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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!

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  • $\begingroup$ What is the method by which it is constructed? $\endgroup$
    – user537667
    Commented Jan 24, 2019 at 15:10
  • $\begingroup$ math.stackexchange.com/questions/3083396/… $\endgroup$
    – Jack J.
    Commented Jan 24, 2019 at 15:14
  • $\begingroup$ (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) $\endgroup$ Commented Jan 24, 2019 at 16:28
  • $\begingroup$ In fact I replied to Praphulla, although I did not apologize for the lack of clarity. I take the opportunity to do it now. $\endgroup$
    – Jack J.
    Commented Jan 24, 2019 at 16:31

3 Answers 3

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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.

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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 $i <k$. The expansions of $a$ and $b$ don't contain any $1$'s, so all the digits are either $0$ or $2$.

Since $a<b$ 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.

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$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 $C_k$.

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