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The following is a sentence from the proof of the theorem 1.2 (P.14-15) in Topological Vector Spaces (Third Printing Corrected 1971) by H.H.Schaefer

Finally, if $K$ is an Archimedean valuated field, then $|2| > 1$ for $2\in K$.

The following is the definition of non-Archimedean absolute value of a field in Handbook of Analysis and its foundation(P.261).

$|n e| \leq 1$ for every $n \in {\mathbb N}$,

where $e$ is the multiplicative identity of the field.

Then, I think that in an Archimedean valuated field all we can say is $|n e| > 1$ for some $n \in {\mathbb N}$. Is the statement in Topological Vector Spaces wrong ?

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Assume $|2e| \le 1$, so $|2^k e| \le 1$ for all $k \ge 0$. By writing $m$ in base $2$, you get from triangle inequality

$$|me| \le 1 + \log_2(m)$$

Take $n \in \mathbb{N}$ such that $|ne| = r > 1$ and apply the inequality to $m = n^k$. For all $k \ge 0$, you have

$$r^k \le 1 + k \ \log_2(n)$$

Which becomes impossible for $k$ sufficiently large.

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    Thank you for your help. Now I can understand.
    Is $(|n^k e|)_{k\geq 0}$ a mistake for $(|2^k e|)_{k\geq 0}$ ?
    2011-10-16
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    @Aki : I don't think there was an error, but to clarify, I removed the extraneous case ($|2e| < 1$).2011-10-16
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    Sorry for the silly question. It is of course correct. It is to show the contradiction to the assumption that $|ne|>1$ for some $n \in {\mathbb N}$.2011-10-16
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    Dear Joel. The new answer is simple and elegant. But the previous two step answer was more easy to understand for me.2011-10-16
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    @Aki : Yes the aim of the proof is to show that assuming both $|2e| \le 1$ and $|n e| > 1$ (for some $n$) yields a contradiction. Basically the two steps in the previous answers used the same method : bound powers of $2$, write any integer $m$ as a sum of power of $2$ (i.e. write it in base $2$), use triangle inequality, and apply with $m = n^k$ for $k$ sufficiently large.2011-10-16