P-adic Integer has Unique Coherent Sequence Representative/Lemma 4

Theorem
Let $p$ be a prime number.

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers as a quotient of Cauchy sequences.

Let $\mathbf a$ be an equivalence class in $\Q_p$ such that $\norm {\mathbf a}_p \le 1$.

Let $\sequence {\alpha_j}$ be a coherent sequence that represents $\mathbf a$.

Then:
 * $\sequence {\alpha_j}$ is the only coherent sequence that represents $\mathbf a$.

Proof
Let $\sequence {\alpha'_j}$ be a coherent sequence not equal to $\sequence {\alpha_j}$.

From Representatives of same P-adic Number iff Difference is Null Sequence, it needs only to be shown that $\sequence {\alpha_j - \alpha'_j}$ is not a null sequence.

Since $\sequence {\alpha'_j} \ne \sequence{\alpha_j}$ then:
 * $\exists i_0 \in \N : \alpha'_{i_0} \ne \alpha_{i_0}$

By definition of coherent sequences:
 * $0 \le \alpha_{i_0}, \alpha'_{i_0} < p^{i_0 + 1}$

From Reduced Residue System Modulo Prime:
 * $\alpha_{i_0} \not \equiv \alpha'_{i_0} \pmod {p^{i_0 + 1} }$

By definition of a coherent sequence, for all $i > i_0$:
 * $\alpha_i \equiv \alpha_{i_0} \pmod {p^{i_0 + 1} }$
 * $\alpha'_i \equiv \alpha'_{i_0} \pmod {p^{i_0 + 1} }$

Then:
 * $\forall i > i_0: \alpha_i \equiv \alpha_{i_0} \not \equiv \alpha'_{i_0} \equiv \alpha'_i \pmod {p^{i_0 + 1} }$

That is:
 * $\forall i > i_0: p^{i_0 + 1} \nmid \alpha_i - \alpha'_i$

By definition of the $p$-adic norm on integers:
 * $\forall i > i_0: \norm {\alpha_i - \alpha'_i} > \dfrac 1 {p^{i_0 + 1} }$

By definition of convergence, $\sequence {\alpha_j - \alpha'_j}$ is not a null sequence.

The result follows.