$1/4$ is in the Cantor set?

Question

I would like to know if $1/4$ is in the Cantor set, I tried a lot without success, I need some hint or help how to proceed in this case. Maybe there is some tool or trick I can use.

Thanks a lot

Answer 1

$1/4$ is in the Cantor set. It is in the lower third. And it is in the upper third of the lower third. And in the lower third of that, and in the upper third of that, and so on.

The quickest way to see this is that it is exactly $1/4$ of the way from $1/3$ down to $0$, and then use self-similarity and symmetry.

A more plodding way to show it is to look at the series $$ \frac29 + \frac{2}{9^2}+\frac{2}{9^3}+\cdots=\frac14. $$ This shows that the base-$3$ expansion of $1/4$ is $0.02020202\ldots$. Since it has a base-$3$ expansion with only $0$s and $2$s, it is in the Cantor set.

Answer 2

Hint: write $\frac{1}{4}$ in base $3$.

To do this, just observe that

$$\frac{1}{4}=\frac{2}{9-1}=\frac{2}{9}\frac{1}{1-\frac{1}{9}}$$

and write the geometric series which has that limit.

Answer 3

We use mathematical induction to show $\frac{1}{4}$ and $\frac{3}{4}$ are in $C_n,\, \forall n\in \mathbb{N}\cup \{0\}$.

Clearly, they are in $[0,1]$.

Suppose they are in $C_N$. We observe the structure of $[0,\frac{1}{3}]$ after $n+1$ cuts is similar to $[0,1]$ after $n$ cuts. Hence if $x \in C_N$, then $\frac{x}{3} \in C_{N+1}$. Since by assumption, $\frac{3}{4} \in C_{N}$, we have $\frac{1}{4} \in C_{N+1}$. Moreover, since the cut is symmetry on $[0,1]$, we know $\frac{3}{4} \in C_{N+1}$.

Hence by induction, $\frac{1}{4}$ and $\frac{3}{4}$ are in $C_n,\, \forall n\in \mathbb{N}$, we have $\frac{1}{4}$ and $\frac{3}{4}$ are in $C$.

Answer 4

From the construction of the Cantor set using "deleted thirds", one sees that each of the numbers $$\textstyle a_1={1\over3},\ \ a_2={1\over3}-{1\over9},\ \ a_3={1\over3}-{1\over9}+{1\over27}, \ \ \ldots$$ is an element of the Cantor set.

Since the sequence $(a_k)$ is the sequence of partial sums of the convergent Geometric series $\sum\limits_{n=1}^\infty(-1)^{n+1} ({1\over3^n}) $, it follows that
$$\lim_{k\rightarrow\infty} a_k= \sum_{n=1}^\infty(-1)^{n+1} ({\textstyle{1\over3^n})}= -\sum_{n=1}^\infty (\textstyle{-1\over3})^n=(-1)\cdot{-1/3\over 1-(-1/3)} ={1/4}.$$

Now, the Cantor set is closed; and, since the Cantor set contains each $a_k$, it must contain the limit of $(a_k)$. Thus, the Cantor set contains the point $1/4$.

(This argument shouldn't seem so mysterious if you determine where $1/4$ lies in relation to the $a_k$; see the first paragraph of Michael Hardy's answer.)

Answer 5

OP wanted "to know if $\frac14$ is in the Cantor set". Let us note $\mathbb K_3$ the Cantor set.

So, first of all, we have to tell what $\mathbb K_3$ is. As @Mickael Hardy said, "$\frac14$ is in the lower third. And $\frac14$ is in the upper third of the lower third. And in the lower third of that, and in the upper third of that, and so on." It is the well known geometrical description of Cantor set.

I propose then to formalize what Mickael said.

Let then $$C_0=D_0:=[0,1]$$ $$C_1=D_1=[0,\frac13]\cup[\frac23,1]$$

The idea I'm going to try to formalize is that if we enlarge three times from step to step, give or take, the non-integer part of $x$ must belong to $C_1$.

Let then$$D_2:=\{x\in [0,1]|9x-\left\lfloor 9x\right\rfloor\in C_1\}$$ $$C_2:=C_1\cap D_2$$ and, by induction, $$\forall n\ge3, D_n:=\{x\in [0,1]|3^nx-\left\lfloor 3^nx\right\rfloor\in C_1\}$$ $$C_n:=C_{n-1}\cap D_n$$ Finally, classically, we define by$$\boxed{\mathbb K_3:=\bigcap_{n=1}^{+\infty}C_n}$$ It is then easy to use the base-three expansion of $x\in [0,1]$ $$x=\sum_{n=1}^{+\infty}\frac{a_n}{3^n}, a_i\in \left\{ 0,1,2 \right\}$$ (by not allowing degenerate expressions) to show that $$\boxed{x\in \mathbb K_3 \iff \forall n \in \mathbb N^*, a_n\in \{0,2\}}(*)$$

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Then, concerning $\frac14=\sum_{n=1}^{+\infty}\frac{a_n}{3^n}, a_i\in \left\{ 0,1,2 \right\}$, (non degenerate development),

1. $\frac14<\frac13$ and $\frac14=\frac{1\times 2}{4 \times 2}=\frac28$ and $\frac19<\frac18$. Then $\frac29<\frac14<\frac13$. So, $a_1=0$ and $a_2=2$

2. If $n$ is even, $3^n \equiv 1 \mod 4$. Let us write $3^n=4q_n+1$(euclidean division). Then $\frac{3^n}{4}=q_n+\frac14=3^{n-1}a_1+3^{n-2}a_2+...+a_n+\frac{a_{n+1}}{3}+\frac{a_{n+2}}{3^2}+...$ Then $\frac14=\frac{a_{n+1}}{3}+\frac{a_{n+2}}{3^2}+...$ and $a_{n+1}=0 $ and $a_{n+2}=2$

3. Then $$\boxed{\frac14=\frac03+\frac{2}{3^2}+\frac{0}{3^3}+\frac{2}{3^4}+...}$$ $$=(0,0:2:0:2:0:2:...)_{three}$$

4. According to $(*)$, $$\boxed{\frac14 \in \mathbb K_3}.\square$$

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Note that it was the search for the proof of the fact that $\frac14\in \mathbb K_3$, that gave me the idea of constructing $\mathbb K_3$ like this.