Separability of a group and its dual Here is the following exercise:

Let $G$ be an abelian topological group. $G$ has a countable topological basis iff its dual $\hat G$ has one. 

I am running into difficulties with the compact-open topology while trying this. Any help?

Let $G$ a topological locally compact abelian group.
If $G$ has a countable topological basis $(U_n)_{n \in \mathbb{N}}$.
We show $\hat{G}$ has a countable topological basis.
For every finite subset $I$ of $\mathbb{N}$, let $O_I=\cup_{i \in I}U_i$.
We define $B:=\{\bar{O_I} | \bar{O_I}$ is compact $\}$.
$B$ is countable, because the cardinality is lower or equal than the cardinality of the finite subset of $\mathbb{N}$.
$U(1)$ has a countable topological basis $(V_n)_{n \in\mathbb{N}}$.
Let $O(K,V)=\{ \chi \in \hat{G} | \chi(K) \subset V\}$, with $K$ compact in $G$, $V$ open in $U(1)$.
$O(K,V)$ is open in the compact-open topology on $\hat{G}$.
Let $B'=\{O(K,V_n)|K \in B, n \in \mathbb{N}\}$. $B'$ is countable and is a topological basis of $\hat{G}$.
If $\hat{G}$ has a countable topological basis, $\hat{\hat{G}}$ too. But $\hat{\hat{G}}=G$ (Pontryagin's duality), so $G$ has a topological basis.
