Real Numbers are Uncountably Infinite/Proof 2 using Ternary Notation

Theorem
The set of real numbers $\R$ is uncountably infinite.

Lemma
Let $\langle d_n \rangle$ and $\langle e_n \rangle$ be sequences in $\{ 0, 1 \}$.

Then the ternary representations $D = 0.d_1 d_2 \ldots$ and $E = 0.e_1 e_2 \dots$ represent distinct real numbers.

Proof
Suppose that $\langle d_n \rangle ≠ \langle e_n \rangle$.

By the Well-Ordering Principle, there is a smallest $n \in \N_{>0}$ such that $d_n ≠ e_n$.

Suppose WLOG that $d_n = 0$ and $e_n = 1$.

Let $K = 0.d_1 d_2 \ldots d_{n-1} = \sum_{i=1}^{n-1} d_i 3^{-i}$.

Then:


 * $D = K + \sum_{i=n+1}^\infty d_i 3^{-i}$
 * $E = K + 3^{-n} + \sum_{i=n+1}^\infty e_i 3^{-i} \ge K + 3^{-n}$

But then $D \le K + \sum_{i=n+1}^\infty 3^{-i} = K + 3^{-n-1} \sum_{i = 0}^\infty 3^{-i} = K + 3^{-n - 1} \dfrac 3 2 = K + 3^{-n}/2$

Thus $D < E$, so $D ≠ E$.