  Let $X$ be a compact metrizable abelian group and $\mathbf{u}=\{u_n\}$ be a
sequence in its dual $X^{\wedge}$. Set $s_{\mathbf{u}} (X)= \{x: (u_n,x)\to
1\}$ and $\mathbb{T}_0^H = \{(z_n)\in \mathbb{T}^{\infty} : z_n\to 1 \}$. Let
$G$ be a subgroup of $X$. We prove that $G=s_{\mathbf{u}} (X)$ for some
$\mathbf{u}$ iff it can be represented as some dually closed subgroup
$G_{\mathbf{u}}$ of ${\rm Cl}_X G \times \mathbb{T}_0^H$. In particular,
$s_{\mathbf{u}} (X)$ is polishable. Let $\mathbf{u}=\{u_n\}$ be a $T$-sequence.
Denote by $(\widehat{X}, \mathbf{u})$ the group $X^{\wedge}$ equipped with the
finest group topology in which $u_n \to 0$. It is proved that $(\widehat{X},
\mathbf{u})^{\wedge} =G_{\mathbf{u}}$ and $\mathbf{n} (\widehat{X}, \mathbf{u})
= s_{\mathbf{u}} (X)^{\perp}$. We also prove that the group generated by a
Kronecker set can not be characterized.
