2
$\begingroup$

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 ?

1 Answers 1

3

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.

  • 1
    @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