This is somewhat related to the question and the answers here:

Is completeness of a field an algebraic property?

My question is (to which I believe the answer must have been known), does **every** extension of of a non-trivial non-archimedean valuation on $\mathbf{Q}$ to $\mathbf{R}$ fail to make $\mathbf{R}$ into a complete field? (Valuations considered here are general Krull valuations.)

I guess the notion of Cauchy sequence should be generalized a little bit, considering that $\mathrm{cf}(2^{\aleph_0}) > \aleph_0$, i.e. one should allow a Cauchy sequence of any well order type.

Thanks!

**Edit:**I apologize that I did not make the question as clear as it should have been. Let me explain it furthermore here.

All the valuations here are Krull's general valuations, which could be of any rank, not necessarily absolute values.

For a valued field with higher-rank value group $\Gamma$, the definition of Cauchy sequences (or completeness) depends on the value group, in the sense that a Cauchy sequence must be of length $\mathrm{cf}(\Gamma)$. (hence my second paragraph above). The definition could be found in Engler and Prestel's book

*Valued Fields*. More specifically, let $\kappa=\mathrm{cf}(\Gamma)$, and a sequence $(a_\nu)_{\nu<\kappa}$ is said to be a Cauchy sequence if for every $\gamma\in \Gamma$, there exists some $\delta <\kappa$ such that for all $\mu, \nu \in (\delta, \kappa)$, one has$v(a_\mu-a_\nu)>\gamma$.

And the valued field is said to be complete if every Cauchy sequence has a limit, meaning there exist some $a$ in the field, such that $v(a_\nu-a)>\gamma$ for any $\gamma$ and sufficiently big $\nu$, which says that $a$ is a limit point of the sequence in the topological sense.

everyextension of a nontrivial Krull valuation...". – Pete L. Clark Nov 14 '10 at 4:29Valued Fieldsthey write: "In the theory of valuations, the notion of a completion had to be replaced by that of the so-called henselization." – Pete L. Clark Nov 14 '10 at 8:33