Let $F$ be a local field with ring of integer $\mathfrak{o}$ and maximal ideal $\mathfrak{p}$. It is clear that $$\mathfrak{o}\supseteq \mathfrak{p}\supseteq \mathfrak{p}^2\supseteq \dots $$ is a fundamental system of open neighborhoods of $0$ in $F$.
It is known that $$\mathfrak{o} \to \lim_{\longleftarrow n\geq 1}\mathfrak{o}/\mathfrak{p}ⁿ$$ is a topological isomorphism.
Now, since $k:=\mathfrak{o}/\mathfrak{p}$ is finite and $\mathfrak{o}/\mathfrak{p}ⁿ$ is a finite dimensional vector space over $k$, we have that $\mathfrak{o}/\mathfrak{p}ⁿ$ is finite and so $\mathfrak{o}$ is compact.
With these informations I want to say that $F$, as additive group, is locally profinite.
I think that this is implicated by the fact that $\mathfrak{o}$ is compact and so it is a profinite group. But in which way $\mathfrak{o}$ profinite implies $F$ locally profinite?
I have the same problem with $F^\times$, in which we have a fundamental system of open neighbourhoods of $1$ in $F$ given by $$1+\mathfrak{o}\supseteq 1+\mathfrak{p}\supseteq 1+\mathfrak{p}^2\supseteq \dots $$