In the paper of Rouquier on the dimension of triangulated categories (found [here](https://www.math.ucla.edu/%7Erouquier/papers/dimension.pdf)) lemma 3.5 says:


**Lemma** Let $\mathcal{T}$ be a triangulated category and let $\mathcal{T}\_1$ and $\mathcal{T}\_2$ be triangulated subcategories of $\mathcal{T}$ such that $\mathcal{T}=\mathcal{T}\_1\diamond\mathcal{T}\_2$. Then $dim\mathcal{T}\leq dim\mathcal{T}\_1+dim\mathcal{T}\_2+1$.


There is no proof given in the text and it says that the lemma is "clear". However, at the moment, I really can't see how one can proceed. My guess is that by assuming that $\mathcal{T}\_1=\langle M\_1\rangle\_{1+dim\mathcal{T}\_1}$ and $\mathcal{T}\_2=\langle M\_2\rangle\_{1+dim\mathcal{T}\_2}$, one can find an object $M$ such that $\mathcal{T}=\langle M\rangle\_{2+dim\mathcal{T}\_1+dim\mathcal{T}\_2}$.


Is there something trivial that I am missing?
Any help would be very much appreciated.



How to prove a lemma of Rouquier on the dimension of triangulated categories?
If $\mathcal{T}\_{1}=\langle M\_{1}\rangle\_{d\_{1}+1}$ and
$\mathcal{T}\_{2}=\langle M\_{2}\rangle\_{d\_{2}+1}$, then
$\mathcal{T}\_{1}\ast\mathcal{T}\_{2}\subseteq\langle M\_{1}\oplus
M\_{2}\rangle\_{d\_{1}+d\_{2}+2}$.


**Sketch proof:** Since $\mathcal{T}\_{i}\subseteq\langle M\_{1}\oplus
M\_{2}\rangle\_{d\_{i}+1}$, it suffices to show that
$$\langle M\_{1}\oplus
M\_{2}\rangle\_{s}\ast\langle M\_{1}\oplus M\_{2}\rangle\_{t}=
\langle M\_{1}\oplus M\_{2}\rangle\_{s+t}.$$


This follows by induction on $s$ and $t$, using the fact that $\ast$
is associative (i.e.,
$\mathcal{A}\ast(\mathcal{B}\ast\mathcal{C})=(\mathcal{A}\ast\mathcal{B})\ast
\mathcal{C}$ for triangulated subcategories
$\mathcal{A},\mathcal{B},\mathcal{C}$ of $\mathcal{T}$), which follows
from the octahedral axiom.


