Sequence of P-adic Integers has Convergent Subsequence/Lemma 2

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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.

Let $\sequence{b_0, b_1, \ldots, b_j}$ be a finite sequence of $p$-adic digits 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 $p$-adic digit $b_{j + 1}$ such that:

there exists infinitely many $n \in \N$ such that the canonical expansion of $x_n$ begins with the $p$-adic digits $b_{j+1}b_j \, \ldots \, b_1 b_0$

Proof

Lemma

there exists a subsequence $\sequence{y_n}$ of $\sequence{x_n}$:

for all $n \in \N$, the canonical expansion of $y_n$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$

$\Box$

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 $b b_j \, \ldots \, b_1 b_0$

Let $b_{j+1} = b$ 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 $y_n$ begins with the $p$-adic digits $b b_j \, \ldots \, b_1 b_0$

So:

$\set{n \in \N : \mathop \exists b \in \closedint{0}{p-2} : \text{ the canonical expansion of } y_n \text{ begins with the } p \text{-adic digits } b b_j \, \ldots \, b_1 b_0}$ is a finite set.

Let:

$N = \max \set{n \in \N : \mathop \exists b \in \closedint{0}{p-2} : \text{ the canonical expansion of } y_n \text{ begins with the } p \text{-adic digits } b b_j \, \ldots \, b_1 b_0}$

Hence for all $n > N$ and $b \in \closedint{0}{p-2}$:

$y_n$ does not begin with the $p$-adic digits $b b_j \, \ldots \, b_1 b_0$.

So for all $n > N$:

the canonical expansion of $y_n$ begins with the $p$-adic digits $\paren{p-1} b_j \, \ldots \, b_1 b_0$

Let $b_{j+1} = p - 1$ and the result follows.

$\blacksquare$