It's shown in cource of topology that metric space is compact iff it satisfies Finite Intersection Axiom iff it is Sequentially Compact.
In cource of calculus it's correspondingly shown that in Archimedean fields Cauchy completeness equivalent to Nested intervals (Archimedean) property, and equivalent to Bolzano–Weierstrass property.
This two chains of equivalence seems to me very similar, and i wandering if it's enough to prove topological one and state some strong relation between compactness and completeness in metrizable fields (Archimedean or not) – Heine-Borel property is not fit (set of reals is complete, but not bounded, so not compact).
So, it's answered here that any locally compact valued field is complete, i'm wandering if converse is true?
Best Answer
No, here is a counter-example: Let $k$ be an infinite field, $A:= k[[x]]$ the ring of formal power series over $k$, $K = k((x))$ the quotient field of $A$. Then the $xA$-adic topology on $K$ is induced by an absolute valuation, hence $K$ with this topology is a metrisable, topological field. It is well-known that $K$ is complete with respect to the metric induced by the absolute valuation. However, $K$ is not locally compact:
Assume it were. Then $A$ is compact. Since its maximal ideal $xA$ is open, the residue field $k \cong A/xA$ is finite. Contradiction.