Letting n,m → ∞ with n−m = 1 mod p we conclude α(d(x n ,x m )) → 1. Butsince α ∈ S this implies d(x n ,x m ) → 0, which is a contradiction. Therefore,given any ε > 0 there exists N ∈ N such that if n,m ≥ N and n − m = 1mod p, d(x n ,x m ) ≤ ε/p. By Step 1 it is possible to choose N 1 ∈ N so thatd(x n ,x n+1 ) ≤ ε/p if n ≥ N 1 . Now let n,m ≥ max{N,N 1 } with m > n.Then there exists k ∈ {1,2,· · ·,p} such that and n − m = k mod p. Thus
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