Lemma 57.17.2. Let $\mathcal{A}$ be a countable abelian category. Then $D^ b(\mathcal{A})$ is countable.

Proof. It suffices to prove the statement for $D(\mathcal{A})$ as the others are full subcategories of this one. Since every object in $D(\mathcal{A})$ is a complex of objects of $\mathcal{A}$ it is immediate that the set of isomorphism classes of objects of $D^ b(\mathcal{A})$ is countable. Moreover, for bounded complexes $A^\bullet$ and $B^\bullet$ of $\mathcal{A}$ it is clear that $\mathop{\mathrm{Hom}}\nolimits _{K^ b(\mathcal{A})}(A^\bullet , B^\bullet )$ is countable. We have

$\mathop{\mathrm{Hom}}\nolimits _{D^ b(\mathcal{A})}(A^\bullet , B^\bullet ) = \mathop{\mathrm{colim}}\nolimits _{s : (A')^\bullet \to A^\bullet \text{ qis and }(A')^\bullet \text{ bounded}} \mathop{\mathrm{Hom}}\nolimits _{K^ b(\mathcal{A})}((A')^\bullet , B^\bullet )$

by Derived Categories, Lemma 13.11.6. Thus this is a countable set as a countable colimit of $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).