Let $(G,+)$ be an Hausdorff topological abelian group. Assume that $G$ is complete, i.e. every Cauchy sequence in $G$ converges in $G$.

Let $\widehat G:=\operatorname{Hom}_{\text{cont}}(G, S^{1})$ be the dual group of $G$ and put on $\widehat G$ the compact-open topology.

Here my question:

Is also $\widehat{G}$ a complete group?

Thanks in advance

  • $\begingroup$ This will most likely be unrelated to the completeness of $G$, but more to things like its local compactness $\endgroup$ Feb 2, 2019 at 18:43

1 Answer 1


I don't know the answer in general, but I suspect the following answer is good enough for your purposes (since completeness just with respect to sequences is not very useful anyways in full generality). Namely, if $G$ is any compactly generated topological abelian group (not necessarily complete), then $\widehat{G}$ is complete. Most spaces you will run into are compactly generated unless you are specifically trying to find pathological examples; in particular, for instance, any metrizable space is compactly generated.

To prove $\widehat{G}$ is complete if $G$ is compactly generated, suppose $(f_n)$ is a Cauchy sequence in $\widehat{G}$. Then in particular $(f_n(x))$ is Cauchy for each $x\in G$, and so we may define a function $f$ as the pointwise limit of $f_n$. We then see that $f$ is a homomorphism since it is a pointwise limit of homomorphisms. Moreover, $(f_n)$ is uniformly Cauchy on each compact subset of $G$, so $f_n\to f$ uniformly on compact sets. Thus $f$ is continuous when restricted to each compact subset of $G$. Since $G$ is compactly generated, this means that $f$ is continuous. Thus $f\in\widehat{G}$, and $f_n\to f$ in $\widehat{G}$ since the convergence is uniform on compact sets.

  • $\begingroup$ Your answer can be adapted for nets anyway, it seems to me. $\endgroup$ Feb 2, 2019 at 22:41
  • $\begingroup$ Sure, but that doesn't eliminate the need for $G$ to be compactly generated. $\endgroup$ Feb 2, 2019 at 22:48
  • $\begingroup$ No of course, but you said "completeness just with respect to sequences is not very useful" : completeness with respect to nets is all we can want, isn't it ? $\endgroup$ Feb 2, 2019 at 22:51
  • $\begingroup$ Right, but my point was kind of meta: if OP is the kind of person who would ask a question that refers to completeness with respect to sequences, they probably don't care about non-compactly generated groups and so this answer is good enough for them. $\endgroup$ Feb 2, 2019 at 22:53
  • $\begingroup$ Oh, ok, I see ! ( I was actually going to write the same answer, but with nets and with locally compact instead of compactly generated - I thought that it would be enough too) $\endgroup$ Feb 2, 2019 at 22:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .