# $1/4$ is in the Cantor set?

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

• Have you tried writing it in base 3? What happens? Jan 29, 2013 at 15:39
• Hint: prove something more (which turns out to be easier to prove), namely prove $1/4$ and $3/4$ are both in the Cantor set. First prove they are both in $[0,1]$, then prove that if they are both in one stage then they are both in the next stage, where the stages are used in construction of the Cantor set. Jan 29, 2013 at 15:40
• en.wikipedia.org/wiki/Cantor_set#Composition
– user940
Jan 29, 2013 at 15:41
• @GiuseppeNegro what do you mean by base 3? thank you for your comment Jan 29, 2013 at 15:49
• with respect to the given answers, see, e.g., this post, which discusses the expression of numbers between 0 and 1 in their ternary (base three) expansion. Jan 29, 2013 at 15:52

$1/4$ 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.

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.

• What do you mean by base 3? thank you for your answer! Jan 29, 2013 at 15:51
• Jan 29, 2013 at 16:02

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 $C$.

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.)