Sequence of P-adic Integers has Convergent Subsequence/Lemma 1

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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 $p$-adic digit $b_0$ such that:

there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digit $b_0$

Proof

Case 1

Let there exist $b \in \set{0, 1, \ldots , p - 2}$:

there exists infinitely many $n \in \N$ such that the canonical expansion of $y_n$ begins with the $p$-adic digits $b$

Let $b_0 = b$ and the result holds.

$\Box$

Case 2

For all $b \in \set{0, 1, \ldots , p - 2}$:

there exists only a finite number of $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digit $b$

So:

$\set{n \in \N : \mathop \exists b \in \set{0, 1, \ldots , p - 2} : \text{ the canonical expansion of } x_n \text{ begins with the } p \text{-adic digit } b}$ is a finite set.

Let:

$N = \max \set{n \in \N : \mathop \exists b \in \set{0, 1, \ldots , p - 2} : \text{ the canonical expansion of } x_n \text{ begins with the } p \text{-adic digits } b}$

Hence for all $n > N$ and $b \in \set{0, 1, \ldots , p - 2}$:

$x_n$ does not begin with the $p$-adic digits $b$.

So for all $n > N$:

the canonical expansion of $y_n$ begins with the $p$-adic digit $p-1$

Let $b_0 = p - 1$ and the result follows.

$\blacksquare$