**Proof.**
The equivalence of (1) and (2) follows immediately from Definition 34.8.4 and the fact that a refinement of a refinement is a refinement. Because of the equivalence of (1) and (2) and since $\{ T_ i \to T\} _{i \in I}$ refines $\{ \coprod _{i \in I} T_ i \to T\} $ we see that (1) implies (3). Finally, assume (3) holds. Let $U \subset T$ be an affine open and let $\{ U_ j \to U\} _{j = 1, \ldots , m}$ be a standard ph covering which refines $\{ U \times _ T \coprod _{i \in I} T_ i \to U\} $. This means that for each $j$ we have a morphism

\[ h_ j : U_ j \longrightarrow U \times _ T \coprod \nolimits _{i \in I} T_ i = \coprod \nolimits _{i \in I} U \times _ T T_ i \]

over $U$. Since $U_ j$ is quasi-compact, we get disjoint union decompositions $U_ j = \coprod _{i \in I} U_{j, i}$ by open and closed subschemes almost all of which are empty such that $h_ j|_{U_{j, i}}$ maps $U_{j, i}$ into $U \times _ T T_ i$. It follows that

\[ \{ U_{j, i} \to U\} _{j = 1, \ldots , m,\ i \in I,\ U_{j, i} \not= \emptyset } \]

is a standard ph covering (small detail omitted) refining $\{ U \times _ T T_ i \to U\} _{i \in I}$. Thus (1) holds.
$\square$

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