# Is the minimal Serre subcategory containing a set also a set?

Let $\mathscr{A}$ be an abelian category, and $S\subset\mathscr{A}$ some set of objects in $\mathscr{A}$ (i.e. $S$ is not a proper class), so we can fairly easily define the category $\mathscr{C}_S$ to be the smallest full subcategory of $\mathscr{A}$ containing $S$ such that $\mathscr{C}_S$ is closed under subobjects, quotients, and extensions, making $\mathscr{C}_S$ the minimal Serre subcategory of $\mathscr{A}$ containing $S$.

But will $\mathscr{C}_S$ be a small category?

The reason I'm asking this is that it is often convenient, when working with Abelian categories, to consider some set $S\subset\mathscr{A}$, and then construct some small Abelian full exact subcategory $\mathscr{C}$ of $\mathscr{A}$ containing $S$ and applying the Freyd-Mitchell embedding theorem to introduce elements (radically simplifying diagram chases).

Not in general, even if by "small" you mean "essentially small" (i.e., there is a small set of isomorphism classes of objects). First of all, it is possible for a single object to have a proper class of subobjects (if this does not happen, the category $\mathscr{A}$ is called well-powered). I don't know an explicit example off the top of my head, but you should be able to build one by taking a category with a proper class of subobjets and then "freely" making it abelian.
It is also not enough just to assume your category is well-powered. It is possible in a well-powered abelian category to have two objects $A$ and $B$ such that there is a proper class of non-isomorphic extensions of $B$ by $A$ (see my answer https://math.stackexchange.com/a/1766337/86856 for an explicit example).
On the other hand, if you know that your category is well-powered and that there is only a set of extensions of any two objects (up to isomorphism), then it is easy to see that $\mathscr{C}_S$ is essentially small. Indeed, define $S_0=S$ and let $S_{n+1}$ be the collection of all subobjects, quotients, and extensions of objects of $S_n$. Since $S_n$ is essentially small and there is (up to isomorphism) only a set of subobjects/quotients of each object of $S_n$ and only a set of extensions of each pair of objects of $S_n$, $S_{n+1}$ is also essentially small. The union $\bigcup_nS_n$ is then essentially small and is closed under subobjects, quotients, and extensions, and so is equal to $\mathscr{C}_S$.
(If you want a category that is literally small and not just essentially small, you can take $S_{n+1}$ to just have one representative of each isomorphism class of subobject/quotient/extension, and then each $S_n$ will be a set and $\bigcup_n S_n$ will be a set of objects which contains a representative of every isomorphism class in $\mathscr{C}_S$.)