Circle Inversion Fractals

Inversion Limit Sets

Limit Sets, Example 1

Example 1 If X is a finite set of points, then L(X) is empty.

Let d₀ be the minimum distance between points in X.

Because X is a finfite set, d₀ > 0.

We do a proof by contradiction, so to show that L(X) is empty, we suppose that there is a point q in L(X).

That is, for every d > 0, there is a point w in X for which 0 < dist(w, q) < d.

Now take d = d₀/3.

Then for some w in X, 0 < dist(w, q) < d, thus w lies in the disc D centered at q and of radius d.

There cannot be any other points of X lying in D, because the diameter of D is 2d₀/3, and d₀ is the smallest distance between any pair of points of X, so D can contain at most one point of X.

Thus w is the only point of X in D, and so w is the point of X closest to q.

Now w is not equal to q, so dist(w, q) = e₁ > 0.

So no point of X is closer than e₁ to q.

That is, no point q can satisfy the conditions for being a limit point of X.

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