Corollary 3

Let H be a subgroup of G. Then $(H^{\bot})^{\bot}= H$.

Since $H^{\bot}$ = $\overline{H}^{\bot}$, we may suppose that H is closed. Let $f: H\rightarrow G$ be the canonical injection and $p: G\rightarrow G/H$ the canonical projection. We have $H^{\bot}=$ Ker($\widehat{f}$) (Lemma 10). Let $k$ be the canonical injection of $H^{\bot}$ into $\widehat{G}$. By Theorem 4, the morphism $\widehat{p}$ induces an isomorphism $\widehat{p}_H:\widehat{G}/H\rightarrow H^{\bot}$ of topological groups and we have $k\circ \widehat{p}_H=\widehat{p}$. Consequently (Corollary 1), we obtain

$(H^{\bot})^{\bot}=$ Ker($\widehat{k}$) $=$ Ker( $\widehat{\widehat{p}}_H\circ \widehat{k}$) $=$ Ker($p$) $= H$.