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My previous question has been answered by YCor; so I am asking a new one with a reasonable additional assumption. See the previous question for the background and motivation.

General question: does there exist a complete, nondiscrete topological group $G$ such that all subgroups of $G$ are closed? Or, does there exist a complete, nondiscrete topological vector space $V$ such that all vector subspaces of $V$ are closed?

More specific question: does there exist a complete, nondiscrete topological abelian group $A$ with linear topology such that all subgroups of $A$ are closed? Or, does there exist a complete, nondiscrete topological vector space $V$ with linear topology such that all vector subspaces of $V$ are closed?

Let me collect here the less obvious definitions appearing in the questions above. A topological abelian group $A$ is said to have linear topology if the open subgroups form a base of neighborhoods of zero in $A$. A topological vector space $V$ is said to have linear topology if open vector subspaces form a base of neighborhoods of zero in $V$. The notion of a topological vector space with linear topology presumes that the topology on the ground field $k$ is discrete.

I define completeness in the case of a linear topology only, as the general case is more involved. A topological abelian group $A$ with linear topology is said to be (Hausdorff and) complete if the natural map $A\longrightarrow \varprojlim_{U\subset A}A/U$ is bijective. Here the projective limit is taken over all the open subgroups $U\subset A$. For topological vector spaces with linear topology, the definition of completeness is similar.

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    $\begingroup$ For a Hausdorff abelian topological group, is complete the same as: for every net $(x_i)$ such that $x_i-x_j$ tends to $0$ when both $i,j$ tend to infinity, the net $(x_i)$ is convergent? $\endgroup$
    – YCor
    Commented Dec 2, 2020 at 19:35
  • $\begingroup$ I edited in a link to @YCor's comments answering the question because, somehow, their answer was not showing up for me. To avoid bumping, I won't edit again, but please feel free to correct my edit if you like. $\endgroup$
    – LSpice
    Commented Dec 2, 2020 at 19:36
  • $\begingroup$ @YCor Yes, I think this is a correct definition. Alternatively, one can define the uniform structure associated with a group topology, and define what it means for a uniform space to be complete. These should be equivalent definitions. $\endgroup$
    – Leonid Positselski
    Commented Dec 2, 2020 at 19:38
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    $\begingroup$ @YCor's definition is the definition of completeness under a uniformity. Under the projective-limit definition, the limit of the Cauchy net $(x_i)$ is $(x_{i(U)})_U$, where $i(U)$ is any element such that $j \ge i(U) \implies x_{i(U)} - x_j \in U$. Alternatively, under the uniform-completeness definition, any element $(x_U)_U$ of the projective limit is a Cauchy net in $A$, and so has as a lift in $A$ its limit. $\endgroup$
    – LSpice
    Commented Dec 2, 2020 at 19:41
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    $\begingroup$ Just a remark: it can't be a Polish abelian group. For a nondiscrete Polish abelian group has cardinal $c$, but always has $2^c$ subgroups, while it has only $c$ closed subsets. $\endgroup$
    – YCor
    Commented Dec 2, 2020 at 21:36

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