About closed subsets of the dual group $\widehat G$

Let $$G$$ be a topological abelian group (linearly topologized), not necessarilly locally compact and define its dual:

$$\widehat{G}:=\operatorname{Hom}(G,S^1)$$

We endow $$\widehat{G}$$ with the compact-open topology. Moreover For any subset $$S\subset G$$, let's introduce the following notation: $$S^\circ:=\{f\in \widehat{G}: f(S)=1\}\subset \widehat{G}\,.$$

My question is the following: Consider the family of subgroups of $$\widehat{G}$$:

$$\mathcal F:=\{C^\circ:C\subset G \text{ is closed and compact}\}$$

Is it a local basis at $$1$$ of closed subgroups? In any case, what is the relationship between closed subsets (or subgroups) of $$\widehat{G}$$ and $$\mathcal F$$?

This answer is partial. I hope it will be useful for others.

I understood that $$G$$ is linearly topologized as there exists a local base $$\mathcal B$$ at $$1$$ consisting of open subgroups of $$G$$. That is $$\bigcap\mathcal B=\{1\}$$ and for each $$H,H’\in\mathcal B$$, $$H\cap H’\in\mathcal B$$. Next, by $$\operatorname{Hom}(G, S^1)$$ I understood a group of continuous homomorphisms from $$G$$ to the unit circle $$S^1=\{|z|\in\Bbb C:|z|=1\}$$ endowed with the topology and multiplication inherited from usual these of $$\Bbb C$$.

Put $$V_0=\{z\in S^1:\operatorname{Re} z>0\}$$. If $$f\in \widehat{G}$$ then there exist $$H\in\mathcal B$$ such that $$f(H)\subset V_0$$. Since $$V_0$$ has no non-trivial subgroups, we have $$f(H)=\{1\}$$. Conversely, any homomorphism $$f:G\to S^1$$ such that $$f(H)=\{1\}$$ for some $$H\in\mathcal B$$, is continuous. Thus for any $$f\in\widehat G$$ we have $$\operatorname{ker} f\in\mathcal B$$.

Then for any nonempty subset $$S$$ of $$G$$, $$S^\circ=\{f\in \widehat{G}: S\subset \operatorname{ker} f\}=\{f\in \widehat{G}: \langle S\rangle\subset \operatorname{ker} f\}=\langle S\rangle^\circ,$$ where $$\langle S\rangle$$ is the subgroup of $$G$$ generated by $$S$$. Thus $$\mathcal F=\{H^\circ: H$$ is a compactly generated subgroup of $$G\}$$.

I assume that the operation of $$\widehat G$$ is defined by putting $$(fg)(x)=f(x)g(x)$$ for each $$f,g\in\widehat G$$ and $$x\in G$$. Then the identity of $$\widehat G$$ is the annulating homomorphism which maps each element of $$G$$ to $$1$$.

Thus $$S^\circ$$ is a subgroup of $$G$$ for any nonempty subset $$S$$ of $$G$$.

Is $$\mathcal F$$ a local basis at $$1$$ of closed subgroups?

The local base $$\widehat{\mathcal B}$$ at $$1$$ of $$\widehat G$$ consists of the sets $$(C,V)$$, where $$C$$ is a compact subset of $$G$$ and $$V\subset S^1$$ is an open neighborhood of $$1$$. Clearly, for any $$(C,V)\in \widehat{\mathcal B}$$ we have $$C^\circ\subset (C,V)$$. I don’t know whether for any non-empty compact subset $$C$$ of $$G$$ there exists a compact subset $$C’$$ of $$G$$ and an open neighborhood $$V\subset S^1$$ of $$1$$ such that $$(C’,V)\subset C^\circ$$.

what is the relationship between closed subsets (or subgroups) of $$\widehat{G}$$ and $$\mathcal F$$?

Clearly, $$S^\circ$$ is a subgroup of $$\widehat{G}$$ for any non-empty subset $$S$$ of $$G$$. Moreover, $$S^\circ$$ is closed. Indeed, suppose to the contrary that there exists $$f\in \overline{S^\circ}$$ and $$x\in S$$ such that $$f(x)\ne 1$$. Then $$(\{x\},S^1\setminus\{1\})$$ is an open neighborhood of $$f$$, so there exists an element $$g\in (\{x\},S^1\setminus\{1\})\cap S^\circ$$. Since $$g\in (\{x\},S^1\setminus\{1\})$$, $$g(x)\ne 1$$. On the other hand, since $$g\in S^\circ$$ and $$x\in S$$, $$g(x)=1$$, a contradiction. I don’t know whether each closed subgroup of $$\widehat G$$ belongs to $$\mathcal F$$.