Sequence of P-adic Integers has Convergent Subsequence/Lemma 3

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.

Then:
 * there exists a sequence $\sequence{b_n}$ of $p$-adic digits:
 * for all $j \in \N$, there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$

Lemma
The proof proceeds using the second principle of recursive definition.

Let $T$ be the set of $p$-adic digits.

From lemma, there exists $b_0 \in {0, 1, \ldots , p-1}$ such that:
 * there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_0$

For each $j \in \N$, let $\map {P_j} {\sequence{b_0, b_1, \ldots, b_j} }$ be the proposition:
 * there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$

For each $j \in \N$, let $G_j : T^j \to T$ be defined by:
 * $\map {G_j} {\sequence{b_0, b_1, \ldots, b_j} } = \begin{cases}

\min \set{b \in T: \map {P_{j+1}} {\sequence{b_0, b_1, \ldots, b_j, b} } } & : \map {P_j} {\sequence{b_0, b_1, \ldots, b_j} } \\ 0 & : \lnot \map {P_j} {\sequence{b_0, b_1, \ldots, b_j} } \end{cases}$

From lemma, for each $j \in \N$:
 * $G_j$ is well-defined

By definition of $G_j$:
 * if $\map {P_j} {\sequence{b_0, b_1, \ldots, b_j} }$ is true, then $\map {P_j} {\sequence{b_0, b_1, \ldots, b_j, \map {G_j} {\sequence{b_0, b_1, \ldots, b_j} } } }$ is true

From second principle of recursive definition:
 * there exists exactly one mapping $b: \N \to T$ such that:


 * $\forall j \in \N: b_j = \begin{cases}

\sequence{ b_0 } & : j = 0 \\ \map {G_j} {\sequence{b_0, \ldots, b_n} } & : j = n + 1 \end{cases}$

Hence:
 * there exists a sequence $\sequence{b_n}$ of $p$-adic digits:
 * for all $j \in \N$, there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$