**Proof.**
In this case the stratification $U = Z_0 \supset Z_1 \supset Z_2 \supset \ldots $ of Lemma 40.14.1 is given by closed immersions $Z_ k \to U$ of finite presentation, see Divisors, Lemma 31.9.6. Part (1) follows immediately from this as $W' \to W$ is locally given by intersecting the open $W$ by $Z_ r$. To see part (2) let $\{ u_1, \ldots , u_ n\} $ be the orbit of $u$. Since the closed subschemes $Z_ k$ are $R$-invariant and $\bigcap Z_ k = \emptyset $, we find an $k$ such that $u_ i \in Z_ k$ and $u_ i \not\in Z_{k + 1}$ for all $i$. The image of $Z_ k \to U$ and $Z_{k + 1} \to U$ is locally constructible (Morphisms, Theorem 29.22.3). Since $u_ i \in U$ is a generic point of an irreducible component of $U$, there exists an open neighbourhood $U_ i$ of $u_ i$ which is contained in $Z_ k \setminus Z_{k + 1}$ set theoretically (Properties, Lemma 28.2.2). In the proof of Lemma 40.14.2 we have constructed $W$ as a disjoint union $\coprod W_ r$ with $W_ r \subset Z_{r - 1} \setminus Z_ r$ such that $U = \bigcup t(s^{-1}(\overline{W_ r}))$. As $\{ u_1, \ldots , u_ n\} $ is an $R$-orbit we see that $u \in t(s^{-1}(\overline{W_ r}))$ implies $u_ i \in \overline{W_ r}$ for some $i$ which implies $U_ i \cap W_ r \not= \emptyset $ which implies $r = k$. Thus we conclude that $u$ is in

\[ W_{k + 1} = Z_ k \setminus \left( Z_{k + 1} \cup \bigcup \nolimits _{r \leq k} t(s^{-1}(\overline{W_ r})) \right) \]

as desired.
$\square$

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