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Hence, we can indeed pick in line 8 a bπœ‡π‘—βˆˆπ‘€πœ‡ that |π‘€πœ‡βˆ©π΅π‘‘ 2πœ€β€²(bπœ‡π‘—)|β‰₯3 5 𝑅> 2 5 𝑅, and by Lemma A.2, there is a𝑗 β€²βˆˆ[π‘˜]thatβˆ₯bπœ‡π‘—βˆ’πœ‡π‘—β€²βˆ₯2≀3πœ€β€²β‰€πœ€

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