Lemma 17.9.8. Let $X$ be a ringed space. There exists a set of $\mathcal{O}_ X$-modules $\{ \mathcal{F}_ i\} _{i \in I}$ of finite type such that each finite type $\mathcal{O}_ X$-module on $X$ is isomorphic to exactly one of the $\mathcal{F}_ i$.

**Proof.**
For each open covering $\mathcal{U} : X = \bigcup U_ j$ consider the sheaves of $\mathcal{O}_ X$-modules $\mathcal{F}$ such that each restriction $\mathcal{F}|_{U_ j}$ is a quotient of $\mathcal{O}_{U_ j}^{\oplus r}$ for some $r_ j \geq 0$. These are parametrized by subsheaves $\mathcal{K}_ i \subset \mathcal{O}_{U_ j}^{\oplus r_ j}$ and glueing data

see Sheaves, Section 6.33. Note that the collection of all glueing data forms a set. The collection of all coverings $\mathcal{U} : X = \bigcup _{j \in J} U_ i$ where $J \to \mathcal{P}(X)$, $j \mapsto U_ j$ is injective forms a set as well. Hence the collection of all sheaves of $\mathcal{O}_ X$-modules gotten from glueing quotients as above forms a set $\mathcal{I}$. By definition every finite type $\mathcal{O}_ X$-module is isomorphic to an element of $\mathcal{I}$. Choosing an element out of each isomorphism class inside $\mathcal{I}$ gives the desired set of sheaves (uses axiom of choice). $\square$

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