P-adic Norm not Complete on Rational Numbers

Theorem
Let $\norm {\,\cdot\,}_p$ be the p-adic norm on the rationals $\Q$.

Let $d_p$ be the p-adic metric; that is, the metric induced by $\norm {\,\cdot\,}_p$.

Then:


 * $\struct {\Q, d_p}$ is not a complete metric space.

Informally, the valued field $\struct {\Q, \norm {\,\cdot\,}_p }$ does not define a complete metric space.

Proof
By definition of the p-adic metric:


 * $\forall x, y \in \Q: d_p \paren {x, y} = \norm {x - y}_p$

To show that $\struct {\Q, d_p}$ is not complete we need to show there exists a Cauchy sequence in $\Q$ which does not converge in $\struct {\Q, d_p}$.

We note that convergence in the metric space $\struct {\Q, d_p}$ is equivalent to convergence in the normed division ring $\struct {\Q, \norm {\,\cdot\,}_p }$.

Consider the sequence $\sequence {x_n} \subseteq \Q$ where $x_n = a^{p^n}$ for some $a \in \Z: 1 \lt a < p-1$.

Let $n \in \N$.

Then:


 * $\norm {a^{p^{n + 1} } - a^{p^n} }_p = \norm {a^{p^n} (a^{p^n \left({p - 1}\right)} - 1) }_p$

From the corollary to Euler's Theorem:
 * $a^{p^n \left({p - 1}\right)} - 1 \equiv 0 \pmod {p^n}$

so:
 * $\norm {a^{p^n} \left({a^{p^n \left({p - 1}\right)} - 1}\right)}_p \le p^{-n} \xrightarrow {n \to \infty} 0$

That is:
 * $\displaystyle \lim_{n \to \infty} \norm {x_{n+1} - x_n } = 0$

By Characterisation of Cauchy Sequence in Non-Archimedean Norm


 * $\sequence {x_n }$ is a cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p }$.

$\sequence {x_n}$ converges to some $x \in \Q$.

That is:
 * $x = \displaystyle \lim_{n \mathop \to \infty} x_n$

By Modulus of Limit/Normed Division Ring:
 * $\displaystyle \lim_{n \mathop \to \infty} \norm {x_n }_p = \norm {x }_p$

Since $\forall n, p \nmid a^{p^n} = x_n$, then:
 * $ \norm {x_n }_p = 1$

So:
 * $\norm {x }_p = \displaystyle \lim_{n \mathop \to \infty} \norm {x_n }_p = 1$

By Axiom (N1) of a normed division ring then:
 * $x \ne 0$.

Since:

and $x \ne 0$ then:


 * $x^{p-1} = 1$.

To arrive at a contradiction we have two cases to consider: $x = a$ and $x \ne a$.

Case: $x = a$
Suppose $x = a$ then:
 * $a^{p-1} = 1$

So:
 * $a = 1$ or $a = -1$

Since by choice of $a$:
 * $a \gt 1$

This is a contradiction.

Case: $x \ne a$
Suppose $x \ne a$.

By Axiom (N1) of a normed division ring then:
 * $\norm {x - a}_p \gt 0$

Since $x_n \to x$ as $n \to \infty$ then:
 * $\exists N: \forall n \gt N: \norm {x_n - x}_p \lt \norm {x - a}_p$

That is:
 * $\exists N: \forall n \gt N: \norm {a^{p^n} - x}_p \lt \norm {x - a}_p$

Let $n \gt N$:

Since $\norm {x - a^{p^n}}_p \lt \norm {x - a}_p$ then:

From:
 * $\norm {x - a}_p = \dfrac 1 {p^{\nu_p \paren {x - a} } } < 1$

it follows that:
 * $p \divides \left({x - a}\right)$

We have that $a \ne 1$ and $a \ne p - 1$.

Hence from $x^p = x$ it follows that $x$ must be a non-trivial $p-1$-th root of unity which is also in $\Q$.

This is a contradiction.

In conclusion:
 * $\sequence {x_n}$ is a Cauchy sequence that does not converge in $\struct {\Q, \norm {\,\cdot\,}_p }$.