Grothendieck group of noetherian scheme - why $\textbf{Coh}(X)$ is a set? According to Hartshorne Exercise II.6.10, the Grothendieck group of a noetherian scheme is defined to be the quotient of the free abelian group generated by all the coherent sheaves on $X$ by the equivalent relation
$$[E]=[E']+[E'']\Longleftrightarrow\exists0\rightarrow E'\rightarrow E\rightarrow E''\rightarrow0$$
So all the coherent sheaves on $X$ form a set, or in other words the category $\textbf{Coh}(X)$ is small.
Maybe it is obvious but I did not see it very clearly. Any hints?
 A: That's not quite correct - the coherent sheaves over any variety form a proper class. Instead, what is true, is that the isomorphism classes of coherent sheaves form a set - so if you start with the free abelian group on isomorphism classes of coherent sheaves and then quotient by the same relation, you get what Hartshorne is after (yes, the presentation is perhaps slightly sloppy, but I think this a pretty minor thing).
Here's one way to see that the isomorphism classes form a set in the case at hand: every point $x\in X$ has a neighborhood $U_x$ which we may assume is affine where our coherent sheaf can be represented as a cokernel of a map $\mathcal{O}^m\to\mathcal{O}^n$. Since $X$ is noetherian, it is quasi-compact, and therefore finitely many of these $U_x$ suffice to cover $X$. Further, the intersections of these sets are again quasi-compact and therefore can be covered by finitely many affine opens. It's now clear that the isomorphism class of our coherent sheaf depends on the product of finitely many sets worth of data: finitely many choices of $m$ and $n$, finitely many choices of entries of $n\times m$ matrices with coefficients in certain rings to define the matrix $\mathcal{O}^m\to\mathcal{O}^n$, and finitely many elements of various other rings to define the gluing data (subject to some compatibility conditions).
