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Suppose is nowhere dense. If $(,<,+,(x \mapsto \lambda x){\lambda \in }, E)$ does not define every bounded Borel subset of every then for every we have for sufficiently large . Then there is an $n \in T : ^n \to T(E^n)E(,<,+,0,(x \mapsto \lambda x){\lambda \in }, E)$ defines every bounded Borel subset of every .

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