Cantor Function
Recall the definition of the Cantor set: Let
= [ 1/3, 2/3]be the middle third of the interval [0, 1]. Let
= [ 1/9, 2/9], = [ 7/9, 8/9 ]be the middle thirds of the intervals remaining after deleting from [0, 1]. Let
= [2/27, 3/27], = [7/27, 8/27],be the middle thirds of the intervals remaining after deleting , , and from [0, 1]. Continue in this fashion so that at the n-th stage we have the intervals
= [19/27, 20/27], = [ 25/27, 26,27]
, ..., , ...,Then the complement of the union of all these intervals is the Cantor set without the endpoints. Now define the following function
if t is in the intervalThen, for example, we have that
if 1/3 t 2/3
Then F(t) is defined everywhere in [0, 1] except at the Cantor set minus the end points 0, 1, 1/3, 2/3, 1/9, 2/9, 7/9, 8/9, ... If t is a number where F is not defined, then there exists an increasing sequence { xn } of these endpoints converging to t, and a decreasing sequence { xn' } of these endpoints converging to t. Since F is defined at those endpoints xn and xn', we define
F(t) = F(xn) = F(xn')
Now we have defined completely the Cantor function. It has the following properties:
- F is defined everywhere in the interval [0, 1]
- F is not constant
- F is increasing
- F is continuous in the interval [0, 1]
- F is differentiable in the interval [0, 1]
- F' is zero at every interior point of the intervals
Proof
Some of these properties are obvious, and some require more thought. In particular, why does the above limit of endpoints exist ? That is a crucial point, because we used this limit to extend the function to the whole interval [0, 1]. For details and hints about the Cantor Function, please consult Kolmogorov and Fomin, p 334 ff.