Lemma 13.36.2. Let $\mathcal{D}$ be a triangulated category. Let $E$ be an object of $\mathcal{D}$. The subcategory

\[ \langle E \rangle = \bigcup \nolimits _ n \langle E \rangle _ n = \bigcup \nolimits _{n, m \geq 1} smd(add(E[-m, m])^{\star n}) \]

is a strictly full, saturated, triangulated subcategory of $\mathcal{D}$ and it is the smallest such subcategory of $\mathcal{D}$ containing the object $E$.

**Proof.**
The equality on the right follows from Lemma 13.36.1. It is clear that $\langle E \rangle = \bigcup \langle E \rangle _ n$ contains $E$, is preserved under shifts, direct sums, direct summands. If $A \in \langle E \rangle _ a$ and $B \in \langle E \rangle _ b$ and if $A \to X \to B \to A[1]$ is a distinguished triangle, then $X \in \langle E \rangle _{a + b}$ by Lemma 13.36.1. Hence $\bigcup \langle E \rangle _ n$ is also preserved under extensions and it follows that it is a triangulated subcategory.

Finally, let $\mathcal{D}' \subset \mathcal{D}$ be a strictly full, saturated, triangulated subcategory of $\mathcal{D}$ containing $E$. Then $\mathcal{D}'[-\infty , \infty ] \subset \mathcal{D}'$, $add(\mathcal{D}) \subset \mathcal{D}'$, $smd(\mathcal{D}') \subset \mathcal{D}'$, and $\mathcal{D}' \star \mathcal{D}' \subset \mathcal{D}'$. In other words, all the operations we used to construct $\langle E \rangle $ out of $E$ preserve $\mathcal{D}'$. Hence $\langle E \rangle \subset \mathcal{D}'$ and this finishes the proof.
$\square$

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