complete ultrametric fieldCompleteUltrametricField2005-01-03 14:29:002008-04-23 11:24:37Theoremultrametric spaceultrametric fieldnon-archimedean fieldconvergence% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}A field $K$ equipped with a non-archimedean valuation\, $|\cdot|$\, is called a {\em non-archimedean field} or also an {\em ultrametric field}, since the valuation \PMlinkescapetext{induces} the ultrametric\, $d(x,\,y) = |x\!-\!y|$\, of $K$.
\begin{thmplain}
\,Let $(K,\,d)$ be a \PMlinkname{complete}{Complete} ultrametric field.\, A necessary and sufficient condition for the convergence of the {\em series}
\begin{align}
a_1\!+\!a_2\!+\!a_3\!+\ldots
\end{align}
in $K$ is that
\begin{align}
\lim_{n\to\infty}a_n = 0.
\end{align}
\end{thmplain}
{\em Proof.} Let $\varepsilon$ be any positive number.\, When (1) converges, it satisfies the Cauchy condition and therefore exists a number $m_\varepsilon$ such that surely
$$|a_{m+1}| = |\sum_{j=1}^{m+1}a_j-\sum_{j=1}^{m}a_j| < \varepsilon$$
for all\, $m \geqq m_\varepsilon$;\, thus (2) is necessary.\, On the contrary, suppose the validity of (2).\, Now one may determine such a great number $n_\varepsilon$ that
$$|a_m| < \varepsilon \quad \forall m \geqq n_\varepsilon.$$
No matter how great is the natural number $n$, the ultrametric then guarantees the inequality
$$|a_m\!+\!a_{m+1}\!+\ldots+\!a_{m+n}| \leqq
\max\{|a_m|,\,|a_{m+1}|,\,\ldots,\,|a_{m+n}|\} < \varepsilon$$
always when\, $m \geqq n_\varepsilon$.\, Thus the partial sums of (1) form a Cauchy sequence, which converges in the complete field.\, Hence the series (1) converges, and (2) is sufficient.