# Cantor Sets

The Cantor set C_{0} is a remarkable construction that is often employed to generate other useful mathematical objects. We constructed the set C_{0} as a limit of an iterative process. Each step of the process led to a union of closed intervals from which on the next step we removed the open middle third. The Cantor set C_{0} was defined as the leftover collection of points. Surprisingly uncountable, C_{0} has no isolated points. It's also ascribed zero length since the total length of the removed intervals is

1/3 + (2/3)(1/3) + (2/3)^{2}(1/3) + ... = 1,

the length of the original interval [0,1]. What if we were to remove some other portion of the set, not just 1/3. What would we get when removing more than 1/3? Would we get an empty set, or a set of negative measure? The latter is absurd: since we only remove from what we have, in the "worst" case we shall remain with nothing not anything "negative".

Let's check this out. Assume we are given a number *k*, 0 < *k* < 1.*k*. Two intervals will remain of the total length *k*).*k*-th part - two intervals whose lengths sums to (1 - *k*)*k*. This leaves four intervals of total length *k*) - (1 - *k*)*k* = (1 - *k*)^{2}.*k*-th parts. On the n-th step we have 2^{n} intervals with the total length of *k*)^{n}.*k* < 1,_{0} *k* = 1/3),*perfect*.

There is a way to visualize those sets. Each is obviously self-similar consisting of two parts, the left part and the right part. Let K_{k} denote the set that corresponds to number *k*. Then _{k} = f_{L}(K_{k})∪f_{R}(K_{k}),_{L}(u) = (1 - *k*)u/2_{R}(u) = (*k* - 1)u/2 + 1._{L}([0,1]) = [0, (1 -*k*)/2]_{R}([0,1]) = [(1 + *k*)/2, 1]*k*)/2]) = length([(1 + *k*)/2, 1]) = (1 - *k*)/2._{R} also reflects intervals in a vertical mirror.)

The two functions f_{L} and f_{R} form a 1-dimensional Iterated Function System with K_{k} as the fixed set. The applet below illustrates the generation of those sets. The *tent* function consists of two parts, the inverses of f_{L} and f_{R}, respectively. *k*).*k* = 1/3.

The sets K_{k} are 1-dimensional and lie on the x-axis at the feet of black lines which are only used to make the result more salient. Due to both the limited precision of the machine arithmetic and the granularity of the display, the diagrams must be taken with a grain of salt - solely as an illustration. Also note that the functions are squashed vertically and are not drawn to scale.

To modify the value of a, drag the vertex of the tent function up and down. The allowed values of a range from a > 2 through a little beyond 4.

As of 2018, Java plugins are not supported by any browsers (find out more). This Wolfram Demonstration, **Cantor Set**, shows an item of the same or similar topic, but is different from the original Java applet, named 'Cantor'. The originally given instructions may no longer correspond precisely.

(image below from deprecated 'Cantor' applet)

For a >2 (*k* < 1), the sets K_{k} are fractals with the similarity dimension equal to log(2)/log(a), although all have "length" 0. When a approaches 2, the dimension tends to 1, and, for

Construction of the sets K_{k} may be compared to the following process. Again start with the unit interval *k* < 1.*k*/2. The length of the two remaining intervals totals *k*/2.*k*/8. The four remaining intervals will have the total length *k*/2 - *k*/4.*k*/32. The remaining length will be *k*/2 - *k*/4 - *k*/8,

1 - *k*/2 - *k*/4 - *k*/8 + ... = 1 - *k* > 0.

These sets are known as *Cantor sets of positive measure*. The Hausdorff-Besicovitch dimension of each is 1! Therefore, strictly speaking, they are not fractal. However, none of them contains an open interval however small, which is to say that the sets are *nowhere dense*. This follows from the fact that Cantor sets of positive measure are continuous 1-1 images of the set C_{0}. The function that performs the job is very similar to the Cantor staircase function (although it's monotone increasing and, therefore, being continuous, is

Consider C_{0} and a Cantor set A of positive measure. Assume on the first step we remove the interval I_{1} from C_{0} and the interval J_{1} from A. Map one onto the other by a linear increasing function. This is easy because the two intervals have the same center. On the second step, we remove intervals I_{21} and I_{22} from C_{0} and J_{21} and J_{22} from A. Map the intervals with the same indices on top of each other with linear increasing functions. Continue this process for all pairs of the removed intervals. Since, so far, the function is monotonic on the complement of C_{0} and C_{0} is nowhere dense it is possible to extend the function by continuity to the whole interval _{0} onto the set A. The function F is monotonic and

## References

- B. R. Gelbaum and J. M. H. Olmsted,
*Counterexamples in Analysis*, Holden-Day, 1964 - B. R. Gelbaum and J. M. H. Olmsted,
*Theorems and Counterexamples in Mathematics*, Springer-Verlag, 1990 - H.-O. Peitgen, H. Jürgens, D. Saupe,
*Chaos and Fractals, New Frontiers of Science*, Springer, 1992

### Cantor Set

- Cantor set and function
- Cantor Sets
- Difference of two Cantor sets
- Difference of two Cantor sets, II
- Sum of two Cantor sets
- Plane Filling Curves: the Lebesgue Curve

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