Now, let $C_1$ be equal to $C_0$ with its middle third removed. Mathematically, $C_1=\left[0,\frac13\right]\cup\left[\frac23,1\right]$. Let the following be an illustration of $C_1$:
If we continue in this manner, always removing the middle thirds of the remaining intervals, then we will obtain the following sequence:
$C_2$:
$C_3$:
$C_4$:
Let $n$ be any number in $\left\{0,1,2,3,\dots\right\}$. Define the Cantor set $C$ as the limit of $C_n$ as $n$ approaches infinity. Mathematically, $C=\lim_{n\to\infty}C_n$.
Clearly, $C_n$ comprises $2^n$ intervals of length $3^{-n}$ each (convince yourself of this fact). It therefore follows that the Cantor set $C$ has a length of
$$\lim_{n\to\infty}\frac{2^n}{3^n}=\lim_{n\to\infty}\left(\frac23\right)^n=0.$$
Intuitively, if an interval has a length of zero, then said interval must contain nothing at all; it must be the empty set, right?
Nevertheless, observe that the endpoints of the intervals are preserved through the sequence. In other words, observe that the points zero and one are preserved; all $C_n$ contain zero and one. It therefore follows that $C_0$ preserves two points, $C_1$ preserves four points, $C_2$ preserves eight points, and so on. Generally, $C_n$ preserves $2^{n+1}$ points. Therefore, the Cantor set $C$ preserves
$$\lim_{n\to\infty}2^{n+1}=\infty$$
points. In other words, a set with a length of zero can have infinitely many points. How can this be?
This exercise clearly demonstrates that our intuition of the mathematical infinity is insufficient to understand it, and that the need for rigorous tools and careful approaches to dissect it are of the utmost essence.