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I was told the following is a fact known in the community of triangulated categories but it has no canonical reference.

Let $T$ be a triangulated category, let $G=\{ g_1, g_2, \dots, g_n \} \subset T$ be a finite set, then we define recursively the following sets

$C_{(0)}=\{ \Sigma^m g_i : m \in \mathbb{Z}, 1 \leq i \leq n \} $ $C_ {(n+1)}= \{ X \in T : \text{there exists a distinguished triangle} \\ \qquad \qquad \qquad \Sigma^m g_i \rightarrow Y \rightarrow X \rightarrow \Sigma^{m+1}g_i \ \text{where} \ m \in \mathbb{Z}, 1 \leq i \leq n, Y \in C_{(n)} \} $

and take $K= \bigcup_{n \in \mathbb{N}} C_{(n)}$. Then this union $K$ coincides with the smallest thick triangulated subcategory of $T$ containing the set $G$.

The proof trivially reduces to showing that $K$ is a thick triangulated subcategory: I managed to prove $K$ is a triangulated subcategory of $T$ but I have no idea how to establish its thickness.

If $X= X_1 \oplus X_2 \in K$ I would argue by induction on $n \in \mathbb{N}$ such that $X \in C_{(n)}$ but I really don't see how to use this induction: for the starting case we don't have any useful information from $X \in C_{(0)}$ and I should get somehow a triangle whose associated cone is $X_1$ (or its suspension) but I cannot see how to present this since all the distinguished triangles I can think of present both $X_1$ and $X_2$.

For the induction step $X \in C_{(n+1)}$ I have a similar problem: even using the triangle given by the definition of $C_{(n+1)}$ and the octahedron axiom I always end up with a distinguished triangle presenting just one object I know to be in $K$ (thus I cannot conclude using the definition of $K$ or the fact it is a triangulated subcategory).

So any idea or suggestion is gladly received, or if you know a reference for this fact feel free to post it.

Thanks in advance for your help.

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It's not true. For example, let $T$ be the derived category of vector spaces over some field, and let $G$ consist of just one object, a $2$-dimensional vector space.

Then every object of $K$ has the property that the alternating sum of the dimensions of its homology is even, but the smallest thck subcategory containing $G$ also contains a $1$-dimensional vector space.

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    Yes, it's clear that the statement it's false in full generality. I think that for the thickness I have to impose additional assumptions like the $g_i \in G$ are not coproducts. Anyway the concrete example I had in mind was $T=\mathcal{SHC}$ the stable homotopy category and $G=\{ \mathbb{S} \}$ the singleton of the sphere spectrum. If someone has any idea to prove the statement in this scenario he is welcomed. – 2017-02-12