We construct the Cantor Set by basically dividing the closed interval $[0,1]$ into equal 3 parts and leaving the middle part out. Then we continue the same process to the remaining two parts, i.e to the intervals $[0,\frac{1}{3}]$ and $[\frac{2}{3},1]$ and iterate for infinitely many times. After several steps the picture looks more or less like this:
Let us name the intervals left in each step as sets $C_0, C_1, C_2, ..., C_n$ where $n$ denote the steps starting with $0$ and runs to $n$.. For instance: $C_0 = [0,1]$ and $C_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$ and so on.. Cantor Set is the intersection of all these sets $C_n$: \[ C = \bigcap_{n} C_n \] We can also see that $C$ is a contracting set, i.e $C_0 \supset C_1 \supset C_2 ... \supset C_n \supset ...$. If we denote the length of the intervals at step $n$ as $\Delta n$, we can see that $\Delta n = \frac{1}{3^n}$. Also let $N_n$ be the number of intervals at step $n$, thus $N_n = 2^n$.
Now we have got all we need to calculate the total length of the interval at the $n^{th}$ step which is given by \[l_n = \Delta n \times N_n = (\frac{2}{3})^n\]Remember that we are doing the procedure of cutting into three and discarding the middle one infinitely many times, hence when $n \to \infty$, $l_n \to 0$, i.e the lengths contracting to end up being zero at the limit $n \to \infty$. Does that mean that our Cantor Set is empty?
Obviously not, since we can see easily see that the end points of the intervals that we create every time stays in the set. As in the case of $0, 1, \frac{1}{3}, \frac{2}{3}, \frac{1}{9}, \frac{2}{9}$ etc... Actually we can define a map \[x_n = \frac{1}{3^n} \,\,\,\,\,\,\, \mathbb{N} \to C \]so that we can map all of the end points with a natural number. This implies that $\mathbb{N} \preceq C$. Since $\mathbb{N}$ is dominated by $C$, cardinality of $C$ is at least $\aleph_0$. In the mean time, since $C \subset [0,1]$, the cardinality of $C$ can not exceed $\aleph_1$, which is the cardinality of $[0,1]$ interval itself (proof omitted). The critical question is whether the cardinality of $C$ is also $\aleph_1$ and the suprising answer to that is YES!
In order to prove that, we can start by labeling the intervals by 0,1 sequences so that after each division, the left remaining part is denoted by $0$ and the right one by $1$. Let us start with naming the interval $[0,1] = I$. Then we divide it by 3 and leave out the middle one, we label the left part $I_0$ and the right one $I_1$. Then we continue the iterations as in the figure below:
All the elements of the Cantor Set can be traced back to upper levels by looking whether it is on the left interval or the right one at each level. For instance if we look for $\frac{1}{4}$, it is in the interval $I$, then in $I_0$ (left), then $I_{01}$ (right), then $I_{010}$(left again)... It alternates between the left interval and the right one. We have denoted the left and right intervals with $0$ and $1$ respectively, thus we can claim that each 0-1 infinite sequence corresponds to an element of our Cantor Set. If we denote the all 0-1 sequences as $l_{\{0,1\}}$, then it can be shown that $card (l_{\{0,1\}}) = \aleph_1$ (proof omitted). Since we have previously said that the cardinality of $C$ can not exceed $\aleph_1$ so we can finally deduce that:
\[l_{\{0,1\}} \preceq C\] thus,
\[ card(C) = \aleph_1\]We showed that the cardinality of the Cantor Set is $\aleph_1$. This means that starting from the interval $[0,1]$, we constructed a set of intervals by removing infinitely many of the them in between, then joining them together we ended up with a set with the same "number of elements" we started with in the first place!
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