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 as a quotient of Cauchy sequences.

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

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

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

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,
 * $\forall j \in \N : \exists \mathop {\map N j} \ge j : \forall m, n \in \N: m, n \ge \map N j : \norm {\beta_n - \beta_m} \le p^{-\paren{j + 1}}$

Let $j \in N$

Suppose $\map N {j + 1} \ge \map N j$

By definition:
 * $\norm{\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren{j + 1}}$

Now suppose $\map N j \ge \map N {j + 1}$

Then:

In either case:
 * $\norm{\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren{j + 1}}$

Since $j$ was arbitrary, then:
 * $\forall j \in \N: \norm{\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren{j + 1}}$

From Leigh.Samphier/Sandbox/Unique Integers Close to Rationals in Valuation Ring of P-adic Norm:
 * for all $j \in \N$ there exists $\alpha_j \in \Z$:


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


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

Then:

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.

It is now shown that $\sequence{\alpha_j}$ is a representative of $a$.

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

It remains to show that $\sequence{\alpha_j}$ is unique.