1st December 2004

Munkres §28

Ex. 28.1 (Morten Poulsen). Let d denote the uniform metric. Choose c ∈ (0, 1]. Let A = {0, c}^{ω} ⊂ [0, 1]^{ω}. Note that if a and b are distinct points in A then d(a, b) = c. For any x ∈ X the open ball B_{d}(x, c/3) has diameter less than or equal 2c/3, hence B_{d}(x, c/3) cannot contain more than one point of A. It follows that x is not a limit point of A.

Ex. 28.6 (Morten Poulsen).

Theorem 1. Let (X, d) be a compact metric space. If f : X → X is an isometry then f is a homeomorphism.

Proof. Clearly any isometry is continuous and injective. If f surjective then f^{−1} is also an isometry, hence it suffices to show that f is surjective.

Suppose f(X) ⊊ X and let a ∈ X − f(X). Note that f(X) is compact, since X compact, hence f(X) closed, since X Hausdorff, i.e. X − f(X) is open. Thus there exists ε > 0 such that a ∈ B_{d}(a, ε) ⊂ X − f(X).

Define a sequence (x_{n}) by

x_{n} = \{ \begin{matrix}a, & n = 1 \\ f(x_{n}), & n > 1. \end{matrix}

If n ≠ m then d(x_{n}, x_{m}) ≥ ε: Induction on n ≥ 1. If n = 1 then clearly d(a, x_{m}) ≥ ε, since x_{m} ∈ f(X). Suppose d(x_{n}, x_{m}) ≥ ε for all m ≠ n. If m = 1 then d(x_{n+1}, x_{1}) = d(f(x_{n}), a) ≥ ε. If m > 1 then d(x_{n+1}, x_{m}) = d(f(x_{n}), f(x_{m−1})) = d(x_{n}, x_{m−1}) ≥ ε.

For any x ∈ X the open ball B_{d}(x, ε/3) has diameter less than or equal to 2ε/3, hence B_{d}(x, ε/3) cannot contain more than one point of A. It follows that x is not a limit point of A. □