For a locally compact abelian (LCA) group $G$ let $\hat{G}$ denotes the dual group (the group of characters on $G$). Now with the compact open topology $\hat{G}$ becomes a LCA group. I need to prove the following -
1) $\hat{\mathbb{Z}} \cong \mathbb{S}^1$
2) $\hat{\mathbb{S}}^1 \cong \mathbb{Z}$
3) $\hat{\mathbb{R}} \cong \mathbb{R}$
as LCA groups. I have absolutely no idea how to do this. Help is appreciated..








This is very basic. The first isomorphism is almost tautological: $\hat{\mathbb{Z}}$ is the group of (continuous) homomorphisms $\mathbb{Z}\to\mathbb{S}^1$ and this is algebraically and topologically isomorphic to $\mathbb{S}^1$.
If you already know Pontryagin duality, then the second statement is again obvious. Without Pontryagin duality, you have to prove
For $\hat{\mathbb{R}}$, it's a bit more difficult. For this it is convenient to consider $\mathbb{S}^1=\mathbb{R}/\mathbb{Z}$, so we have the canonical map $\pi\colon \mathbb{R}\to\mathbb{R}/\mathbb{Z}$.
For $r\in\mathbb{R}$, define $\alpha_r\colon\mathbb{R}\to\mathbb{R}/\mathbb{Z}$ by $$ \alpha_r(x)=\pi(rx) $$ It is clear that $\alpha_r\in\hat{\mathbb{R}}$ and it's not really difficult to show that $r\mapsto\alpha_r$ is a topological isomorphism.