# Are localizing subcategories thick?

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.

$$(X\oplus Y)\oplus(X\oplus Y)\oplus(X\oplus Y)\oplus\dots\cong X\oplus (Y\oplus X)\oplus(Y\oplus X)\oplus\dots,$$ so there is a triangle $$X\to(X\oplus Y)^{(\mathbb{N})}\to(X\oplus Y)^{(\mathbb{N})}\to\Sigma X,$$ where $$(X\oplus Y)^{(\mathbb{N})}$$ denotes the coproduct of countably many copies of $$X\oplus Y$$.
• The second map in the triangle is the one twisting the sum $X \oplus Y$ in $Y \oplus X$ and then injecting infinitely many such copies in $X \oplus (Y \oplus X) \oplus (Y \oplus X) \dots$, right?
Using the language of $$\alpha$$-localising subcategories (Take a triangulated category $$T$$ with arbitrary coproducts and let $$\alpha$$ be a regular cardinal, $$\alpha$$-localising subcategories are closed under coproducts of fewer than $$\alpha$$ objects), I believe it is said that $$\alpha$$-localising subcategories are thick if $$\alpha > \aleph_0$$.