Let $\alpha\in (0,1)$ and $\beta=\frac{1-\alpha}{2}$.

Define $T_0(x) = \beta x$ and $T_1(x) = (1-\beta) + \beta x$ , $\forall x\in [0,1]$.

Recursively define $I_0 =[0,1]$ and $I_{n+1}= T_0(I_n) \cup T_1(I_n)$.

The Middle-$\alpha$ Cantor Set is defined as $\bigcap_{n\in \omega} I_n$.

I have proved that $I_n$ is a disjoint union of $2^n$ intervals, each of length $\beta^n$. That is, $I_n=\bigcup_{i=1}^{2^n} [a_i,b_i]$

My question is that how do i prove that every endpoint $a_i,b_i$ in $I_n$ is in $\bigcap_{n\in\omega} I_n$?

It seems trivial, but i don't know how to prove this..


Almost entirely irrelevant to the question

That the Middle-$\alpha$ Cantor set is closed is easy as it is the intersection of closed sets.

The rest will be a fairly broad outline, and there are some details to fill in. Let $C$ denote the Middle-$\alpha$ Cantor set, and let $x \in C$ be arbitrary. We need to show that for all $\epsilon > 0$ there is a $y \in C \cap ( x - \epsilon , x + \epsilon )$ distinct from $x$.

Note that there must be an $n$ such the the unique closed interval containing $x$ in the $n$th stage of the construction of $C$ is entirely contained within the $( x - \epsilon , x + \epsilon )$. (Remember how I said some details are missing? This is where they would go. one has to determine the lengths of the intervals at each stage of the construction, but it is not overly difficult.) Note that the endpoints of this interval will be elements of $C$, and (at least) one of them is distinct from $x$. Clearly each endpoint of $I$ is an endpoint of either $I_0$ or $I_1$.

Perhaps slightly relevant to the question

I think your problem might come down to notational issues. Perhaps a better way of attacking this problem is to determine the endpoints of the open middle-$\alpha$ interval removed given an arbitrary closed interval $[a,b]$. Relatively simple calculation shows that this open interval is $\left( \frac{(b-a)(1-\alpha)}{2} , \frac{(b-a)(1+\alpha)}{2}\right)$, meaning that the subintervals remaining are $\left[ a , \frac{(b-a)(1-\alpha)}{2} \right]$ and $\left[ \frac{(b-a)(1+\alpha)}{2} , b \right]$. From here the result you are looking for is easy.

As it stands, your functions $T_0$ and $T_1$ seem to really mix up the intervals, and it will make it quite difficult to find for each interval remaining in the $(n+1)$st stage which interval from the $n$th stage generated it. (You would have to play around with how these interact, and you could come up with a formula, but it won't be pretty.)

  • $\begingroup$ As i said in my post, I have proved that $I_n$ is a disjoint union of $2^n$ closed intervals, each of length $\beta^n$, and just like you said, i know how this argument should go like... My main problem is, even though i know all these facts, i cannot prove that every endpoint of these intervals is in $C$. $\endgroup$ – Katlus Dec 3 '12 at 17:30
  • $\begingroup$ @Katlus: Please see my addition. $\endgroup$ – user642796 Dec 3 '12 at 18:06

Define $I_n^* = I_0 - I_n, \forall n\in \omega$.

Note that (i);

$I_{n+1}^*\\=I_0 \setminus I_{n+1} \\=I_0 \setminus (T_0(I_n)\cup T_1(I_n)) \\=(I_0\setminus (T_0(I_0)\setminus T_0(I_n^*)))\cap (I_0\setminus (T_1(I_0)\setminus T_1(I_n^*)) \\=T_0(I_n^*)\cup I_1^* \cup T_1(I_n^*)$.

Also(ii), it can be found that, $\forall x\in I_n, \beta x\in I_n$ and $(1-\beta)+\beta x \in I_n$.

Let $E_n$ be a set of endpoints of $I_n$.

Let $G=\{n\in \omega | n<m \Rightarrow E_n\subset E_m\}$

Then, it can be shown that $n<m\Rightarrow E_n \subset E_m$.

  • $\begingroup$ This is false, but since i'm not a registered user, i cannot delete this.. Relation(ii) is wrong $\endgroup$ – Katlus Dec 3 '12 at 18:22
  • $\begingroup$ Edited proof seems correct to me. $\endgroup$ – Katlus Dec 3 '12 at 18:58

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