The Stacks project

Proof. The implication $(\gamma ) \Leftrightarrow (\alpha ) + (\beta )$ is immediate. The implications in the diamond on the left are clear from the definitions.

Assume $(\zeta )$, i.e., that $X$ is Zariski locally quasi-separated. Then $(\epsilon )$ holds by Properties of Spaces, Lemma 66.6.6.

Assume $(\epsilon )$. By Lemma 68.4.7 there exists a Zariski open covering $X = \bigcup X_ i$ such that for each $i$ there exists a scheme $U_ i$ and a quasi-compact surjective étale morphism $U_ i \to X_ i$. Choose an $i$ and an affine open subscheme $W \subset U_ i$. It suffices to show that $W \to X$ has universally bounded fibres, since then the family of all these morphisms $W \to X$ covers $X$. To do this we consider the diagram

Since $W \to X$ factors through $X_ i$ we see that $W \times _ X U_ i = W \times _{X_ i} U_ i$, and hence $q$ is quasi-compact. Since $W$ is affine this implies that the scheme $W \times _ X U_ i$ is quasi-compact. Thus we may apply Morphisms, Lemma 29.57.9 and we conclude that $p$ has universally bounded fibres. From Lemma 68.3.4 we conclude that $W \to X$ has universally bounded fibres as well.

Assume $(\delta )$. Let $U$ be an affine scheme, and let $U \to X$ be an étale morphism. By assumption the fibres of the morphism $U \to X$ are universally bounded. Thus also the fibres of both projections $R = U \times _ X U \to U$ are universally bounded, see Lemma 68.3.3. And by Lemma 68.3.2 also the fibres of $R \to X$ are universally bounded. Hence for any $x \in X$ the fibres of $|U| \to |X|$ and $|R| \to |X|$ over $x$ are finite, see Lemma 68.3.6. In other words, the equivalent conditions of Lemma 68.4.5 hold. This proves that $(\delta ) \Rightarrow (\gamma )$. $\square$