How to show that $\frac14$ is in the standard Cantor set?

Question

I have studied the method that shows $\dfrac14$ belongs to the standard Cantor set, proving that it has only $0$ and $2$ as digits in its ternary expansion.

But how can I do this in some other way?

Answer 1

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

Clearly, $\frac{1}{4}$ and $\frac{3}{4}$ are in $[0,1]=C_0$.

Suppose $\frac{1}{4}$ and $\frac{3}{4}$ 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 symmetrical on $[0,1]$, we know that $\frac{3}{4} \in C_{n+1}$.

By induction, $\frac{1}{4}$ and $\frac{3}{4}$ are in $C_n$ for every $n\in \mathbb{N}$, hence $\frac{1}{4}$ and $\frac{3}{4}$ are in $C$.

Answer 2

This does not answer the question but since I cannot comment, I am writing this here just in case this view helps someone. Here's an easy way to see what the ternary expansion of the cantor set means (and also the Cantor function):

The cantor set is formed iteratively by dividing each interval of the previous iteration into three parts and keeping only the first and third part. The ternary representation of a point is therefore the address of the point. A $0$ in the $k^{th}$ position means at the $k^{th}$ iteration the point falls in the first (left most) division and a $2$ means it falls in the third(right most) division. And a $1$ means it falls on the boundary of the first part. The moment a $1$ is followed by any other number in the expansion then at that iteration the point has fallen in the middle third interval and is no longer a part of the Cantor set.