Open and Closed Balls in P-adic Numbers are Compact Subspaces

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Theorem

Let $p$ be a prime number.

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

Let $a \in \Q_p$.

Let $n \in \Z$.


Then the open ball $\map {B_{p^{-n} } } a$ and closed ball $\map {B^-_{p^{-n}}} a$ are compact.


Corollary

Then the set of $p$-adic integers $\Z_p$ is compact.


Proof

We begin by proving the theorem for the closed ball $\map {B^-_{p^{-n} } } a$.

From Open Ball in P-adic Numbers is Closed Ball then the theorem will be proved.


Let $d$ denote the subspace metric induced on $\map {B^-_{p^{-n}}} a$ by the $p$-adic Metric.


From Open and Closed Balls in P-adic Numbers are Totally Bounded:

the closed ball $\map {B^-_{p^{-n}}} a$ is totally bounded in $d$.

From P-adic Numbers form Completion of Rational Numbers with P-adic Norm:

$\struct {\Q_p, \norm {\,\cdot\,}_p}$ is complete.

From Open and Closed Balls in P-adic Numbers are Clopen in P-adic Metric:

the closed ball $\map {B^-_{p^{-n}}} a$ is closed in the $p$-adic metric.

From Subspace of Complete Metric Space is Closed iff Complete:

the closed ball $\map {B^-_{p^{-n}}} a$ is complete in $d$.

From Complete and Totally Bounded Metric Space is Sequentially Compact:

the closed ball $\map {B^-_{p^{-n}}} a$ is sequentially compact in $d$.

From Sequentially Compact Metric Space is Compact:

the closed ball $\map {B^-_{p^{-n} } } a$ is compact in $d$.


Hence the closed ball $\map {B^-_{p^{-n} } } a$ is a compact subspace in the $p$-adic metric by definition.

The result follows.

$\blacksquare$