Equivalence Class in P-adic Integers Contains Unique Coherent Sequence/Lemma 3

Theorem

Let $p$ be a prime number.

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

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers as a quotient of Cauchy sequences.

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

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

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

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

Then:

$\sequence{\alpha_n}$ and $\sequence{\beta_n}$ are representatives of the same equivalence class in $\Q_p$.

Proof

From Representatives of same P-adic Number iff Difference is Null Sequence, it needs only to be shown that $\sequence{\alpha_j - \beta_j}$ is a null sequence.

Let $\epsilon \in \R_{> 0}$.

From Sequence of Powers of Number less than One, the sequence $\sequence{p^n}$ is a null sequence.

Then there exists $j \in \N$ such that $p^{-j} < \epsilon$.

Then for all $i \ge \map N j$:

 $\displaystyle \norm{\alpha_i - \beta_i}_p$ $=$ $\displaystyle \norm{\alpha_i - \beta_{\map N i} + \beta_{\map N i} - \beta_i}_p$ $\displaystyle$ $\le$ $\displaystyle \max \set{ \norm{\alpha_i - \beta_{\map N i} }_p, \norm{\beta_{\map N i} - \beta_i}_p}$ $\displaystyle$ $\le$ $\displaystyle \max \set{ \norm{p^{-\paren{i + 1} } }_p, \norm{\beta_{\map N i} - \beta_i}_p }$ $\displaystyle$ $\le$ $\displaystyle \max \set{ \norm{p^{-\paren{i + 1} } }_p, \norm{p^{-\paren{i + 1} } }_p }$ $\displaystyle$ $=$ $\displaystyle \norm{p^{-\paren{i + 1} }_p }_p$ $\displaystyle$ $\le$ $\displaystyle \norm{p^{-j }_p }_p$ $\displaystyle$ $<$ $\displaystyle \epsilon$

It follows that $\sequence{\alpha_n - \beta_n}$ is a null sequence by definition.

$\blacksquare$