Definition:Cantor Set/Limit of Intersections

Definition
Define:


 * $\displaystyle A_n := \bigcup_{i=1}^{\frac{3^n - 1} 2} \left({\dfrac{2i-1}{3^n} .. \dfrac{2i}{3^n}}\right)$

Since $3^n$ is always odd, $\dfrac{3^n-1} 2$ is always an integer, and hence this union will always be perfectly defined.

Consider the closed interval $\left[{0 .. 1}\right] \subset \R$.

Define:
 * $\mathcal C_n := \left[{0 .. 1}\right] \setminus A_n$

The Cantor Set $\mathcal C$ is defined as:
 * $\displaystyle \mathcal C = \bigcap_{i=1}^\infty \ \mathcal C_i$

Thus $\mathcal C$ can be formed by deleting a sequence of open intervals occupying the middle third of the resulting sequence of the closed intervals resulting from that deletion.

From the closed interval $\mathcal C_0 = \left[{0 .. 1}\right]$, the open interval $\left({\dfrac 1 3 .. \dfrac 2 3}\right)$ is removed.

This leaves:
 * $\mathcal C_1 = \left[{0 .. \dfrac 1 3}\right] \cup \left[{\dfrac 2 3 .. 1}\right]$

from which the open intervals $\left({\dfrac 1 9 .. \dfrac 2 9}\right)$ and $\left({\dfrac 7 9 .. \dfrac 8 9}\right)$ are removed.

This leaves:
 * $\mathcal C_2 = \left[{0 .. \dfrac 1 9}\right] \cup \left[{\dfrac 2 9 .. \dfrac 1 3}\right] \cup \left[{\dfrac 2 3 .. \dfrac 7 9}\right] \cup \left[{\dfrac 8 9 .. 1}\right]$

And so on.

Then:
 * $\displaystyle \mathcal C = \bigcap_{i=i}^\infty \ \mathcal C_i$