Most of the content here is adapted from (Edgar, chapter 1.1). # Definitions ###### Recursive defintion We start with $C_{0}=[0,1]$; The next set is constructed by removing the middle third: $C_{1}= C_{0} \setminus \left( \frac{1}{3}, \frac{2}{3} \right) = \left[ 0, \frac{1}{3} \right] \cup \left[ \frac{2}{3}, 1 \right]$; The next set is constructed by removing the middle thirds of the two intervals: $C_{2}= \left[ 0, \frac{1}{9} \right] \cup \left[ \frac{2}{9}, \frac{1}{3} \right] \cup \left[ \frac{2}{3}, \frac{7}{9} \right] \cup \left[\frac{8}{9}, 1 \right] ;$and so on ... See that $C_{k}$ is a union of $2^{k}$ intervals of length $\frac{1}{3^{k}}$. The sets decrease: $C_{0} \supset C_{1} \supset C_{2} \dots$. Cantor's dust is defined as their limit, that is, their intersection given by $C = \bigcap_{k\in \mathbb{N}} C_{k}.$ Which points belong to $C$? It can be seen inductively from this definition that the boundary of $C_{k}$ is a subset of $C_{k+l}$ for all $l\geq 1$. So, for example, the boundary of $C_{0}=\{ 0,1 \}$ is a subset of $C$. But these are not the only points belonging to $C$. We can show for example that $\frac{1}{4}$ belongs to $C$, that is, it is never removed at each stage of our construction. Observe that $\frac{1}{4}$ belongs to the lower third of $C_{0}$; then the upper third of $\left[ 0, \frac{1}{3} \right]$; the lower third of $\left[ \frac{2}{9}, \frac{1}{3} \right]$, etc. This should continue by the self-similarity and symmetry of our constructions. Formally, for our inductive step, we want to show that if $\frac{1}{4}$ is a closed interval $I_{k}\subset C_{k}$, then - $\frac{1}{4}$ is in the lower third of $I_{k}$ if $k$ is even; - and $\frac{1}{4}$ is in the upper third of $I_{k}$ if $k$ is odd - showing that $\frac{1}{4}\in C_{k+1}$. ###### Explicit formula for $C_{k}$ ###### Iterated function system # Properties # Applications