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For a profinite group $G={\underleftarrow{\lim }}_j{G}_j$, a subset $S\subseteq G$ is a topological generating set exactly when $⟨p_j(S)⟩=p_j(G)$ in every finite quotient. Indeed, a subgroup of a profinite group is dense exactly when its image in every finite continuous quotient is surjective.

Apply part i toThe element $\alpha$ generates the additive cyclic group $\mathbb{Z}/p^n\mathbb{Z}$ exactly when it is coprime to $p$, which is equivalent to its reduction modulo $p$ being nonzero. Hence $\{\alpha \}$ is a topological generating set of the additive group ${\mathbb{Z}}_p$ if and only if $\alpha \not\equiv 0(\mathrm{mod}p)$.

A single topological generator of a topological group $G$ is an element $g$ whose cyclic subgroup is dense:More generally, a tuple $g_1,\dots ,g_d$ consists of topological generators when the subgroup they jointly generate is dense. Membership in such a tuple does not imply that one element alone generates densely.

In the additive group ${\mathbb{Z}}_p$ of p-adic integers, the element one is a topological generator: ordinary integers surject onto every finite quotient $\mathbb{Z}/p^n\mathbb{Z}$. The element $p$ generates only the proper closed subgroup $p{\mathbb{Z}}_p$. The profinite group ${\mathbb{Z}}_p^2$ has a two-element topological generating set but no single topological generator, since its quotient $(\mathbb{Z}/p\mathbb{Z}{)}^2$ is not cyclic. ^[ref]

A topological group is topologically finitely generated when it has a finite topological generating set.