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 $\mathcal F$?
 A: 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 $\mathcal F$.  
