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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?

nagvalhm
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  • If you have a metric space $(X,d)$, then it is locally compact if and only if every point has a compact neighborhood (i.e. open ball with radius $>0$ such that the closure is compact). Complete is not a topological property, so you cannot use it in the definition of locally compactness. – Amirreza Hashemi Oct 13 '22 at 15:37
  • @Amirreza Hashemi, complete is metric space property. I'm asking if field, which is metric space, complete, then topology space induced by metric is locally compact. In general it's not true - irrational numbers - complete but not locally compact, but they don't form a field (even ring) – nagvalhm Oct 13 '22 at 16:26

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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.

Ulli
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