ScalingStacks

Proof. As $\phi$ is continuous, $\mathbb{D}^{\vee}(L,\phi)\setminus\mathbb{O}^{\mathrm{an}}$ is an open subset of $Tot(L^{\vee})^{\mathrm{an}}\setminus\mathbb{O}^{\mathrm{an}}$. By Lemma 3.19, under the map $p(\boldsymbol{0})^{\mathrm{an}}$, the image of $\mathbb{D}^{\vee}(L,\phi)\setminus\mathbb{O}^{\mathrm{an}}$ is an open subset of $(\spec V_{{\scriptscriptstyle\bullet}}(L))^{\mathrm{an}}\setminus\boldsymbol{0}^{\mathrm{an}}$, so it is also an open subset of $(\spec V_{{\scriptscriptstyle\bullet}}(L))^{\mathrm{an}}$. It suffices to treat $\boldsymbol{0}^{\mathrm{an}}$ which is the image of $\mathbb{O}^{\mathrm{an}}$.

As $L$ is ample, there exist $n\in\mathbb{N}$ such that $L^{\otimes n}$ is globally generated. Let $\{s_{n,j}\}_{j\in\{0,\dots,d_{n}\}}$ be a basis and let $\psi$ be the Fubini-Study metric associated with some ultrametric norm $\lVert\mathord{\cdot}\rVert_{n}$ for which this basis is orthogonal. As both $\frac{1}{n}\psi$ and $\phi$ are continuous and $X^{\mathrm{an}}$ is compact, there exist $\alpha\in\mathbb{R}$ such that

$$\forall x\in X^{\mathrm{an}},\ \frac{1}{n}\psi(\alpha)(x)\leq\phi(x),$$

so

$$p(\boldsymbol{0})^{\mathrm{an}}(\mathbb{D}^{\vee}(L,\frac{1}{n}\psi(\alpha)))\subseteq p(\boldsymbol{0})^{\mathrm{an}}(\mathbb{D}^{\vee}(L,\phi)).$$

the left hand side is an open subset of $(\spec V_{{\scriptscriptstyle\bullet}}(L))^{\mathrm{an}}$ by Lemma 3.21. Then

$$p(\boldsymbol{0})^{\mathrm{an}}(\mathbb{D}^{\vee}(L,\phi))=p(\boldsymbol{0})^{\mathrm{an}}(\mathbb{D}^{\vee}(L,\phi)\setminus\mathbb{O}^{\mathrm{an}})\cup p(\boldsymbol{0})^{\mathrm{an}}(\mathbb{D}^{\vee}(L,\frac{1}{n}\psi(\alpha)))$$

is an open set in $(\spec V_{{\scriptscriptstyle\bullet}}(L))^{\mathrm{an}}$. ∎