Cantor Set is Uncountable/Proof 1

Proof
From the definition as a ternary representation, $\CC$ consists of all the elements of $\closedint 0 1$ which can be written without using a $1$.

So let $x \in \CC$. Then in base $3$ notation, we have (as $0 \le x \le 1$):


 * $\ds x = \sum_{i \mathop = 1}^\infty r_j 3^{-j}$

From the definition of the Cantor set, we have $\forall j: r_j \in \set {0, 2}$.

Furthermore, from Representation of Ternary Expansions, the $r_j$ are unique.

Now define the following function:


 * $\ds f: \CC \to \closedint 0 1, \quad \map f {\sum_{i \mathop = 1}^\infty r_j 3^{-j} } = \sum_{i \mathop = 1}^\infty \frac {r_j} 2 2^{-j}$

Observe that $\forall j: \dfrac {r_j} 2 \in \set {0, 1}$.

That the expression is in fact an element of $\closedint 0 1$ now follows from binary notation.

Furthermore by Existence of Base-N Representation, any element $y$ of $\closedint 0 1$ may be written in the following form (where $\forall j: b_j \in \set {0, 1}$):


 * $\ds y = \sum_{i \mathop = 1}^\infty b_j 2^{-j}$

Obviously, $y = \map f x$, where $x \in \CC$ is defined as follows:


 * $\ds x = \sum_{i \mathop = 1}^\infty 2 b_j 3^{-j}$

It follows that $f$ is surjective.

From Closed Interval in Reals is Uncountable, the closed interval $\closedint 0 1$ is uncountable.

The result follows.