Example 5.2.13(b): Properties of the Cantor Set

The Cantor set is perfect and hence uncountable.
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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.

Interactive Real Analysis, ver. 1.9.5
(c) 1994-2007, Bert G. Wachsmuth
Page last modified: Mar 28, 2007