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.
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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 $d$.
Recall that by definition, the $p$-adic numbers $\struct {\Q_p, \norm {\,\cdot\,}_p}$ are complete.
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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.
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From Subspace of Complete Metric Space is Closed iff Complete, the closed ball $\map {B^-_{p^{-n}}} a$ is complete in $d$.
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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.
$\blacksquare$
