Let $G$ be a locally compact abelian group and $(G_n)_{n \in \mathbb{Z}}$ a strictly increasing family of open compact subgroups of $G$ with $\cup G_n=G$ and $\cap G_n=\{0\}$.
We denote by $H_n=G_n^\perp$ the annihilator of $G_n$ in the dual $\hat{G}$.
How prove that $\cup H_n=\hat{G}$ ?
I am able to prove that $\overline{\cup H_n}=\hat{G}$ but removing the closure is a mystery for me.
Let $f\in\hat{G}$ and let $I_n\subseteq S^1$ be the image $f(G_n)$. Since $G_n$ is compact, $I_n$ is a closed subgroup of $S^1$. Any descending sequence of closed subgroups of $S^1$ stabilizes (they are either finite or all of $S^1$), so there exists $N\in\mathbb{Z}$ such that $I_n$ is the same for all $n\leq N$. Let $I$ be this common value of $I_n$. If $I$ is trivial, then $f$ annihilates $G_n$ for $n\leq N$, so we are done.
So suppose $I$ is not trivial; let $i\in I\setminus\{1\}$ and let $C_n=f^{-1}(\{i\})\cap G_n$. Then the sets $C_n$ are nonempty for all $n$ and form a decreasing sequence of compact subsets of $G$. The intersection $C=\bigcap C_n$ is thus nonempty. But $C\subseteq\bigcap G_n=\{0\}$, so this means $0\in C$. This is a contradiction, since $i\neq 1$ so $0\not\in f^{-1}(\{i\})$.