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Take the field $\mathbb{Q}$, If we complete it topologically with respect to the Euclidean norm, we get $\mathbb{R}$, then if we complete it algebraically, we get $\mathbb{C}$. On the other hand, the algebraic completion of $\mathbb{Q}$ is $\overline{\mathbb{Q}}$, and if we complete it with respect to the Euclidean norm now, we get $\mathbb{C}$. So it indeed terminates and in fact the result is the same field/space.

Now, another example is the topological completion of $\mathbb{Q}$ with respect to the p-adic norm, giving us $\mathbb{Q}_p$ the p-adic numbers, then the algebraic completion $\overline{\mathbb{Q}_p}$, and then once more with topological completion under p-adic norm, we get $\mathbb{C}_p$ which is the end.

Maybe this question can be better phrased in terms of ascending chains/homology but if anyone has any insight, that would be great.

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    $\begingroup$ I think the sentence "On the other hand, the algebraic completion of $\mathbb{Q}$ is $\overline{\mathbb{Q}}$ and if we complete it with respect to the Euclidean norm now, we get $\mathbb{C}$" is a bit confusing, because for a finite extension of $\mathbb{Q}$, there is no unique extension of the norm (the Euclidean norm on $\mathbb{Q}[\sqrt{2}]$ is not invariant under the Galois action). $\endgroup$ – YCor Feb 8 at 18:05
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    $\begingroup$ I think this answer of Laurent Moret-Bailly's at least addresses YCor's important objection (by observing that all extensions of the norm are conjugate) and gives some insight on the question. Does it make a difference if we consider valuations in an ordered abelian group versus norms valued in $\mathbb R$? $\endgroup$ – Tim Campion Feb 8 at 18:10
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    $\begingroup$ In fact, the problem does not make much sense if we don't restrict it to valued fields, because in general there is no preferred choice of a topology on the algebraic closure (not even up to isomorphism). $\endgroup$ – Laurent Moret-Bailly Feb 10 at 14:36
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    $\begingroup$ Another general comment: the (topological) completion of a topological field is not always a field. For instance, $\mathbb{Q}$ is dense in $\mathbb{Q}_2\times\mathbb{Q}_3$, which is therefore the completion of $\mathbb{Q}$ for the induced topology. (This topology is generated by the 2-adic and 3-adic ones). $\endgroup$ – Laurent Moret-Bailly Apr 3 at 9:26
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    $\begingroup$ @FrançoisG.Dorais It is the product ring, with the product topology. Density is just (a special case of) the approximation theorem for independent valuations. $\endgroup$ – Laurent Moret-Bailly Apr 3 at 18:35

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