Let $X = \text{Spec}A$ be an affine scheme. It is well known that if $U$ is an open subset which contains $\text{SpecMax}A$, then $U\supseteq X$. The previous statement generalizes to arbitrary schemes, due to the nature of the Zariski topology of a scheme, being locally affine.


When talking about stacks, one typically replaces the notion of closed points with points of finite type. Again, say $\mathcal{X}$ is a stack, and assume $U\subseteq \mathcal{X}$ is an open substack which contains all points of finite type in $\mathcal{X}$. Is it true that $U\supseteq \mathcal{X}$?


I am not really sure how to approach such a question, as my intuition from the Zariski topology, having a basis of basic open subsets of the topology doesn't generalize at all.


Thanks in advance!



Covering a stack by an open substack that contains all points of finite type
Assuming you mean that $\mathscr X$ is an *algebraic* stack (e.g. in the sense of [Tag [026N](https://stacks.math.columbia.edu/tag/026N)]), then this is true, and follows relatively straightforwardly from the case of schemes.


Although you did not make a precise statement in the scheme case, it is indeed true that if $U \subseteq X$ is an open subscheme containing all *locally closed points*, then $U = X$. Moreover, locally closed points are the same thing as points that are (locally) of finite type [Tag [01TA](https://stacks.math.columbia.edu/tag/01TA)].


Now let $\mathscr U \subseteq \mathscr X$ be an open substack, i.e. $j\colon\mathscr U \to \mathscr X$ is representable by open immersions. In particular, if $x \colon \operatorname{Spec} k \to \mathscr X$ is a point of finite type, the pullback along $\mathscr U \to \mathscr X$ gives an open immersion $U \hookrightarrow \operatorname{Spec} k$, and we say $\mathscr U$ *contains $x$* if $U \neq \varnothing$. If this holds for all finite type points, the question is whether $\mathscr U \to \mathscr X$ is an isomorphism.


By assumption, there exists a smooth surjection $f \colon X \to \mathscr X$ where $X$ is a scheme. Checking that a morphism is an isomorphism is smooth-local, so it suffices to check that the base change $U \to X$ of $j$ along $f$ is an isomorphism. Note that $U \to X$ is again an open immersion, so by the above it suffices to check that $U$ contains all points of finite type of $X$. If $x \colon \operatorname{Spec} k \to X$ is a point that is (locally) of finite type, then so is the composition $\operatorname{Spec} k \to X \to \mathscr X$, since $f$ is locally of finite type [Tags [06MH](https://stacks.math.columbia.edu/tag/06MH) and [03YJ](https://stacks.math.columbia.edu/tag/03YJ)]. By assumption, the pullback of $U \to X$ along $x$ is an isomorphism, i.e. $x \in U$. $\square$


