Definition:Cantor Set/Limit of Intersections

Definition
Define, for $n \in \N$, subsequently:


 * $k \left({n}\right) := \dfrac {3^n - 1} 2$


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

Since $3^n$ is always odd, $k \left({n}\right)$ is always an integer, and hence the 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:
 * $\displaystyle \mathcal C = \bigcap_{n \mathop = 1}^\infty \ \mathcal C_n$

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

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

Then from $\left[{0 \,.\,.\, 1}\right]$, the open intervals $\left({\dfrac 1 9 \,.\,.\, \dfrac 2 9}\right)$, $\left({\dfrac 3 9 \,.\,.\, \dfrac 4 9}\right)$, $\left({\dfrac 5 9 \,.\,.\, \dfrac 6 9}\right)$ and $\left({\dfrac 7 9 \,.\,.\, \dfrac 8 9}\right)$ are removed.

This leaves:
 * $\mathcal C_2 = \left[{0 \,.\,.\, \dfrac 1 9}\right] \cup \left[{\dfrac 2 9 \,.\,.\, \dfrac 1 3}\right] \cup \left[{\dfrac 4 9 \,.\,.\, \dfrac 5 9}\right]\cup \left[{\dfrac 2 3 \,.\,.\, \dfrac 7 9}\right] \cup \left[{\dfrac 8 9 \,.\,.\, 1}\right]$

And so on.

Then:
 * $\displaystyle \mathcal C = \bigcap_{i \mathop = 1}^\infty \ \mathcal C_i$