Proposition 8

If the additive uniform structure of a topological field K is a structure of a complete Hausdorff space, then the multiplicative structure on K is a structure of a complete space.*

We shall show that if $\mathfrak{F}$ is a Cauchy filter with respect to the multiplicative structure on K*, then $\mathfrak{F}$ is a Cauchy filter with respect to the additive structure on K, and does not converge to 0; this will establish the result. Let U be any neighbourhood of 0 in K, let V be a closed neighbourhood of 0 such that $V \subset U, VV \subset U$ [axiom (AV$_\text{II}$)] and $-1 \notin V$; then by hypothesis there is a set $A \in \mathfrak{F}$ such that, for all $x \in A$ and $y \in A$, we have $x^{-1}y \in 1 + V$. Let $a \in A$, then $A \subset a + aV$, and $a + aV$ is a closed set which does not contain 0; hence 0 is not in the closure of A and is therefore not a cluster point of $\mathfrak{F}$. Let W be a neighbourhood of 0 such that $aW \subset V$ [axiom (AV$_\text{I}$)]; then by hypothesis there is a set $B \in \mathfrak{F}$ such that $B \subset A$ and, for all $x \in B$ and $y \in B$, we have $x^{-1}y \in 1 + W$; hence $y - x \in xW \subset AW \subset aW + aVW$; but K is commutative, and therefore $aVW = aWV \subset VW \subset U$; consequently $y - x \in U + U$ and the Proposition is proved.

The same proof shows that Proposition 8 can be extended to the case in which every Cauchy filter with respect to one of the multiplicative structures of K is also a Cauchy filter with respect to the other multiplicative structure.