Example 5.2.13(b): Properties of the Cantor Set
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The Cantor set is perfect and hence uncountable.
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The definition of the Cantor set is as follows: let
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:
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.