Leigh.Samphier/Sandbox/P-adic Norm of p-adic Number is Power of p/Lemma

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $p$ be a prime number.

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

Let $\sequence {x_n}$ be a Cauchy sequence in $\struct{\Q, \norm {\,\cdot\,}_p}$ such that $\sequence {x_n}$ does not converge to $0$.


Then:

$\exists v \in \Z: \displaystyle \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$


Proof

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


Since $\sequence {x_n}$ does not converge to $0$, from Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary:

$\exists N \in \N: \forall n, m > N: \norm {x_n}_p = \norm {x_m}_p$

By definition of the $p$-adic norm on $\Q$:

$\exists v \in \Z: \norm {x_{N + 1} }_p = p^{-v}$

Therefore:

$\forall n > N: \norm {x_n}_p = \norm {x_{N + 1} }_p = p^{-v}$


Let $\sequence {y_n}$ be the subsequence of $\sequence {\norm {x_n}}$ defined by:

$\forall n: y_n = \norm {x_{N + n} }_p$

Then $\sequence {y_n}$ is the constant sequence $\tuple {p^{-v}, p^{-v}, p^{-v}, \dotsc}$.


Therefore:

\(\displaystyle \displaystyle \lim_{n \mathop \to \infty} \norm {x_n}_p\) \(=\) \(\displaystyle \displaystyle \lim_{n \mathop \to \infty} y_n\) Limit of Subsequence equals Limit of Real Sequence
\(\displaystyle \) \(=\) \(\displaystyle p^{-v}\) Constant Sequence Converges to Constant in Normed Division Ring

$\blacksquare$