\begin{definition}[Definition:Cantor Set/Limit of Intersections]
Define, for $n \in \N$, subsequently:
:$\map k n := \dfrac {3^n - 1} 2$
:$\ds A_n := \bigcup_{i \mathop = 1}^{\map k n} \openint {\frac {2 i - 1} {3^n} } {\frac {2 i} {3^n} }$
Since $3^n$ is always odd, $\map k n$ is always an integer, and hence the union will always be perfectly defined.
Consider the closed interval $\closedint 0 1 \subset \R$.
Define:
:$\CC_n := \closedint 0 1 \setminus A_n$
The '''Cantor set''' $\CC$ is defined as:
:$\ds \CC = \bigcap_{n \mathop = 1}^\infty \CC_n$
Thus $\CC$ 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 $\closedint 0 1$, the open interval $\openint {\dfrac 1 3} {\dfrac 2 3}$ is removed.
This leaves:
:$\CC_1 = \closedint 0 {\dfrac 1 3} \cup \closedint {\dfrac 2 3} 1$
Then from $\closedint 0 1$, the open intervals $\openint {\dfrac 1 9} {\dfrac 2 9}$, $\openint {\dfrac 3 9} {\dfrac 4 9}$, $\openint {\dfrac 5 9} {\dfrac 6 9}$ and $\openint {\dfrac 7 9} {\dfrac 8 9}$ are removed.
This leaves:
:$\CC_2 = \closedint 0 {\dfrac 1 9} \cup \closedint {\dfrac 2 9} {\dfrac 1 3} \cup \closedint {\dfrac 4 9} {\dfrac 5 9} \cup \closedint {\dfrac 2 3} {\dfrac 7 9} \cup \closedint {\dfrac 8 9} 1$
And so on.
Then:
:$\ds \CC = \bigcap_{i \mathop = 1}^\infty \CC_i$
\end{definition}