$ and $N$) are the left and right endpoints of some basic intervals in $E_{k_{0}}$ for some $k_{0}\geq1$, respectively. Then $K_{\lambda}\cap[A,B]=\cap_{n={k_{0}}}^{\infty}G_{n}$ ($K_{\lambda}\cap[M,N]=\cap_{n={k_{0}}}^{\infty}G_{n}^{\prime}$). Moreover, if for any $n\geq k_{0}$ and any basic intervals $I_{1}\subset G_{n}$, $ I_{2}\subset G_{n}^{\prime}$ $$F(I_{1}, I_{2})=F(\widetilde{I_{1}}, \widetilde{I_{2}}),$$ then $F(K_{\lambda}\cap[A,B],K_{\lambda}\cap[M,N] )=F(G_{k_{0}}, G_{k_{0}}^{\prime}).$

Let $G_{n}=\cup_{i=1}^{t_{n}}I_{n,i}$ for some $t_{n}\in\mathbb{N}^{+}$, where $I_{n,i}\in E_{n}$ and $I_{n,i}\subset[A,B]$. Then by the construction of $G_{n}$, i.e. $G_{n+1}\subset G_{n}$ for any $n\geq k_{0}$, it follows that $$K_{\lambda}\cap[A,B]=\cap_{n=k_{0}}^{\infty}G_{n}.$$ Analogously, we can prove that $$K_{\lambda}\cap[M,N]=\cap_{n=k_{0}}^{\infty}G_{n}^{\prime}.$$ By the continuity of $F$, we conclude that $$\begin{aligned}
\label{identity}F(K_{\lambda}\cap[A,B],K_{\lambda}\cap[M,N])=\cap_{n=k_{0}}^{\infty}F(G_{n}, G_{n}^{\prime}).\end{aligned}$$ By virtue of the relation $G_{n+1}=\widetilde{G_{n}}$ ($G_{n+1}^{\prime}=\widetilde{G_{n}^{\prime}}$) and the condition in the lemma, we have $$\begin{aligned}
F(G_{n}, G_{n}^{\prime}) & = \cup_{1\leq i\leq t_{n},1\leq j\leq t_{n}^{\prime}}F(I_{n,i}, I_{n,j}^{\prime})\\
& = \cup_{1\leq i\leq t_{n},1\leq j\leq t_{n}^{\prime}}F(\widetilde{I_{n,i}},\widetilde{I_{n,j}^{\prime}})\\
& = F(\cup_{1\leq i\leq t_{n}}\widetilde{I_{n,i}},\cup_{1\leq j\leq t_{n}^{\prime}}\widetilde{I_{n,j}^{\prime}})\\
& =F(G_{n+1}, G_{n+1}^{\prime}).\end{aligned}$$ Therefore, $F(K_{\lambda}\cap[A,B],K_{\lambda}\cap[M,N] )=F(G_{k_{0}},
G_{k_{0}}^{\prime})$ follows immediately from identity (\[identity\]) and $F(G_{n},
G_{n}^{\prime})=F(G_{n+1}, G_{n+1}^{\prime})$ for any $n\geq k_{0}.$

Recall that $\delta=\lambda+g$ and $g=\frac{1-m\lambda}{m-1}$ with $m\geq2.$

\[pro1\] Let $\lambda\in\lbrack\frac{1}{m+1},\frac{1}{m})$. Suppose $I\in\mathcal{A}$ and $I^{(0)},\cdots,I^{(m-1)}$ are basic subintervals of $I$ from left to right such that $|I^{(i)}|/|I|=\lambda$ for all $0\leq i\leq
m-1.$ Then for any $p<q,$ we have $$(I^{(p)}\cap K_{\lambda})\cdot(I^{(q)}\cap K_{\lambda})=I^{(p)}\cdot I^{(q)}.$$

Given basic subintervals $I^{\ast}\subset I^{(p)}$ and $J^{\ast}\subset
I^{(q)}$ with $I^{\ast}=[a,a+t]$ and $J^{\ast}=[b,b+t]$ ($b\geq a$), we suppose that $A^{(0)},\cdots,A^{(m-1)}$ $(B^{(0)},\cdots,B^{(m-1)})$ are basic subintervals of $I^{\ast}$ ($J^{\ast}$ respectively) from left to right.

For rectangles $R_{1}=[a_{1},b_{1}]\times\lbrack a_{2},b_{2}]$ and $R_{2}=[c_{1},d_{1}]\times\lbrack c_{2},d_{2}]$ with $a_{1},a_{2},c_{1},c_{2}\geq0,$ we denote by $R_{1}\rightarrow R_{2}$ if $b_{1}b_{2}\geq
c_{1}c_{2}.$ By Lemma \[key1\], it suffices to verify that $$\begin{aligned}
& \ \ \ \ \ (A^{(0)}\times B^{(0)})\\
& \rightarrow(A^{(0)}\times B^{(1)})\rightarrow(A^{(1)}\times B^{(0)})\\
& \rightarrow(A^{(0)}\times B^{(2)})\rightarrow(A^{(1)}\times B^{(1)})\rightarrow(A^{(2)}\times B^{(0)})\\
& \rightarrow(A^{(0)}\times B^{(3)})\rightarrow(A^{(1)}\times B^{(2)})\rightarrow(A^{(2)}\times B^{(1)})\rightarrow(A^{(3)}\times B^{(0)})\\
& \cdots\\
& \rightarrow(A^{(0)}\times B^{(m-1)})\rightarrow(A^{(1)}\times B^{(m-2)})\rightarrow\cdots\rightarrow(A^{(m-2)}\times B^{(1)})\rightarrow
(A^{(m-1)}\times B^{(0)})\\
& \rightarrow(A^{(1)}\times B^{(m-1)})\rightarrow(A^{(2)}\times B^{(m-2)})\rightarrow\cdots\rightarrow(A^{(m-1)}\times B^{(1)})\\
& \cdots\\
& \rightarrow(A^{(m-1)}\times B^{(m-1)}).\end{aligned}$$ For example, please see Fig. 1 when $m=5.$

![$m=5$ and $(i,j)$ represents rectangle $A^{(i)}\times B^{(j)}$](1.eps){width="34.00000%"}

In fact, we need to verify $$(A^{(i)}\times B^{(j)})\rightarrow(A^{(i+1)}\times B^{(j-1)}),\label{xi1}$$ and $$(A^{(k)}\times B^{(0)})\rightarrow(A^{(0)}\times B^{(k+1)})\mbox{ and }(A^{(m-1)}\times B^{(i)})\rightarrow(A^{(i+1)}\times
B^{(m-1)}).\label{xi2}$$

**(1)** For inequality (\[xi1\]), let $(a,b)+(i,j)\delta t=(x,y),$ then $(a,b)+(i+1,j-1)\delta t=(x+\delta t,y-\delta t).$ We only need to show$$g(x,y)=(x+\lambda t)(y+\lambda t)-(x+\delta t)(y-\delta t)\geq0.\label{xi3}$$ Note that $x\geq a$, $y\leq b+(1-\lambda)t$ and$$\frac{\partial g(x,y)}{\partial x}=\lambda t+\delta t>0\mbox{ and }\frac
{\partial g(x,y)}{\partial y}=-gt<0.$$ We obtain that $$\begin{aligned}
g(x,y)  & \geq g(a,b+(1-\lambda)t)\\
& =t\left(  a\lambda+a\delta+t\lambda^{2}+t\delta^{2}-bg-gt+gt\lambda\right)
=t\cdot u(\lambda,t).\end{aligned}$$ Since $a\geq\delta$ and $b\leq1$ we have$$\begin{aligned}
u(\lambda,t)  & \geq\delta\lambda+\delta^{2}+t\lambda^{2}+t\delta
^{2}-g-gt+gt\lambda\\
& =t\left(  \lambda^{2}+g\lambda+\delta^{2}-g\right)  +(\delta^{2}+\lambda\delta-g).\end{aligned}$$ By $\delta=\lambda+g$ and $g=\frac{1-m\lambda}{m-1},$ one can get$$\begin{aligned}
& \lambda^{2}+g\lambda+\delta^{2}-g\\
& =\frac{m^{2}\lambda-m\lambda^{2}-m+2\lambda^{2}-3\lambda+2}{\left(
m-1\right)  ^{2}}=\frac{h(\lambda)}{\left(  m-1\right)  ^{2}}.\end{aligned}$$ Note that $\lambda<\frac{1}{m}\leq1/2,$ then$$h^{\prime}(\lambda)=(m^{2}-3)-2\lambda\left(  m-2\right)  \geq m(m-1)-1\geq0.$$ Hence for any $\lambda\geq\frac{1}{m+1}$ and any $m\geq2,$ $$\frac{h(\lambda)}{\left(  m-1\right)  ^{2}}\geq\frac{h(\frac{1}{m+1})}{\left(
m-1\right)  ^{2}}=\frac{1}{\left(  m^{2}-1\right)  ^{2}}\left(  m^{2}-m+1\right)  >0.$$ On the other hand, we obtain$$\begin{aligned}
\delta^{2}+\lambda\delta-g  & =\frac{1}{\left(  m-1\right)  ^{2}}\left(
\left(  2-m\right)  \lambda^{2}+\left(  m^{2}-3\right)  \lambda+\left(
2-m\right)  \right) \\
& =\frac{1}{\left(  m-1\right)  ^{2}}v(\lambda).\end{aligned}$$ If $m=2$, then $\delta^{2}+\lambda\delta-g \geq 0$. If $m=3$, then we note that the symmetric axis of the graph $w=v(\lambda)$ is $$\lambda=\frac{m^{2}-3}{2(m-2)}\geq\frac{1}{2}\frac{m^{2}-4}{m-2}=\frac{m+2}{2}\geq2.$$ which implies$$\begin{aligned}
\inf_{\lambda\in\lbrack\frac{1}{m+1},\frac{1}{m})}v(\lambda)  & \geq
\min\left(  v(\frac{1}{m}),v(\frac{1}{m+1})\right) \\
& =\min\left(  \frac{2\left(  m-1\right)  ^{2}}{m^{2}},\frac{m^{2}-m+1}{\left(  m+1\right)  ^{2}}\right)  >0.\end{aligned}$$ Then (\[xi3\]) follows.

**(2)** For inequality (\[xi2\]), we will need the following

\[claim00\]If $x_{1}+y_{1}\geq x_{2}+y_{2}$ with $x_{1}<y_{1},$ $x_{2}<y_{2}$ and $x_{1}>x_{2},$ then we have $x_{1}y_{1}\geq x_{2}y_{2}.$

In fact, let $H(x)=x(c-x)$ with $c=x_{1}+y_{1},$ then we have$$x_{2}y_{2}\leq x_{2}(y_{2}+(x_{1}+y_{1})-(x_{2}+y_{2}))=H(x_{2})<H(x_{1})$$ since $x_{2}<x_{1}<c/2.$

Let $(x_{1},y_{1})$ be the upper right corner of the rectangle $A^{(k)}\times
B^{(0)},$ and $(x_{2},y_{2})$ the lower left corner of $A^{(0)}\times
B^{(k+1)},$ then $$(x_{1}+y_{1})-(x_{2}+y_{2})=2\lambda t-\delta t=t\frac{(2m-1)\lambda-1}{m-1}\geq0$$ due to $\lambda\geq\frac{1}{m+1}\geq\frac{1}{2m-1}.$ Using Claim \[claim00\], we have $x_{1}y_{1}\geq x_{2}y_{2},$ i.e., $$(A^{(k)}\times B^{(0)})\rightarrow(A^{(0)}\times B^{(k+1)}).$$ In the same way, we have $(A^{(m-1)}\times B^{(i)})\rightarrow(A^{(i+1)}\times
B^{(m-1)}).$

The following lemma is from the fact $$K_{\lambda}= \{0\}\cup\left( \bigcup_{n=0}^{\infty}\lambda^{n} ( K_{R})\right)
.$$

\[translate\] $$K_{\lambda}\cdot K_{\lambda}= \{0\}\cup\left( \bigcup_{n=0}^{\infty}\lambda^{n} (K_{R}\cdot K_{R})\right) .$$

\[**Proof of Theorem \[main1\]**\]$\ $ By Lemma \[translate\], it remains to prove $$K_{R}\cdot K_{R}=\left( \cup_{I\in\mathcal{A}}\widehat{I}\right) \cup
\{x^{2}:x\in K_{R}\}.$$ By Proposition \[pro1\], it follows that $$K_{R}\cdot K_{R}\supset\left( \cup_{I\in\mathcal{A}}\widehat{I}\right)
\cup\{x^{2}:x\in K_{R}\}.$$ Conversely, for any $x, y\in K_{R}$, if $x=y$, then $xy\in\{x^{2}:x\in
K_{R}\}$. If $x\neq y$, then there exists some basic interval, denoted by $I$, such that $x\in I^{(p)}, y\in I^{(q)}$ for $p\neq q$. Therefore, by Proposition \[pro1\], we have that $x\cdot y\in\cup_{I\in\mathcal{A}}\widehat{I}.$

Proof of Theorem \[main2\]
--------------------------

\[c1\] Let $\lambda\in\left[  1/(m+1),1/m\right)  $. For any $n\geq1$, we have $$\mathcal{L}\left(  (K_{R}\cdot K_{R})\backslash\bigcup\nolimits_{I\in
\mathcal{A},\text{\emph{rank}}(I)\leq k}\widehat{I}\right)  \leq3
\dfrac{(m\lambda)^{k+1}}{1-m\lambda}.$$

First, note that for any basic interval $I=[a,a+t]$, by the definition of $\widehat{I}$, we have $$\mathcal{L}(\widehat{I})\leq(a+t)^{2}-a^{2}=t(2a+t)\leq3t=3\mathcal{L}(I).$$ Now we have the following estimation: $$\begin{aligned}
& \mathcal{L}\left( K_{R}\cdot K_{R}\setminus\left( \bigcup\nolimits_{I\in
\mathcal{A}\text{,\textrm{rank}}(I)\leq k}\widehat{I}\right) \right) \\
& \leq\mathcal{L}\left( \bigcup\nolimits_{I\in\mathcal{A}\text{,\textrm{rank}}(I)>k}\widehat{I}\right) \\
& \leq \sum_{I\in\mathcal{A}\text{,\textrm{rank}}(I)>k}3(m\lambda)^{k+1}\\
& =3\dfrac{(m\lambda)^{k+1}}{1-m\lambda}.\end{aligned}$$

Let $$\phi_{n}(\lambda)=\mathcal{L}\left(  \bigcup\nolimits_{k=0}^{n}\lambda
^{k}\left( \bigcup\nolimits_{I\in\mathcal{A}\text{,\textrm{rank}}(I)\leq
n}\widehat{I}\right) \right)  ,$$ which is a continuous function for $\lambda\in\lbrack1/(m+1),1/m).$ Lemma \[c1\] implies the following result.

\[uni\] \[c3\] If $\alpha>0$, then $\phi_{n}(\lambda)$ uniformly converges to $\Phi(\lambda)=\mathcal{L}(K_{\lambda}\cdot K_{\lambda})$ for any $\lambda\in\lbrack1/(m+1),1/m-\alpha).$

Note that $$\Phi(\lambda)=\mathcal{L}\left(  \bigcup\nolimits_{k=0}^{\infty}\lambda
^{k}(K_{R}\cdot K_{R})\right)  .$$ It follows from Lemma \[c1\] that $$\begin{aligned}
& |\Phi(\lambda)-\phi_{n}(\lambda)|\\
& \leq\left\vert \Phi(\lambda)-\mathcal{L}\left(  \bigcup\nolimits_{k=0}^{n}\lambda^{k}(K_{R}\cdot K_{R})\right)  \right\vert +\left\vert
\mathcal{L}\left(  \bigcup\nolimits_{k=0}^{n}\lambda^{k}(K_{R}\cdot
K_{R})\right)  -\phi_{n}(\lambda)\right\vert \\
& \leq\mathcal{L}\left(  \bigcup\nolimits_{k=n+1}^{\infty}\lambda^{k}(K_{R}\cdot K_{R})\right)  +\sum_{k=0}^{n}\lambda^{k}3 \dfrac{(m\lambda
)^{n+1}}{1-m\lambda}\\
& \leq\sum\limits_{k=n+1}^{\infty}\lambda^{k}+3\dfrac{1-\lambda^{n+1}}{1-\lambda} \dfrac{(1-m\alpha)^{n+1}}{1-m\lambda}\leq\frac{m ^{-n-1}}{1-\lambda}+\dfrac{3}{1-\lambda} \dfrac{(1-m\alpha)^{n+1}}{1-m\lambda}.\end{aligned}$$

\[**Proof of Theorem \[main2\]**\]$\ $

By the continuity of $\phi_{n}(\lambda)$ and Lemma \[uni\], it follows that $\Phi(\lambda)$ is continuous.

Final remarks
=============

From the graph of function $\mathcal{L}(K_{\lambda}\cdot K_{\lambda})$ (Fig. 2), we pose a natural

The function $\mathcal{L}(K_{\lambda}\cdot K_{\lambda})$ is an increasing function for any $\lambda\in[1/(m+1),1/m).$

![The graph of $\mathcal{L}(K_{\lambda}\cdot K_{\lambda})$ with $m=2$](2.eps){width="50.00000%"}

[99]{} Jayadev S. Athreya, Bruce Reznick, and Jeremy T.Tyson. Cantor set arithmetic. , 126(1):4–17, 2019.

Carlos Gustavo T. de A. Moreira and Jean-Christophe Yoccoz. Stable intersections of regular [C]{}antor sets with large [H]{}ausdorff dimensions. , 154(1):45–96, 2001.

Bárány, Balázs On some non-linear projections of self-similar sets in [$\mathbb{R}^{3}$]{}. , 237(1):83–100, 2017.

Jose Barrionuevo, Robert M. Burton, Karma Dajani, and Cor Kraaikamp. Ergodic properties of generalized [L]{}üroth series. , 74(4):311–327, 1996.

M. Bond, I. [Ł]{}aba, and J. Zahl. Quantitative visibility estimates for unrectifiable sets in the plane. , 368(8):5475–5513, 2016.

Karma Dajani and Cor Kraaikamp. , volume 29 of *Carus Mathematical Monographs*. Mathematical Association of America, Washington, DC, 2002.

Karma Dajani and Martijn de Vries. Measures of maximal entropy for random [$\beta$]{}-expansions. , 7(1):51–68, 2005.

Karma Dajani and Martijn de Vries. Invariant densities for random [$\beta$]{}-expansions. , 9(1):157–176, 2007.

Karma Dajani and Margriet Oomen. Random [$N$]{}-continued fraction expansions. , 227:1–26, 2018.

Kenneth Falconer. . John Wiley & Sons, Ltd., Chichester, 1990. Mathematical foundations and applications.

De-Jun Feng, Su Hua, and Zhi-Ying Wen. The pointwise densities of the [C]{}antor measure. , 250(2):692–705, 2000.

Michael Hochman and Pablo Shmerkin. Local entropy averages and projections of fractal measures. , 175(3):1001–1059, 2012.

John E. Hutchinson. Fractals and self-similarity. , 30(5):713–747, 1981.

Kan Jiang and Lifeng Xi. Interiors of continuous images of the middle-third cantor set. , 2018.

Kan Jiang, Xiaomin Ren, Jiali Zhu, and Li Tian. Multiple representations of real numbers on self-similar sets with overlaps. , 27(4):1950051, 17, 2019.

Pedro Mendes and Fernando Oliveira. On the topological structure of the arithmetic sum of two [C]{}antor sets. *Nonlinearity*, 7(2):329–343, 1994.

Hugo Steinhaus. Mowa W[ł]{}asność Mnogości Cantora. 1985.

Li Tian, Jiangwen Gu, Qianqian Ye, Lifeng Xi, and Kan Jiang. Multiplication on self-similar sets with overlaps. , 478(2):357–367, 2019.

Jacob Palis and Floris Takens. , volume 35 of *Cambridge Studies in Advanced Mathematics*. Cambridge University Press, Cambridge, 1993. Fractal dimensions and infinitely many attractors.

Yuval Peres and Pablo Shmerkin. Resonance between [C]{}antor sets. , 29(1):201–221, 2009.

[^1]: Kan Jiang is the corresponding author. Supported by National Natural Science Foundation of China (Nos. 11701302, 11831007, 11771226, 11371329, 11301346, 11671147) and K.C. Wong Magna Fund in Ningbo University.