By Michel Hervé

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3 of the previous section. 1 (The Mazur–Ulam theorem) If L : E → F is an isometry of a real normed space (E, . E ) onto a real normed space (F, . F ) with L(0) = 0, then L is a linear mapping. In order to prove this, we introduce some ideas concerning the geometry of metric spaces, of interest in their own right. First, suppose that x, y, z are elements of a metric space. We say that y is between x and z if d(x, y) + d(y, z) = d(x, z), and we say that y is halfway between x and z if d(x, y) = 324 Metric spaces and normed spaces d(y, z) = 12 d(x, z).

3 Suppose that (X, d), (Y, ρ) and (Z, σ) are metric spaces, that f is a continuous surjective mapping of (X, d) onto (Y, ρ) and that g : (Y, ρ) → (Z, σ) is continuous. Show that if g ◦ f is a homeomorphism of (X, d) onto (Z, σ) then f is a homeomorphism of (X, d) onto (Y, ρ) and g is a homeomorphism of (Y, ρ) onto (Z, σ). 4 Show that the punctured unit sphere {x ∈ Rd : x = 1} \ {(1, 0, . . , 0)} of Rd , with its usual metric, is homeomorphic to Rd−1 . 5 Give an example of three metric subspaces A, B and C of R such that A ⊂ B ⊂ C, A and C are homeomorphic, and B and C are not homeomorphic.

If x ∈ Nδ (a), then f (x) E ≤ f (x) − f (a) E + f (a) E ≤ η + M , so that λ(x)f (x) − λ(a)f (a) E = (λ(x) − λ(a))f (x) + λ(a)(f (x) − f (a)) ≤ |λ(x) − λ(a)|. f (x) E E + |λ(a)|. f (x) − f (a) E ≤ η(η + M ) + M η ≤ . (v) Suppose that > 0. Let η = |λ(a)|2 /2. There exists δ > 0 such that |λ(x) − λ(a)| < max(|λ(a)|/2, η) for x ∈ Nδ (a). If x ∈ Nδ (a), then |λ(x)| ≥ |λ(a)|/2, and so 1 2η 1 λ(a) − λ(x) − = ≤ = . 4 (The sandwich principle) Suppose that f , g and h are real-valued functions on a metric space (X, d), and that there exists η > 0 such that f (x) ≤ g(x) ≤ h(x) for all x ∈ Nη (a), and that f (a) = g(a) = h(a).