# C_{0} + C_{0} = [0, 2]

Elsewhere I suppplied 3 proofs of the fact that the differences of numbers from 2 triadic Cantor sets C_{0} fills entirely the closed segment

Here I'll prove the sister question, viz., that the sum of numbers from two Cantor sets fills the interval

For two sets A and B,

A + B = {a + b : a∈A and b∈B}

I wish to prove that

C_{0} + C_{0} = [0, 2].

### Proof #1

We'll show that every line x + y = a, where a belongs to _{0}×C_{0}. If

The product C_{0}×C_{0} can be constructed similarly to the Cantor set itself. Draw a 1×1 square, and divide it into 9 squares of side 1/3. Remove 5 "central" squares. Repeat the process in each of the remaining four squares. That is, divide each of them into 9 equal squares of size 1/9 and remove 5 "central" squares. And so on.

It's not hard to convince oneself that the line x + y = a intersects at least one of the 4 "corner" squares with side 1/3. Denote any of these as S_{1}. By the same token, any line with slope 1 that intersects the square S_{1} intersects at least one of the corner squares of side length 1/9 that are contained in S_{1}. Denote one of them S_{2}. S_{2}⊂S_{1}.

In this manner we obtain a decreasing sequence {S_{i}} of (closed) squares. Since the sequence is decreasing, the intersection of any finite subsequence is not empty. (It's simply the smallest square in the subsequence.) This is another facet of compactness. A sequence of closed sets is said to possess a *finite intersection property*, if any finite subsequence has a non-empty intersection. A topological space is compact iff the intersection of every sequence of closed sets with the finite intersection property is not empty.

The original unit square being closed and bounded is compact. Therefore, the squares {S_{i}} have a common point. This point belongs to both C_{0}×C_{0} and the line

### Proof #2

Cantor set C_{0} is symmetric in the midpoint of the inderval

C_{0} - 1/2 = 1/2 - C_{0}.

We thus can write,

C_{0} + C_{0} - 1 =

C_{0} - 1/2 + C_{0} - 1/2 =

(C_{0} - 1/2) + (1/2 - C_{0}) =

C_{0} - C_{0}.

So that C_{0} + C_{0} = C_{0} - C_{0} + 1 =

## References

- B. R. Gelbaum and J. M. H. Olmsted,
*Counterexamples in Analysis*, Dover, 2003 - I. Stewart,
*Game, Set and Math*, Penguin Books, 1989

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