# Leigh.Samphier/Sandbox/Sequence of P-adic Integers has Convergent Subsequence

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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 convergent subsequence $\sequence {x_{n_r} }_{r \mathop \in \N}$ of $\sequence{x_n}$

## Proof 1

### Lemma 5

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 $p$-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 $b_{j+1}b_j \, \ldots \, b_1 b_0$

$\Box$

### Lemma 1

- 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 $b_j \, \ldots \, b_1 b_0$

$\Box$

### Lemma 2

- there exists a subsequence $\sequence{x_{n_rj}}_{j \mathop \in \N}$ of $\sequence{x_n}$:
- for all $j \in \N$, the canonical expansion of $x_{n_j}$ begins with the $p$-adic digits $b_j \, \ldots \, b_1 b_0$

$\Box$

### Lemma 3

- the canonical expansion $\ldots \, b_n \, \ldots \, b_1 b_0$ converges to some $x$ in the $p$-adic integers $\Z_p$

$\Box$

### Lemma 4

- the subsequence $\sequence{x_{n_r}}_{r \mathop \in \N}$ converges to $x \in \Z_p$

$\blacksquare$

## Proof 2

From P-adic Integers are Compact Subspace:

- $\Z_p$ is a compact subspace in the metric space induced by $\norm{\,\cdot\,}_p$

From Compact Subspace of Metric Space is Sequentially Compact in Itself:

- $\Z_p$ is sequentially compact in itself

By definition of sequentially compact in itself:

- every sequence in $\Z_p$ has a subsequence which converges in the topology to a point in $\Z_p$

From Equivalence of Definitions of Convergence in Normed Division Rings:

- every sequence in $\Z_p$ has a subsequence which converges in the norm to a point in $\Z_p$

$\blacksquare$