# Cantor Set Construction through Ternary Expansions

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

where and for is obtained by removing the middle-thirds of each of the disjoint closed intervals making up , i.e. each disjoint closed interval making up gets mapped to .

For example, to obtain , gets mapped to , and this latter set is . To obtain , gets mapped to , and gets mapped to , and is the union of these disjoint intervals.

The purpose of this post is to show that in fact

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

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

the latter series being geometric.

Next, it will be shown that . Let . It is sufficient to construct a sequence corresponding to such that

By definition of , for all . First, . Second, . Since is the result of removing the middle-third of , therefore necessarily is in the left or right remaining third. If it was the left, define , and otherwise define . Now let . Since , therefore it is in one of the disjoint closed intervals making up . Since and has its left and right thirds mapped into , therefore necessarily is in the left third () or the right third ().

Inductively this defines a sequence corresponding to . Define the partial sums

To show (*) holds for our choice of , it is sufficient to show that , and are in one of the disjoint closed intervals making up with . This is sufficient because a property of is that the length of the , excluding , is (e.g. by an inductive argument), so as . This would imply

We proceed by induction. For , if , then by definition and . Otherwise in which case and . Hence the claim holds for .

Now suppose that and are both in one of the disjoint closed intervals making up and . Then is mapped to for . If is in the left interval, then , in which case . Otherwise is in the right interval in which case , so

This completes showing (*) so that indeed .

To show that as well, let . Then

and . The latter fact and an earlier argument show that the partial sums of are still a left endpoint of one of the disjoint closed intervals making up for all . Consequently . Since is (topologically) closed, it then follows that

is in . This completes the proof.