Lemma 57.18.2. 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

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

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