Let $K^{\bullet}$ be a bounded complex of abelian étale sheaves on a quasi-compact and quasi-separated scheme $X$.


* For any étale cover $\mathcal{U} :=\{ U\_i\to X\}\_{i\in I}$, can we find a refinement $\mathcal{U}'$ of $\mathcal{U}$ with $\mathcal{U}'$ finite (the answer to this should be easily yes, but I just want to double check)
* We have a Cech-to-cohomology spectral sequence


$$E\_2^{p,q} := H^p(\text{Tot}(\check{C}(\mathcal{U},\underline{H}^q(K^{\bullet})))\Rightarrow H^{p+q}(X\_{\rm ét},K^{\bullet})$$
for any $\mathcal{U}$.


I must be misunderstanding the meaning of convergence of this spectral sequence for **any** $\mathcal{U}$. Zariski covers are, in particular, étale covers. But if I choose $\mathcal{U}$ a Zariski cover, the spectral sequence should converge to **Zariski** hypercohomology of $K^{\bullet}$, which won't be **étale** hypercohomology.


What is the meaning of **for any $\mathcal{U}$**, then?


Thanks a lot.



Étale hypercohomology of complexes
1. Any étale morphism is flat and locally of finite presentation [Tag [02GR](https://stacks.math.columbia.edu/tag/02GR), Tag [02GS](https://stacks.math.columbia.edu/tag/02GS)], hence open [Tag [01UA](https://stacks.math.columbia.edu/tag/01UA)]. Quasi-compactness now immediately shows that étale covers have finite subcovers.
2. The presheaves $\underline{H}^q(K^i)$ are given by $U \mapsto H^q\_{\text{ét}}(U,K^i|\_U)$, which is not the same as the Zariski version $U \mapsto H^q\_{\text{Zar}}(U,K^i|\_U)$. Thus, there is no reason this should converge to the Zariski cohomology (e.g. think of the one-object covering $\{X \to X\}$).


