P-adic Integer has Unique Coherent Sequence Representative

Theorem
Let $p$ be a prime number.

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.

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

Then $\mathbf a$ has exactly one representative that is a coherent sequence.

Proof
Let $\sequence{\beta_n}$ be a sequence representing $\mathbf a$.

That is, $\sequence{\beta_n}$ is a Cauchy sequence in the $p$-adic norm $\norm{\,\cdot\,}_p$ on the rational numbers $\Q$.

By definition of a Cauchy sequence:
 * $\forall j \in \N : \exists \mathop {\map M j} \in \N: \forall m, n \in \N: m, n \ge \map M j : \norm {x_n - x_m} < p^{-j}$

For all $j \in \N$, let:
 * $\map N j = \max \set{j, \map M j}$

From P-adic Norm of p-adic Number is Power of p,
 * $\forall j \in \N : \exists \mathop {\map N j} \ge j : \forall m, n \in \N: m, n \ge \map N j : \norm {\beta_n - \beta_m} \le p^{-\paren{j + 1}}$

Lemma 1
From Unique Integer Close to Rational in Valuation Ring of P-adic Norm:
 * for all $j \in \N$ there exists $\alpha_j \in \Z$:


 * $(1): \quad \norm{\beta_{\map N j} - \alpha_j}_p \le p^{-\paren{j + 1}}$


 * $(2): \quad 0 \le \alpha_j \le p^{j + 1} - 1$

Lemma 2
By definition of the $p$-adic norm,
 * $\forall j \in \N : \alpha_{j + 1} \equiv \alpha_j \pmod {p^{j + 1}}$

Then $\sequence{\alpha_j}$ is a coherent sequence by definition.

From Corollary of Characterisation of Cauchy Sequence in Non-Archimedean Norm, $\sequence{\alpha_j}$ is a Cauchy sequence.

Lemma 3
Then $\sequence{\alpha_j}$ is a representative of $\mathbf a$.

Also see

 * Coherent Sequence Converges to P-adic Integer
 * Equivalence Class in P-adic Integers Contains Unique P-adic Expansion
 * Equivalence Class in P-adic Numbers Contains Unique P-adic Expansion