Real Numbers are Uncountably Infinite/Proof 1 using Ternary Notation

Proof
It is sufficient to show that the real interval $I = \left\{{x \in \R: 0 < x \le 1}\right\}$ is uncountable.

Let $x \in I$.

From Existence of Base-N Representation, $x$ has a unique representation of the form:
 * $x = \dfrac {\epsilon_1} 3 + \dfrac {\epsilon_2} {3^2} + \dfrac {\epsilon_3} {3^3} + \cdots$

where $\epsilon_k = 0, 1$ or $2$ and an infinite number of $\epsilon_k$ are different from $0$.

Let $S \subseteq I$ be countably infinite.

Let $S = \left\{{x_1, x_2, \ldots}\right\}$.

Let $\epsilon_{k1}, \epsilon_{k2}, \epsilon_{k3}, \ldots$ be the ternary digits of $x_k$.

Let $\epsilon_k = 1 + 2 \epsilon_{kk} - \epsilon_{kk}^2$ so that:
 * $\epsilon_k = 1$ if $\epsilon_{kk} = 0$ or $\epsilon_{kk} = 2$
 * $\epsilon_k = 2$ if $\epsilon_{kk} = 1$

Then:
 * $(1): \quad \forall k: \epsilon_k \ne 0$
 * $(2): \quad \forall k: \epsilon_k \ne \epsilon_{kk}$

We have that $0 < x \le 1$ so $x \in I$.

But the real number $\displaystyle x = \sum \epsilon_k 3^{-k}$ is different from every $x_k \in I$.

Thus we have found an element of $I$ which is not an element of $S$.

Therefore $S$ is a proper subset of $I$.

It follows by definition that $I$ is uncountable.