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Proof. Let X be a compact with metric d, let f be a continuous, one- to-one, positively expansive map of X into (not necessarily onto) itself, and let ǫ> 0 be an expansive constant. Consider the following condition (∗) there exists n ≥ 1 such that if d(f i(x),f i(y)) ≤ ǫ for i = 1, 2,...,n, then d(x, y) ≤ ǫ, too.

Suppose that (∗) is not true. Then for every n ≥ 1, there exist xn,yn such that i i d(f (xn),f (yn)) ≤ ǫ for i =1, 2,...,n, but d(xn,yn) >ǫ. Choose convergent subse- i i quences xnk → x and ynk → y. Then x 6= y and for every i ≥ 1, d(f (xnk ),f (ynk )) → i i i i i i d(f (x),f (y)). Now d(f (xnk ),f (ynk )) ≤ ǫ for i =1, 2,...,nk, so d(f (x),f (y)) ≤ ǫ for all i ≥ 1. But f is one-to-one, so f(x) 6= f(y). This contradicts positive ex- pansiveness (with f(x) and f(y) in place of x and y), so the condition holds. Fix k ≥ 0, and apply (∗) consecutively for j = k, k − 1,..., 1, 0, with f j (x) and f j (y) in place of x and y. We get (∗∗) for every k ≥ 0, if d(f i(x),f i(y)) ≤ ǫ for i = k +1, k +2,...,k + n, then d(f i(x),f i(y)) ≤ ǫ for i =0, 1,...,k, too. Now cover X by finitely many, say N, open sets of the form

U(z) := {x ∈ X : d(f i(x),f i(z)) ≤ ǫ/2 for i =1, 2,...,n}, where n is as in (∗). If X contains more then N points, consider a finite subset containing N+1 points. k k For every k ≥ 0, there exist xk 6= yk in this subset such that f (xk) and f (yk) lie i i in the same set U(z). Then by (∗∗), d(f (xk),f (yk)) ≤ ǫ for i = 0, 1,...,k + n. Since there are only finitely many pairs of these N + 1 points, there exist x 6= y such that d(f i(x),f i(y)) ≤ ǫ for infinitely many i ≥ 0, and hence for all i ≥ 0. This contradicts positive expansiveness.

References

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