Lemma 25.12.6. Let $\mathcal{C}$ be a site. Let $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be a subset. Assume

1. $\mathcal{C}$ has fibre products,

2. for all $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ there exists a finite covering $\{ U_ i \to X\} _{i \in I}$ with $U_ i \in \mathcal{B}$,

3. if $\{ U_ i \to X\} _{i \in I}$ is a finite covering with $U_ i \in \mathcal{B}$ and $U \to X$ is a morphism with $U \in \mathcal{B}$, then $\{ U_ i \to X\} _{i \in I} \amalg \{ U \to X\}$ is a covering.

Then for every $X$ there exists a hypercovering $K$ of $X$ such that each $K_ n = \{ U_{n, i} \to X\} _{i \in I_ n}$ with $I_ n$ finite and $U_{n, i} \in \mathcal{B}$.

Proof. This lemma is the analogue of Lemma 25.11.4 for sites. To prove the lemma we follow exactly the proof of Lemma 25.12.2 paying attention to the following two points

1. We choose our initial covering $\{ U_{0, i} \to X\} _{i \in I_0}$ with $U_{0, i} \in \mathcal{B}$ such that the index set $I_0$ is finite, and

2. in choosing the coverings (25.12.2.1) we choose $J_{i'}$ finite.

The reader sees easily that with these modifications we end up with finite index sets $I_ n$ for all $n$. $\square$

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