Recall that the Cantor set is defined as the intersection where

and is obtained by removing from each closed interval that makes up its open middle third, so

,

,

etc. Each is the union of closed intervals, each of length .

Let’s prove that is the interval . (Cf. Abbott, Understanding analysis, Exercise 3.3.6.)

1.

The usual proof consists in showing inductively that for all . This is easy: Note first that

,

where

and

.

This equality is verified by induction. Using this, we can use induction again to verify that, indeed, for all .

We clearly have that . To prove the converse, for each and each , pick such that . The sequence of is bounded, so it has a convergent subsequence . The corresponding subsequence has itself a convergent subsequence . One argues that their limit values belong to , because they belong to each , since these sets are nested and closed. Finally, it follows immediately that as well.

2.

A very elegant different argument is obtained by using an alternative characterization of : Note that each can be written in base three as

where each is , , or . By induction, one easily verifies that iff it admits such an expansion with . It follows that iff it admits an expansion where no is .

Given , we have , so we can write where the ternary expansion of has only s and s (so ), and the expansion of has only s and s: If

,

we can set where unless , in which case as well, and similarly where unless , in which case as well.

We then have that , and both and are in .

This construction has the further advantage of making clear that the typical admits continuum many () representations as sum of two members of : If we can split (where the expansions of only have s and s), we can set

.

This gives us as many representations as subsets of .

3.

The related problem of describing appears to be much more complicated. See here and here.

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