# The cantor set $K$ contains no intervals

Let $$K$$ the Cantor set. I already proved the following properties

Properties.$$\begin{equation} \begin{split} (1)&\quad|K|=|\mathbb{R}|\\ (2)&\quad\lambda(K)=0,\text{where}\;\lambda\;\text{is the Lebesgue measure};\\ (3)&\quad\text{If}\;E\in 2^{K}\Rightarrow E\;\text{is Lebesgue measurable;} \end{split} \end{equation}$$

I want to prove that $$K$$ does not contain intervals.

Now, if $$I=[a,b]\subseteq K$$ is an interval, $$a\ne b$$, then for $$(3)$$ it is measurable and $$\lambda(I)>0$$, absurd. Therefore, $$K$$ does not contain intervals.

Question. How can I show that $$K$$ contains no intervals using only the method by which it is constructed, ie without using the monotony of the measure?

Thanks!

• What is the method by which it is constructed?
– user537667
Jan 24, 2019 at 15:10
• math.stackexchange.com/questions/3083396/… Jan 24, 2019 at 15:14
• (there are many different methods to construct the Cantor set as evidenced by the different answers, so I believe Praphulla was asking which method /you/ considered to be the method by which it is constructed) Jan 24, 2019 at 16:28
• In fact I replied to Praphulla, although I did not apologize for the lack of clarity. I take the opportunity to do it now. Jan 24, 2019 at 16:31

Given any interval $$[a,b]$$ you can prove it is not in $$K$$ by finding an interval within it that is removed. Find an $$n$$ such that $$3^{-n} \lt b-a$$ and you will remove a segment of length $$3^{-(n+1)}$$ from it at stage $$n+1$$ unless you have removed some sooner. This shows $$[a,b]$$ is not in the set.
The Cantor set can be described as the set of numbers in $$[0,1]$$ that do not require the digit $$1$$ to be written in base $$3$$. You can prove this is equivalent to the definition you provided - in the first step you remove all numbers of the form $$0.1\dots$$ (except $$0.1$$ itself which can also be written as $$0.0222\cdots$$). In the second step you remove all numbers of the form $$0.x1\cdots$$ where $$x$$ is $$0$$ or $$2$$ (except $$0.01$$ and $$0.21$$ which can be written as $$0.00222\cdots$$ and $$0.20222\ \cdots$$, respectively). And so on.
If $$[a,b] \subseteq K$$ with $$a \neq b$$, then consider the base $$3$$ expansions of $$a$$ and $$b$$: $$a=0.a_1a_2a_3\dots$$, $$b=0.b_1b_2b_3\dots$$. These numbers are distinct, so consider the first digit at which they differ. Say $$a_k \neq b_k$$ but $$a_i=b_i$$ for all $$i . The expansions of $$a$$ and $$b$$ don't contain any $$1$$'s, so all the digits are either $$0$$ or $$2$$.
Since $$a and they agree in their first $$k-1$$ digits, we must have $$a_k=0$$ and $$b_k=2$$. Then the number $$c=0.a_1a_2\cdots a_{k-1}111 \cdots$$ is between $$a$$ and $$b$$, and it requires a $$1$$ in its ternary expansion, so $$c \notin K$$, a contradiction.
$$C=\bigcap_k C_k$$, where each $$C_k$$ is the union of $$2^k$$ closed intervals in $$[0,1]$$ of length $$3^{-k}.$$ Therefore, as soon as $$\mathbb N \ni N>-\log_3(b-a),\ (a,b)\nsubseteq C_N$$ and so of course, $$(a,b)$$ cannot be contained in the intersection of the $$C_k$$.