Alternative description of the compact-open topology Let $G$ be a locally compact abelian topological group and let $\widehat{G}$ be its Pontryagin dual (I assume that $G$ is Hausdorff). In other words:
$$\widehat{G}:=\operatorname{Hom}_{\text{cont}}(G,\mathbb C^\times)\,.$$
Usually $\widehat G$ is endowed with the compact-open topology (which is defined by giving a subbase).
For any subset $S\subset G$ we define the annihilator $S^\perp\subset \widehat{G}$ in the following way:
$$S^\perp:=\{f\in\widehat{G}\colon f(S)=1\}\,.$$

Here the question:
Consider the following topology on $\widehat{G}$:

$U\subseteq \widehat{G}$ is open if there exisists a compact subgroup $W\subset G$ such that $W^\perp\subseteq U$.

Is it true that this topology is exactly the compact-open topology on $\widehat{G}$?
 A: Well, I have not checked that defines a topology on $\hat{G}$, but even if it did, it would not be the compact-open topology in general. Let us see it with $G=S^1$.
It is well known that the Pontrjagin dual of $S^1$ is the group of integers $\mathbb{Z}$ with the discrete topology. In fact, the character corresponding to the integer $n$ is the map $z \mapsto z^n$. On the other hand the only compact subgroups of $S^1$ are itself, the trivial subgroup and the subgroups of roots of unity, which are isomorphic to $\mathbb{Z}/m$ for some $m$. And with your definition of annihilator, we have
$$ (S^1)^{\perp} = \{ 0 \} $$
$$ \{ 1 \}^{\perp} = \mathbb{Z} $$
$$ (\mathbb{Z}/m)^{\perp} = m\mathbb{Z} $$
The first two equalities are obvious. For the third one, think that any $m$th root of unity by definition will be $1$ when exponentiated to $m$. That gives you $m\mathbb{Z} \subseteq (\mathbb{Z}/m)^{\perp}$. For the other contention, consider a primitive $m$th root of unity $\xi$. If $\xi^k=1$, then $m$ divides $k$. 
Now take any $n \neq 0$ in $\mathbb{Z} = \hat{S^1}$. Since $\mathbb{Z}$ has the discrete topology, the subset $\{ n \}$ is open, but it does not contain the annihilator of any compact subgroup of $S^1$.
