Page 100 - DMTH401_REAL ANALYSIS

Real Analysis

Notes 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 is
n
n
indeed comprised of 2 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.
Hence, C is the intersection of closed, bounded sets, and therefore C is also closed and bounded.
But then C is compact.

Example: The Cantor set has length zero, but contains uncountably many points.
Solution: The definition of the Cantor set is as follows: let

A = [0, 1]
0
and define, for each n, the sets A recursively as
n
+
+
¥ æ 1 3k 2 3k ö
A = A \  ç , ÷
n n – 1 k 0è 3 n 3 n ø
=
Then the Cantor set is given as:
C = A
n
To be more specific, we have:
A = [0, 1]
0
A = [0, 1] \ (1/3, 2/3)
1
A = A \ [(1/9, 2/9)(7/9, 8/9)] =
2 1
[0, 1] \ (1/3, 2/3) ) \ (1/9, 2/9) \ (7/9, 8/9)
...
That is, at the n-th stage (n > 0) we remove 2 n – 1 intervals from each previous set, each having
n
length 1/3 . Therefore, we will remove a total length from the unit interval [0, 1]. Since we
remove a set of total length 1 from the unit interval, the length of the remaining Cantor set must
be 0.
The Cantor set contains uncountably many points because it is a perfect set.

Example: The Cantor set does not contain any open set
The definition of the Cantor set is as follows: let
A = [0, 1]
0
and define, for each n, the sets A recursively as
n
+
+
¥ æ 1 3k 2 3k ö
A = A \  ç , ÷
n n – 1 k 0è 3 n 3 n ø
=
Then the Cantor set is given as:
C = A
n

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