Cantor Set is Uncountable

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The Cantor set $\mathcal C$ is uncountable.

Proof 1

From the definition as a ternary representation, $\mathcal C$ consists of all the elements of $\left[{0 \,.\,.\, 1}\right]$ which can be written without using a $1$.

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

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

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

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

Now define the following function:

$\displaystyle f: \mathcal C \to \left[{0 \,.\,.\, 1}\right],\quad f \left({\sum_{i \mathop = 1}^\infty r_j 3^{-j} }\right) = \sum_{i \mathop = 1}^\infty \frac {r_j} 2 2^{-j}$

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

That the right hand side expression is in fact an element of $\left[{0 \,.\,.\, 1}\right]$ now follows from binary notation.

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

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

Obviously, $y = f \left({x}\right)$, where $x \in \mathcal C$ is defined as follows:

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

It follows that $f$ is surjective.

From Closed Interval in Reals is Uncountable, the closed interval $\left[{0 \,.\,.\, 1}\right]$ is uncountable.

The result follows.


Proof 2

It follows from Representation of Ternary Expansions that every string of the form $0.nnnnn \ldots$ where $n \in \left\{{0, 2}\right\}$ is an element of $\mathcal C$.

We also have that every string of the form $0.nnnnn \ldots$ where $n \in \left\{{0, 1}\right\}$ is an element of $\left[{0 \,.\,.\, 1}\right] \subset \R$ expressed in binary notation.

Let $f: \mathcal C \to \left[{0 \,.\,.\, 1}\right]$ be the function defined by:

$\forall x \in \mathcal C: f \left({x}\right) = \text{ the number obtained by replacing every } 2 \text { in } x \text { with a } 1$

where $x$ is expressed in base $3$ notation.

It is clear from the above that $f$ is a surjection.