A valuation satisfying the triangle inequality gives a metric for the
topology on defining the distance from
to
to be
.
Assume for the rest of this section that we only consider valuations
that satisfy the triangle inequality.
Proof.
It suffices to prove that if
![$ \left\vert x_n\right\vert _1\to 0$](img1654.png)
then
![$ \left\vert x_n\right\vert _2\to 0$](img1655.png)
, since the proof of the
other implication is the same.
Let
![$ \varepsilon >0$](img330.png)
. The topologies induced by the two absolute
values are the same, so
![$ B_2(0,\varepsilon )$](img1656.png)
can be covered by
open balls
![$ B_1(a_i,r_i)$](img1657.png)
. One of these open balls
![$ B_1(a,r)$](img1658.png)
contains
0. There is
![$ \varepsilon '>0$](img1659.png)
such that
Since
![$ \left\vert x_n\right\vert _1\to 0$](img1654.png)
, there exists
![$ N$](img5.png)
such
that for
![$ n\geq N$](img1661.png)
we have
![$ \left\vert x_n\right\vert _1 <\varepsilon '$](img1662.png)
.
For such
![$ n$](img19.png)
, we have
![$ x_n\in B_1(0,\varepsilon ')$](img1663.png)
, so
![$ x_n\in B_2(0,\varepsilon )$](img1664.png)
,
so
![$ \left\vert x_n\right\vert _2<\varepsilon $](img1665.png)
. Thus
![$ \left\vert x_n\right\vert _2\to 0$](img1655.png)
.
Proof.
If
![$ x\in K$](img655.png)
and
![$ i=1,2$](img1667.png)
, then
![$ \left\vert x^n\right\vert _i \to 0$](img1668.png)
if and only if
![$ \left\vert x\right\vert _i^n\to 0$](img1669.png)
, which is the
case if and only if
![$ \left\vert x\right\vert _i<1$](img1670.png)
. Thus
Lemma
16.1.3 implies that
![$ \left\vert x\right\vert _1<1$](img1671.png)
if and only if
![$ \left\vert x\right\vert _2<1$](img1672.png)
.
On taking reciprocals we see that
![$ \left\vert x\right\vert _1>1$](img1673.png)
if and only if
![$ \left\vert x\right\vert _2>1$](img1674.png)
, so finally
![$ \left\vert x\right\vert _1 = 1$](img1675.png)
if and only if
![$ \left\vert x\right\vert _2=1$](img1676.png)
.
Let now
be nonzero elements with
and
. On
applying the foregoing to
we see that
if and only if
Dividing through by
![$ \log\left\vert z\right\vert _i$](img1683.png)
, and rearranging,
we see that for every rational number
![$ \alpha=-n/m$](img1684.png)
,
Thus
so
Since this equality does not depend on the choice of
![$ z$](img1061.png)
,
we see that there is a constant
![$ c$](img352.png)
(
![$ =\log\left\vert z\right\vert _1/\log \left\vert z\right\vert _2$](img1688.png)
)
such that
![$ \log\left\vert w\right\vert _1/\log \left\vert w\right\vert _2 = c$](img1689.png)
for all
![$ w$](img1690.png)
.
Thus
![$ \log\left\vert w\right\vert _1 = c\cdot \log\left\vert w\right\vert _2$](img1691.png)
, so
![$ \left\vert w\right\vert _1 = \left\vert w\right\vert _2^c$](img1692.png)
, which implies that
![$ \left\vert \cdot \right\vert _1$](img1645.png)
is equivalent to
![$ \left\vert \cdot \right\vert _2$](img1646.png)
.