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Unit 8: Completeness and Compactness
The Cantor set gives an indication of the complicated structure of closed sets in the real line. It Notes
has the following properties:
Example: The Cantor set is compact.
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
+
+
¥ æ 1 3k 2 3k ö
Each set ç , ÷ is open. Since A is closed, the sets A are all closed as well, which
k 0è 3 n 3 n ø 0 n
=
can be shown by induction. Also, each set A is a subset of A , so that all sets A are bounded.
n 0 n
Example: The Cantor set is perfect and hence uncountable.
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 n n
k 0è 3 3 ø
=
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.
n
One way to do this is to note that each of the sets A can be written as a finite union of 2 closed
n
intervals, each of which has a length of 1/3 , as follows:
n
A = [0, 1]
0
A = [0, 1/3] [2/3, 1]
1
A = [0, 1/9][2/9, 3/9] [6/9, 7/9] [8/9, 1]
2
...
Note that all endpoints of every subinterval will be contained in the Cantor set. Now take any
x C =A . Then x is in A for all n. If x is in A , then x must be contained in one of the 2 intervals
n
n n n
that comprise the set A . Define x to be the left endpoint of that subinterval (if x is equal to that
n n
endpoint, then let x be equal to the right endpoint of that subinterval). Since each subinterval
n
has length 1/3 , we have:
n
|x – x | < 1/3 n
n
Hence, the sequence {x } converges to x, and since all endpoints of the subintervals are contained
n
in the Cantor set, we have found a sequence of numbers contained in C that converges to x.
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