We work in $T$ triangulated category admitting small coproducts, we say $S$ a full subcategory is exact or triangulated iff it is closed under suspensions and triangles. Moreover such $S$ is called localising whenever it is closed under infinite coproducts and thick if it is closed under summands, i.e. if $X \oplus Y \in S$ then both $X$ and $Y$ are in $S$.
These are classical definitions in the theory of triangulated category which I think are familiar to anyone who studied it.
I was told that every localizing subcategory is thick but I cannot see this from the definitions. My first thought was using the triangle $X \rightarrow X \oplus Y \rightarrow Y \rightarrow \Sigma X$ to get the claim but I only know $X \oplus Y \in S$ so I cannot conclude using the closure under triangles.
Does this follows from the definitions alone or do I need some result? My intuition tells me that if the claim is true it should be an easy fact which does not require a complicate proof.
Maybe I need an additional assumption on $T$?
Edit: I recalled now that if $T$ is compactly generated then Brown representability theorem implies the existence of a localization functor $L \colon T \rightarrow T$ with kernel $S$ and such $L$ is left adjoint to the inclusion $S^\perp \rightarrow T$. Thus $L(X \oplus Y)\cong LX \oplus LY$ and this is zero iff both $LX$ and $LY$ are zero. This gives the claim I wanted but the proof is more complex that what I thought at first and I need the additional assumption that $T$ must be compactly generated. Can you provide an easier proof or one which does not require the compactly generated hypothesis? Or give a counterexample if such exists.