Bourbaki

Exercise 12, § 3

General Topology, IX

Bourbaki marks this one as harder than the rest.

Let KK be a non-discrete, Hausdorff, topological division ring, and let φ:K→R+\varphi : K \to \mathbf{R}_+ be a mapping such that φ(0)=0,φ(xy)=φ(x)φ(y)\varphi(0) = 0, \varphi(xy) = \varphi(x)\varphi(y) and such that if VnV_n denotes the set of all x∈Kx \in K such that φ(x)≤1/n\varphi(x) \leq 1/n, the sets VnV_n form a fundamental system of neighbourhoods of 00 in KK.

a) Show that there exists a real number a>0a > 0 such that, for all x∈Kx \in K,

φ(1+x)≤a(1+φ(x)) \varphi(1 + x) \leq a(1 + \varphi(x))
[if not, there would exist a sequence (xn)(x_n) of points of KK such that both (1+xn)−1(1 + x_n)^{-1} and xn(1+xn)−1x_n(1 + x_n)^{-1} tended to zero]. Hence show that if ψ(x)=(φ(x))α\psi(x) = (\varphi(x))^{\alpha} we have
ψ(x+y)≤2sup⁡(ψ(x),ψ(y)) \psi(x + y) \leq 2 \sup (\psi(x), \psi(y))
for α\alpha sufficiently small.

b) Show that if n=2pn = 2^p we have ψ(∑i=1nxi)≤nsup⁡(ψ(xi))\psi \left( \sum_{i=1}^n x_i \right) \leq n \sup (\psi(x_i)), and deduce that for every integer m>0m > 0 we have ψ(m)≤2m\psi(m) \leq 2m.

c) Deduce from b) that for every n=2pn = 2^p and every x∈Kx \in K we have ψ((1+x)n−1)≤2n(1+ψ(x))n−1\psi((1 + x)^{n-1}) \leq 2n(1 + \psi(x))^{n-1}, and hence that ψ\psi is an absolute value on KK which defines the topology of KK.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.