91Ó°ÊÓ

The Cantor set \(\mathbb{F}\) is obtained by starting with the closed interval \([0,1]\) and removing the open middle third (\(\frac{1}{3}, \frac{2}{3}\)) to get two closed intervals \([0, \frac{1}{3}]\) and \([\frac{2}{3}, 1]\). Then remove the open middle third of each remaining intervals, and so on. The process continues indefinitely, so the Cantor set consists of all points in \([0,1]\) that are not removed at any step.

A cluster point of a set is a point that can be 'approached' by points in the set other than the point itself. Formally, a point \(x\) in the plane is a cluster point of a set \(\mathbb{F}\) if for any given distance \(\epsilon > 0\), there exists some point \(y\) in \(\mathbb{F}\) such that \(0 < |x - y| < \epsilon\).

Let \(x\) be a point in \(\mathbb{F}\). We show that for any \(\epsilon > 0\), there exists \(y \neq x\) in \(\mathbb{F}\) such that |x - y| < \(\epsilon\). Since \(x\) belongs to \(\mathbb{F}\), it is not removed at any stage of the construction of \(\mathbb{F}\). Consider a stage at which an interval \(I\) containing \(x\) has length less than \(\epsilon\). Since \(I\) is divided into two closed intervals with \(x\) in one of those, we can always take \(y\) in the other interval and close to the end-point common with \(x\)'s interval. Then, |x - y| < \(\epsilon\) and \(y \neq x\) and \(y \in \mathbb{F}\). Thus, \(x\) is a cluster point of \(\mathbb{F}\). Since \(x\) was any point in \(\mathbb{F}\), each point of the Cantor set is a cluster point.