Sequence of P-adic Integers has Convergent Subsequence/Lemma 6

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

Let $\sequence{x_n}$ be a sequence of $p$-adic integers.

Let $\sequence{b_0, b_1, \ldots, b_j}$ be a finite sequence of $p$-adic digits, possibly empty, such that:
 * there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$

Then there exists a subsequence $\sequence{y_n}$:
 * for all $n \in \N$, the canonical expansion of $y_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$

Proof
Let $g:\N \to \N$ be the mapping defined by:
 * $\map g n = \min \set{n > j : \text{ the canonical expansion of } x_n \text{ begins with the } p \text{-adic digits } b_j \, \ldots \, b_1 b_0}$

Let:
 * $n_0 = \min \set{n \in \N : \text{ the canonical expansion of } x_n \text{ begins with the } p \text{-adic digits } b_j \, \ldots \, b_1 b_0}$

From Well-Ordering Principle:
 * $g$ and $n_0$ are well-defined

From Principle of Recursive Definition:
 * there exists exactly one mapping $r: \N \to \N$ such that:


 * $\forall m \in \N: \map r m = \begin{cases}

a & : m = 0 \\ \map g {\map r n} & : m = n + 1 \end{cases}$

It is noted that:

Hence $r$ is a strictly increasing sequence in $\N$.

For all $n \in \N$, let $m_n = \map r n$.

Then $\sequence{x_{m_n}}$ is a subsequence of $\sequence{x_n}$ by definition.

Let $\sequence{y_n} = \sequence{x_{m_n}}$ and the result follows