I am reading Andrew Baker's notes on p-adic number theory. I have a question regarding a result he is using. In Lemma 2.19 he says in the third line the limit $$\lim_{n\to\infty}N(a_n-b)$$ is positive, where $\{a_n\}$ is Cauchy, $N$ is a non-archimedean norm, and $b$ is not the limit of $a_n$ in the $N$-norm. I am not quite sure why this is true. The notes are obtainable here.
If ${a_n}$ is Cauchy and $N$ is a nonarchimedean norm, why is the limit of $N(a_n-b)$ positive
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0Please try to make the body of your post self-contained, not relying on the title/subject. – 2010-12-11
1 Answers
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I think you are missing the fact that $b$ is not the $N$-limit of the $a_n$, which you did not include in your original post (I just added it).
By definition, $N(a_n-b)$ is greater than or equal to $0$ (it is a norm). Since you are also assuming that $b\neq{\lim\limits_{n\to\infty}}^{(N)}a_n$, that means that $N(a_n-b)\not\to 0$ as $n\to\infty$; that is, the distance from $a_n$ to $b$ is bounded away from $0$; and since the terms are positive and bounded away from zero, the limit (which exists since we are dealing with a Cauchy sequence) must be strictly positive.
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0Thanks. The positivity condition was just a restatement of the fact that $b$ is not the limit, which as you rightly guessed, I missed. I will try to make my posts self contained. – 2010-12-11