Circle Inversion Fractals

Mandelbrot's Algorithm

The limit set L(C₁, C₂, C₃, C₄, C₅) lies in the disc bounded by S₁ = A.

That is, L(1, 2, 3, 4, 5) lies in the disc bounded by A.

To see this, note the outside of the disc bounded by A (on the left) inverts to the half of the plane above A' (on the right).

Next, following the orbit of any point outside the discs bounded by 1, 2, 3, 4, or 5, as soon as inversion in 5' is applied, the orbit drops below A'.

This is because each successive inversion acts on a point outside the inverting circle, and so gives a point inside the inverting circle.

No subsequent inversion in 1' or 2' gives a point above A'.

This is because these inversions are reflections across the vertical lines 1' and 2', and these do not change y-coordinates.

No subsequent inversion in 3' or 4' gives a point above A'.

This is because the centers of 3' and 4' lie on A', so points below A' invert to points below A'.

No subsequent inversion in 5' gives a point above A'.

Again, inversion in 5' is always applied to points outside of 5', so the inverted point lies inside 5', hence below A'.

Because any finite set of orbit points has empty limit set, we can ignore all finite collections of orbit points before inversion in 5' is applied.

The only remaining case is infinite orbits with no applications of 5'.

These generate limit points in L(1', 2', 3', 4'), which is the circle A'.

Consequently, no limit points lie above A'.

Return to Mandelbrot's method.