Lemma 37.70.2. Let $X$ be a scheme. Let $X = \coprod _{i \in I} X_ i$ be a finite affine stratification. There exists an affine stratification with index set $\{ 0, \ldots , n\}$ where $n$ is the length of $I$.

Proof. Recall that we have a partial ordering on $I$ such that the closure of $X_ i$ is contained in $\bigcup _{j \leq i} X_ j$ for all $i \in I$. Let $I' \subset I$ be the set of maximal indices of $I$. If $i \in I'$, then $X_ i$ is open in $X$ because the union of the closures of the other strata is the complement of $X_ i$. Let $U = \bigcup _{i \in I'} X_ i$ viewed as an open subscheme of $X$ so that $U_{red} = \coprod _{i \in I'} X_ i$ as schemes. Then $U$ is an affine scheme by Schemes, Lemma 26.6.8 and Lemma 37.2.3. The morphism $U \to X$ is affine as each $X_ i \to X$, $i \in I'$ is affine by the same reasoning using Lemma 37.3.1. The complement $Z = X \setminus U$ endowed with the reduced induced scheme structure has the affine stratification $Z = \bigcup _{i \in I \setminus I'} X_ i$. Here we use that a morphism of schemes $T \to Z$ is affine if and only if the composition $T \to X$ is affine; this follows from Morphisms, Lemmas 29.11.9, 29.11.7, and 29.11.11. Observe that the partially ordered set $I \setminus I'$ has length exactly one less than the length of $I$. Hence by induction we find that $Z$ has an affine stratification $Z = Z_0 \amalg \ldots \amalg Z_{n - 1}$ with index set $\{ 1, \ldots , n\}$. Setting $Z_ n = U$ we obtain the desired stratification of $X$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).