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## Example 5.2.13(b): Properties of the Cantor Set

The Cantor set is perfect and hence uncountable.
The definition of the Cantor set is as follows: let
• A 0 = [0, 1]
and define, for each n, the sets A n recursively as
• A n = A n-1 \
Then the Cantor set is given as:
• C = A n
From this representation it is clear that C is closed. Next, we need to show that every point in the Cantor set is a limit point.

One way to do this is to note that each of the sets A n can be written as a finite union of 2 n closed intervals, each of which has a length of 1 / 3 n, as follows:

• A 0 = [0, 1]
• A 1 = [0, 1/3] [2/3, 1]
• A 2 = [0, 1/9] [2/9, 3/9] [6/9, 7/9] [8/9, 1]
• ...
Note that all endpoints of every subinterval will be contained in the Cantor set. Now take any x C = A n Then x is in A n for all n. If x is in A n, then x must be contained in one of the 2 n intervals that comprise the set A n. Define x n to be the left endpoint of that subinterval (if x is equal to that endpoint, then let x n be equal to the right endpoint of that subinterval). Since each subinterval has length 1 / 3 n, we have:
• | x - x n | < 1 / 3 n
Hence, the sequence { x n } converges to x, and since all endpoints of the subintervals are contained in the Cantor set, we have found a sequence of numbers contained in C that converges to x. Therefore, x is a limit point of C. But since x was arbitrary, every point of C is a limit point. Since C is also closed, it is then perfect.

Note that this proof is not yet complete. One still has to prove the assertion that each set A n is indeed comprised of 2 n closed subintervals, with all endpoints being part of the Cantor set. But that is left as an exercise.

Since every perfect set is uncountable, so is the Cantor.

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