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'complete ultrametric field'
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| Title of object: |
complete ultrametric field |
| Canonical Name: |
CompleteUltrametricField |
| Type: |
Theorem |
| Created on: |
2005-01-03 14:29:00 |
| Modified on: |
2005-01-03 15:59:08 |
| Classification: |
msc:54E35, msc:12J10 |
| Keywords: |
convergence |
| Defines: |
ultrametric field, non-archimedean field |
Preamble:
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\usepackage{amssymb}
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Content:
A field $K$ together with its ultrametric $d$ is an {\em ultrametric field}; we can also speak of a {\em non-archimedean field} if the ultrametric is \PMlinkescapetext{induced} by a non-archimedean valuation of $K$.
\textbf{Theorem.} \,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+...
\end{align}
in $K$ is that
\begin{align}
\lim_{n\to\infty}a_n = 0.
\end{align}
{\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
$$d(a_{m+1},\,0) = d(\sum_{j=1}^{m+1}a_j-\sum_{j=1}^{m}a_j,\,0) < \varepsilon$$
for all \,$m \geqq m_\varepsilon$; thus (2) is necessary. \,On the contrary, suppose the validity of (2). \,Now one may determine a number $n_\varepsilon$ so that
$$d(a_m,\,0) < \varepsilon \quad \forall m \geqq n_\varepsilon.$$
No matter how great is the natural number $m$, one has the inequality
$$d(a_m+a_{m+1}+...+a_{m+n},\,0) \leqq
\max\{d(a_m,\,0),\,d(a_{m+1},\,0),\,...,\,d(a_{m+n},\,0)\} < \varepsilon$$
always when \,$n \geqq n_\varepsilon$. \,Thus the partial sums of (1) form a Cauchy sequence, which converges in the complete field. \,So the series (1) converges, and (2) is sufficient. |
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