Let $D$ be a triangulated category (the triangulated category in my mind is $D^{b}(X)$, that is the derived category of bounded complex of coherent sheaves on a smooth projective variety), $A \subset D$ is a subset of $D$. By $\langle A \rangle$ we mean the triangulated category generated by $A$ (i.e. the smallest triangulated category contain $A$). Let 
$$ A ^{\perp} : = \{ b \in D\mid \forall a \in A, \forall i \in \mathbb{Z}, \operatorname{Hom}(a, b[i])=0\}.$$


Then I was wondering if the following statement is true:



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I saw this result used implicitly in several places when $D = D^{b}(X)$, but even in this case, I still do not know the validity of the statement.



A statement for a triangulated category generated by a subset
Edited based on Sasha's answer:


I will assume that we are interested in the *thick* subcategory generated by $A$.


Under this assumption, the desired statement is closely related to a theorem of Neeman and Ravenel. See for instance the paper by Bondal and van den Bergh. The theorem of Neeman and Ravenel says that a subset $A$ of compact objects in a compactly generated triangulated category $D$ generates $D$ if and only if it classically generates $D^c$. Here, to say that $A$ classically generates $D^c$ means that $D^c$ is the smallest thick (closed under summands and equivalences) subcategory of $D^c$ containing $A$, while to say that $A$ generates $D$ means that if $Hom(a,b[i])=0$ for all $i\in\mathbb{Z}$ and $a\in A$, then $b=0$. Recall that an object $a$ is compact if $Hom(a,-)$ commutes with coproducts. This really only makes sense in the big derived category $D$, not in the small derived category $D^c$.


In the setting of (quasi-compact and quasi-separated) schemes, $D=D\_{qc}(X)$, the derived category of complexes of $O\_X$-modules with quasi-coherent cohomology, and $D^c=D\_{perf}(X)$, the triangulated category of perfect complexes. When $X$ is smooth and projective, as in the original question, $D\_{perf}(X)=D^b(X)$.


Thus, to finish answering the question, if the right-orthogonal to $A$ in $D$ vanishes, then the right-orthogonal to $A$ in $D^c$ vanishes (this is where compactness and compact generation is used). In this case $A$ generates $D$ by definition, which by the theorem implies that $A$ classically generates $D^c$, as desired.


In my original answer, I mistakenly went the other way, concluding that the right-orthogonal to $A$ in $D$ vanishes from the assumption that the right-orthogonal to $A$ in $D^c$ vanishes. This is just not true in general, as I interpret Sasha's example to show. For instance, consider $D^b(\mathbb{Z})$, the bounded derived category of the integers (I realize it's not projective, but the projective case is similar). Then, any complex right-orthogonal to $\mathbb{Z}/p$ for all $p$ must be zero. This can be seen because any complex has closed support in $Spec\mathbb{Z}$. So, if it contains no closed point it contains no point at all. But, if the support of a perfect complex is $0$, the complex is quasi-isomorphic to $0$. Thus, if $A=\{\mathbb{Z}/p\}$, then $A^\perp=0$ in $D^b(\mathbb{Z})$. It is pretty clear that the complex $\mathbb{Z}$ cannot be generated from these. One reason is as follows: if it could be, then it support would not contain the generic point. But, it does.


To summarize, if $X$ is a scheme and if $\langle A\rangle$ denotes the smallest thick subcategory of $D\_{perf}(X)$ containing $A$, then $\langle A\rangle=D\_{perf}(X)$ if and only if the right-orthogonal to $A$ in $D\_{qc}(X)$ vanishes. This is not what was asked in the original post. But, it is at least good to know.


