# Equivalence Class in P-adic Integers Contains Unique Coherent Sequence/Lemma 1

## Theorem

Let $p$ be a prime number.

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

Let $\sequence{\beta_n}$ be a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ such that:

$\forall j \in \N : \exists \mathop {\map N j} \ge j : \forall m, n \in \N: m, n \ge \map N j : \norm {\beta_n - \beta_m} \le p^{-\paren{j + 1}}$

Then:

$\forall j \in \N: \norm{\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren{j + 1}}$

## Proof

Let $j \in N$

Suppose $\map N {j + 1} \ge \map N j$

By definition:

$\norm{\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren{j + 1}}$

Now suppose $\map N j \ge \map N {j + 1}$

Then:

 $\displaystyle \norm{\beta_{\map N {j + 1} } - \beta_{\map N j} }_p$ $\le$ $\displaystyle p^{-\paren{j + 2} }$ $\displaystyle$ $<$ $\displaystyle p^{-\paren{j + 1} }$ Power Function on Integer between Zero and One is Strictly Decreasing

In either case:

$\norm{\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren{j + 1}}$

Since $j$ was arbitrary, then:

$\forall j \in \N: \norm{\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren{j + 1}}$

$\blacksquare$