Is $\mathbb{A}_k^n(k)$ dense in the Berkovich analytification of $\mathbb{A}_k^n$?

Question

Let $k$ be a non-archimedean field and denote by $\mathbb{A}_k^n$ the analytic affine space of $n$ dimensions over $k$ (analytic in the sense of Berkovich). There is a natural injective map of sets $\mathbb{A}_k^n(k) \to \mathbb{A}_k^n$. Is $\mathbb{A}_k^n(k)$ mapped onto a dense subset of $\mathbb{A}_k^n$? Clearly, this is only possible if the value group of $k$ is dense. I know that it is true if $k$ is non-trivially valued and algebraically closed (by Proposition 2.1.15 in Berkovich's monograph). So it remains to treat the densely valued, but not necessarily algebraically closed case.

Answer 1

If $k$ is not algebraically closed, then $\mathbb A^n_k(k)$ is not doing to be dense with $\mathbb A_k^n$. I'll show this for $n=1$ for simplicity. Take any point $P$ in $\mathbb A^1_k$ with a residue field $K$ which is a proper finite extension $k$. Since $k$ is complete, $k$ is closed as a subspace of $K$, and hence the element $\alpha\in K\setminus k$ corresponding to $P$ (unique up to Galois conjugates) will have an open neighbourhood which doesn't intersect $P$. This neighbourhood gives rise to a nonempty neighbourhood of $P$ in $\mathbb A_k^1$ which does not intersect $\mathbb A_k^1(k)$.

Edit: in hindsight I admit the construction of a neighbourhood is not that straightforward, so here are the details. Let $\alpha_1=\alpha,\alpha_2,\dots,\alpha_n$ be the conjugates of $\alpha$ and let $f(x)=(x-\alpha_1)\dots(x-\alpha_n)$. By the argument above there is some $\varepsilon>0$ such that $||x-\alpha_i||\geq\varepsilon$ for all $x\in k$, where $||\cdot||$ is the norm on the algebraic closure of $k$, and hence $||f(x)||\geq\varepsilon^n$. Now we can consider the open subset $U$ of the Berkovich space $\mathbb A_k^1$ consisting of seminorms $|\cdot|$ such that $|f|<\varepsilon^n$. By the above, no point in $\mathbb A_k^1(k)$ lies in $U$.