smallest localising subcategory that contains a set of objects

Question

Let $D$ a triangulated category closed under arbitrary coproducts and $\mathcal{R}$ a localising triangulated full subcategory that contains a set of objects $R$,which $R$ is closed under the translation functor of the triangulated category.Just a reminder,localising means closed under arbitrary coproducts.So,i am trying to prove the following: $$\mathcal{R}\ \text {is the smallest localising triangulated subcategory that contains}\ R\ \Leftrightarrow \ R^{\perp}=0 $$ where $R^{\perp}$ is the full subcategory such that $Hom(r,A)=0,\forall \ r\in R$,for $A\in \mathcal{R}$.It is easily seen that $R^{\perp}$ is localising triangulated subcategory.But how can i prove the desired result?

Answer 1

It seems that it works when the set $R$ consists of compact objects(at least the $\impliedby$ implication)
$"\implies"$ Let the class of objects $\mathscr{X}$ of all $X\in \mathcal{R}$ such that $Hom\left(X,Y\right)=0,\hspace{0.1cm}\forall\hspace{0.1cm}Y\in R^{\perp}$ Then,we note that $\mathscr{X}$ is a triangulated subcategory and in particular is localising because $$Hom\left(\bigoplus X_{i},Y\right)\cong \prod Hom\left(X_{i},Y\right)\cong0$$ for $X_{i}\in\mathscr{X}$.Also,since $Hom\left(r,Y\right)=0$ $\hspace{0.1cm}\forall\hspace{0.1cm}Y\in R^{\perp}$,where $r\in R$,the $\mathscr{X}$ contains $R$.By assumption,it'll be true that $\mathcal{R}=\mathscr{X}$ and hence $R^{\perp}=0$
$"\impliedby"$Let $\mathcal{A}$ be the smallest localising subcategory that contains $R$.Then lemma 1.7 of Neeman's paper is true for $\mathcal{A}$.Let $X\in\mathcal{R}$.Then according to the lemma there exists a $R-$local object $Y$ and morphism $f:X\rightarrow Y$ such that $Cone(f)\in\mathcal{A}$.By assumption $R^{\perp}=0\implies Y=0$.So $$X\cong Cone(f)\in\mathcal{A}\implies X\in \mathcal{A}\implies \mathcal{A}=\mathcal{R}$$