Dual of a complete topological group

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?

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

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.
• Sure, but that doesn't eliminate the need for $G$ to be compactly generated. Feb 2, 2019 at 22:48