If X is a set of points in the plane, a point z is a limit point of X, if for every e > 0 there is a point wÎ X with 0 < d(w, z) < e . (Note that 0 < d(w, z) implies that w ¹ z) [R, pg. 32] Here d(w, z) denotes the Euclidean distance between w and z. The set of all limit points of X is called the Limit Set of X denoted be L (X).

Proposition 2.1: If X is a finite set of points, then L (X) = Æ .

Proof: Let e _o be the minimum distance between points in X. We will do a proof by contradiction. So, to show that L (X) is empty, we’ll suppose that there is a point p in L (X). That means, for every e > 0, there is a point q in X for which 0 < d(p, q) < e . Take . Then for some q in X, 0 < d(p, q) < e . This means q lies in the disc D centered at p and of radius e . There cannot be any other points of X lying in D, because the diameter of D is , and e _o is the smallest distance between any pair of points of X. Thus q is the only point of X in D, and so q is the point of X closest to p. Now q is not equal to p, so d(p, q) = e ₁ > 0. Take . Then d(p, q) > e , and q is the point of X closest to p, so d(p, q) < e is false for all q in X. Therefore, if X is a finite set of points, then L (X) = Æ .

With Proposition 2.1, we can now see that if X = {(0, 0)}, then L (X) = Æ .

If X = {(x, y): x and y are rational} then L (X) =Â ². Then to show L (X) =Â ², we must show that for any (a, b)Î Â and any e >0, there is (x, y)Î X with d((a, b), (x, y)) < e . In a square with one corner on (a, b) and the diagonally opposite corner on the circle, d((a, b), (x, y)) = e . By the density of the rationals in the reals, there is a rational point w along a horizontal side on the box, and there is a rational point v along a vertical side of the box. Then (w, v)Î X and it lies inside the box. So d((a, b), (w, v)) < e .

The Cantor Middle Thirds Set is obtained by starting with the unit interval and removing the open middle third interval, leaving . Then repeat infinitely often this process of removing the open middle third interval. See Rudin, pg. 41. Every point of the Cantor Middle Thirds Set is a limit point of that set; in fact, it is its own limit set.

A set is Perfect if it is closed, non-empty, and every point is a limit point. A general Cantor Set is a totally disconnected, perfect set. So, for every Cantor Set X, L (X) = X.

Most limit sets in this thesis will be constructed with transformations of the plane. Let G be a group of transformations of  ². Then the orbit of a point pÎ Â ² is G(p) = {g(p): gÎ G}. The limit set L (p) is the limit set of the orbit.