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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?

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Are you sure $G$ is not supposed to be locally compact ? –  francis-jamet Sep 22 '11 at 14:55
    
Yes, locally compact, and even compact if you want. –  Chera Sep 23 '11 at 6:38

1 Answer 1

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.

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