Circle Inversion Fractals

Inversion Limit Sets

Limit Sets, Example 3

Example 3 If X is the Cantor set, then L(X) = X.

The usual construction of the Cantor set is to start with the closed unit interval [0, 1] and remove the open middle third (1/3, 2/3), then remove the open middle thirds (1/9, 2/9), (7/9, 8/9) of the remaining intervals, and so on.

An equivalent approach is to focus on the sequence of intervals remaining at each stage:

E	=	[0, 1]
E0 U E1	=	[0, 1/3] U [2/3, 1]
E00 U E01 U E10 U E11	=	[0, 1/9] U [2/9, 1/3] U [2/3,7/9] U [8/9,1]

We can think of the union of the intervals at each level as representing an approximation of the Cantor set. Say

C0	=	E
C1	=	E0 U E1
C2	=	E00 U E01 U E10 U E11
	...

Then it is easy to see

C₀ contains C₁ contains C₂ contains ...

We see C₀, C₁, C₂, ... is a nested sequence of sets.

The Cantor set is the intersection of all the C_i.

The set C_i consists of intervals of length 1/3ⁱ.

Note that the endpoints of every interval in every C_i belongs to all the C_i, and so belongs to the Cantor set.

Next, every point of the Cantor set is a limit point of the Cantor set.

To see this, take any point q of the Cantor set.

Then q belongs to C_i for each i. Now pick any distance d > 0.

We must show there is a point w of the Cantor set with 0 < dist(w, q) < d.

Two cases need to be considered.

First, suppose q is an endpoint of an interval of one of the C_j.

Then q is an endpoint of an interval in C_k for all k >= j.

No matter how small d is, there is a k >= j large enough that 1/3^k < d.

Suppose q is an endpoint of an interval L in this C_k.

Then take w to be the other endpoint of L.

Certainly, w belongs to the Cantor set, and dist(w, q) = 1/3^k < d.

So q is a limit point of the Cantor set.

Second, suppose q is not an endpoint of an interval in any of the C_j.

(It is not obvious that there are such points. That there are is most easily established by showing the set of all endpoints is countable, while the Cantor set is uncountable.)

Again, take k large enough that 1/3^k < d.

Because q belongs to C_i for all i, q belongs to C_k, hence to an interval L of C_k.

Let w be either of the endpoints of L.

Then w is not q (the hypothesis of this second case) so 0 < dist(w, q).

Because both w and q belong to L, dist(w, q) < 1/3^k < d.

Consequently, q is a limit point of the Cantor set.

Combining these two steps, we have shown that every point of the Cantor set is a limit point of the Cantor set.

To complete this example, we must show that every limit point of the Cantor set belongs to the Cantor set.

Suppose q is a limit point of the Cantor set.

For the purpose of contradiction, suppose q does not belong to the Cantor set.

Then q must belong to (-infinity, 0), to (1, infinity), or to one of the open intervals removed in going from some C_i to C_i+1.

That is, q belongs to some open interval (a, b) disjoint from the Cantor set.

Let d be the smaller of the distances dist(q, a) and dist(q, b).

Then no point of the Cantor set lies within a distance d of q, so q cannot be a limit point of the Cantor set.

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