Sequence is Cauchy in P-adic Norm iff Cauchy in P-adic Numbers

Theorem
Let $p$ be a prime number.

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

Let $\sequence{x_n}$ be a sequence in the rational numbers $\Q$.

Then $\sequence {x_n}$ is a Cauchy sequence in $\struct{\Q, \norm{\,\cdot\,}_p}$ $\sequence {x_n}$ is a Cauchy sequence in $\struct{\Q_p, \norm{\,\cdot\,}_p}$

Proof
By definition, the $p$-adic norm is non-Archimedean on the $p$-adic numbers $\Q_p$.

From P-adic Norm is Non-Archimedean Norm, the $p$-adic norm is non-Archimedean on $\Q$.

By Characterisation of Cauchy Sequence in Non-Archimedean Norm:
 * $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$ $\displaystyle \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$.

Similarly:
 * $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q_p, \norm {\,\cdot\,}_p}$ $\displaystyle \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$.

By definition, the $p$-adic norm on the $p$-adic numbers is an extension of the $p$-adic norm on the rational numbers.

Hence:
 * $\displaystyle \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$ in $\struct {\Q, \norm {\,\cdot\,}_p}$ $\displaystyle \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$ in $\struct {\Q_p, \norm {\,\cdot\,}_p}$

The result follows.