Let $R$ be a Noetherian ring and $n\ge 1$ an integer. Consider the set $S_n^R:=\left\{\mathfrak p \in \text{Spec}(R) \text{ }| \text{ } \text{depth } R_{\mathfrak p} \ge \inf \{n, \dim R_{\mathfrak p}\}\right\}$.
Then, is it true that $\text{Spec}(R)\smallsetminus S_n^R$ is a specialization closed subset of $\text{Spec}(R)$? ,i.e., if $\mathfrak p \subseteq \mathfrak q$ are prime ideals of $R$ such that $\mathfrak p\notin S_n^R$, then must it hold that $\mathfrak q\notin S_n^R$ ?
Meta-answer: this is elementary- and reasonable-sounding but not in Stacks so it’s probably false.
Actual answer: let $R=k[x,y,z]/(yz,z^2)$. Then $x$ is a regular element so $(x,y,z) \in S_1^R$.
But $(y,z)$ (which isn’t minimal) is the annihilator of $z$ so $(y,z) \notin S_1^R$.