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.)
The usual proof consists in showing inductively that for all . This is easy: Note first that
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.
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 .
The related problem of describing appears to be much more complicated. See here and here.
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