Cantor Set Construction through Ternary Expansions

Recall the Cantor (middle-third) set \Lambda  which is equal to

\displaystyle{\bigcap_{k=0}^{\infty}{E_k}} 

where E_0 = [0,1]  and E_k  for k \geq 1  is obtained by removing the middle-thirds of each of the disjoint closed intervals making up E_{k-1}  , i.e. each disjoint closed interval [a,b]  making up E_{k-1}  gets mapped to [a, a + 1/3^k] \cup [a+2/3^k, b]  .

For example, to obtain E_1  , [0,1]  gets mapped to [0,1/3] \cup [2/3,1]  , and this latter set is E_1  . To obtain E_2  , [0,1/3]  gets mapped to [0,1/9] \cup [2/9,1/3]  , and [2/3,1]  gets mapped to [2/3, 7/9] \cup [8/9, 1]  , and E_2  is the union of these disjoint intervals.

The purpose of this post is to show that in fact

\displaystyle{S := \left\{ \sum_{k=1}^{\infty}a_{k}3^{-k}:a_{k}\in\left\{ 0,2\right\}\right\} } = \Lambda 

This set S  is the collection of all ternary expansions with coefficients 0  and 2  .

First, it is important to note that the elements of S  all coverge to real numbers. This is true because

\displaystyle{\sum_{k=1}^{\infty}{a_k 3^{-k}} \leq 2 \sum_{k=1}^{\infty}{3^{-k}} <\infty} 

the latter series being geometric.

Next, it will be shown that \Lambda \subseteq S  . Let x \in \Lambda  . It is sufficient to construct a sequence a_k \subseteq \{0,2\}  corresponding to x  such that

\displaystyle{x = \sum_{k=1}^{\infty}{a_k 3^{-k}}} (*) 

By definition of \Lambda , x \in E_k  for all k  . First, x \in E_0 = [0,1]  . Second, x \in E_1  . Since E_1  is the result of removing the middle-third of [0,1]  , therefore necessarily x  is in the left or right remaining third. If it was the left, define a_1 = 0  , and otherwise define a_1 = 2  . Now let k \geq 1  . Since x \in E_{k-1}  , therefore it is in one of the disjoint closed intervals [a,b]  making up E_{k-1}  . Since x \in E_k  and [a,b]  has its left and right thirds mapped into E_k  , therefore necessarily x  is in the left third (a_k = 0  ) or the right third (a_k = 2  ).

Inductively this defines a sequence a_k \subseteq \{0,2\}  corresponding to x  . Define the partial sums

\displaystyle{S_k = \sum_{i=1}^k{a_i 3^{-i}}} 

To show (*) holds for our choice of a_k  , it is sufficient to show that \forall{k}  , a_k  and S_k  are in one of the disjoint closed intervals [a,b]  making up E_k  with a = S_k  . This is sufficient because a property of \Lambda  is that the length of the [a,b]  , excluding E_0  , is 3^{-k}  (e.g. by an inductive argument), so |S_k - x| \leq |a-b| = 3^{-k} \rightarrow 0 as k \rightarrow \infty  . This would imply

\displaystyle{\sum_{k=1}^\infty{a_k 3^{-k}} = \lim S_k = x} 

We proceed by induction. For k=1  , if a_1 = 0  , then by definition x \in [0,1/3]  and S_1 = 0  . Otherwise a_1 = 2  in which case x \in [2/3,1]  and S_1 = 2/3  . Hence the claim holds for k=1  .

Now suppose that x  and S_k  are both in one of the disjoint closed intervals [a,b]  making up E_k  and S_k = a  . Then [a,b]  is mapped to [a,a+1/3^{k+1}] \cup [a+2/3^{k+1},b]  for E_{k+1}  . If x  is in the left interval, then a_{k+1} = 0  , in which case S_{k+1} = S_k = a . Otherwise x  is in the right interval in which case a_{k+1} = 2  , so

S_{k+1} = S_k + 2/3^{k+1} = a + 2/3^{k+1} 

This completes showing (*) so that indeed \Lambda \subseteq S  .

To show that S \subseteq \Lambda  as well, let x \in S  . Then

\displaystyle{\sum_{k=1}^\infty{a_k 3^{-k}} = x} 

and a_k \in \{0,2\}  . The latter fact and an earlier argument show that the partial sums S_k  of x are still a left endpoint of one of the disjoint closed intervals making up E_k  for all k  . Consequently S_k \in \Lambda  . Since \Lambda  is (topologically) closed, it then follows that

\lim S_k = x 

is in \Lambda  . This completes the proof.

Advertisements

Tags: ,

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: