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The restriction to A of any trace in (D) is a tracial state, so (A)≠∅ . Given a positive integer n , apply Lemma 3 ([Full positive elements with small tracial support]) to obtain b∈A_+ with b=1 and ρ(b)≤ρ((b))<1n (ρ∈(A)). Choose a finite-spectrum self-adjoint a∈D with a-b<1/8 , and set g=1_ [1/2,∞) (a) . Then gbg≥(3/8)g , so restriction of any τ∈(D) to A gives τ(g)≤83τ(b)≤(8)/(3n). We claim that g is full in D . Otherwise, let π:D→D/I be the quotient by the proper closed ideal generated by g . The inequalities a≥-1/8 and a(1-g)≤(1/2)(1-g) imply π(a)≤1/2 , and hence π(b)≤5/8 . On the other hand, π|_A is a unital homomorphism from the simple algebra A to a nonzero unital algebra. It is injective and therefore isometric, contradicting b=1 . Thus g is full, and choosing n with 8/(3n)<δ finishes the proof.
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