The Stacks project

Proof. Since $X$ is quasi-compact, there is a reduction to the case where $I$ is finite (details omitted). We prove the result for $I$ finite by induction on $p$. The base case $p = 0$ is immediate by taking a covering as in Lemma 5.13.4 refining the open covering $X = \bigcup W_{i_0, k}$.

Induction step. Assume the lemma proven for $p - 1$. For all $p$-tuples $i'_0, \ldots , i'_{p - 1}$ of $I$ let $U_{i'_0} \cap \ldots \cap U_{i'_{p - 1}} = \bigcup W_{i'_0 \ldots i'_{p - 1}, k}$ be a common refinement of the coverings $U_{i_0} \cap \ldots \cap U_{i_ p} = \bigcup W_{i_0 \ldots i_ p, k}$ for those $(p + 1)$-tuples such that $\{ i'_0, \ldots , i'_{p - 1}\} = \{ i_0, \ldots , i_ p\} $ (equality of sets). (There are finitely many of these as $I$ is finite.) By induction there exists a solution for these opens, say $X = \bigcup V_ j$ and $\alpha : J \to I$. At this point the covering $X = \bigcup _{j \in J} V_ j$ and $\alpha $ satisfy $\overline{V_ j} \subset U_{\alpha (j)}$ and each $V_{j_0} \cap \ldots \cap V_{j_ p}$ is contained in $W_{\alpha (j_0) \ldots \alpha (j_ p), k}$ for some $k$ if there is a repetition in $\alpha (j_0), \ldots , \alpha (j_ p)$. Of course, we may and do assume that $J$ is finite.

Fix $i_0, \ldots , i_ p \in I$ pairwise distinct. Consider $(p + 1)$-tuples $j_0, \ldots , j_ p \in J$ with $i_0 = \alpha (j_0), \ldots , i_ p = \alpha (j_ p)$ such that $V_{j_0} \cap \ldots \cap V_{j_ p}$ is not contained in $W_{\alpha (j_0) \ldots \alpha (j_ p), k}$ for any $k$. Let $N$ be the number of such $(p + 1)$-tuples. We will show how to decrease $N$. Since

we find a finite set $K = \{ k_1, \ldots , k_ t\} $ such that the LHS is contained in $\bigcup _{k \in K} W_{i_0 \ldots i_ p, k}$. Then we consider the open covering

The first open on the RHS intersects $V_{j_1} \cap \ldots \cap V_{j_ p}$ in the empty set and the other opens $V_{j_0, k}$ of the RHS satisfy $V_{j_0, k} \cap V_{j_1} \ldots \cap V_{j_ p} \subset W_{\alpha (j_0) \ldots \alpha (j_ p), k}$. Set $J' = J \amalg K$. For $j \in J$ set $V'_ j = V_ j$ if $j \not= j_0$ and set $V'_{j_0} = V_{j_0} \setminus (\overline{V_{j_1}} \cap \ldots \cap \overline{V_{j_ p}})$. For $k \in K$ set $V'_ k = V_{j_0, k}$. Finally, the map $\alpha ' : J' \to I$ is given by $\alpha $ on $J$ and maps every element of $K$ to $i_0$. A simple check shows that $N$ has decreased by one under this replacement. Repeating this procedure $N$ times we arrive at the situation where $N = 0$.

To finish the proof we argue by induction on the number $M$ of $(p + 1)$-tuples $i_0, \ldots , i_ p \in I$ with pairwise distinct entries for which there exists a $(p + 1)$-tuple $j_0, \ldots , j_ p \in J$ with $i_0 = \alpha (j_0), \ldots , i_ p = \alpha (j_ p)$ such that $V_{j_0} \cap \ldots \cap V_{j_ p}$ is not contained in $W_{\alpha (j_0) \ldots \alpha (j_ p), k}$ for any $k$. To do this, we claim that the operation performed in the previous paragraph does not increase $M$. This follows formally from the fact that the map $\alpha ' : J' \to I$ factors through a map $\beta : J' \to J$ such that $V'_{j'} \subset V_{\beta (j')}$. $\square$