Real Numbers are Uncountably Infinite/Proof 2 using Ternary Notation/Lemma

Lemma to Real Numbers are Uncountably Infinite: Proof 2 using Ternary Notation
Let $\sequence {d_n}$ and $\sequence {e_n}$ be infinite sequences in $\set {0, 1}$ such that:
 * $\exists m \in \N: d_m \ne e_m$

That is, the sequences $\sequence {d_n}$ and $\sequence {e_n}$ are different in at least one term.

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

Proof
Let $\sequence {d_n} \ne \sequence {e_n}$.

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

, suppose that $d_n = 0$ and $e_n = 1$.

Let:
 * $\ds K = 0.d_1 d_2 \ldots d_{n - 1} = \sum_{i \mathop = 1}^{n - 1} d_i 3^{-i}$
 * $\ds D := K + \sum_{i \mathop = n + 1}^\infty d_i 3^{-i}$
 * $\ds E := K + 3^{-n} + \sum_{i \mathop = n + 1}^\infty e_i 3^{-i} \ge K + 3^{-n}$

But then:

Thus $D < E$, so $D \ne E$.