By William Anthony Granville

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13 (Urysohn metrization theorem) A second countable topological space X is metrizable iff X is normal. Proof. Clearly a metric space is normal. For the converse, let (Bi ) be a countable base for X. For each pair with Bi ⊂ Bj , construct a continuous function fij : X → [0, 1] with fij = 0 on Bi and fij = 1 outside Bj . Let F be the collection of all such functions, and consider the natural continuous map f : X → [0, 1]F , sending x to (fij (x)). Since F is countable, f (X) is metrizable; we need only show that the inverse map f (X) → X is defined and continuous.

Construct a measure-preserving map within distance r of the identity, such that g(yi ) = xi+1 . This map g is supported close to Pi . ; so that ultimately xn is in the forward orbit of x0 , and hence f moves B1 into B2 . Using a countable base for X, we can now conclude that a generic f ∈ M (X) has the property that for any two nonempty open sets U, V ⊂ X, there exists an n > 0 such that f n (U ) ∩ V = ∅. We claim any such f has a dense orbit. Indeed, consider for any open ball B the set U (B) of x such that f n (x) ∈ B for some n > 0.

Define G(x) = φ(χ[a,x] ). |ai − bi | < δ, we Then for any collection of disjoint intervals (ai , bi ) with have |G(ai ) − G(bi )| = |φ(χ(ai ,bi ) )| = φ 1/p |ai − bi | ≤ M ±χ(ai ,bi ) ≤ M δ1/p , since φ(f ) ≤ M f p . Thus G(x) is absolutely continuous, and thus there is an integrable function g(x) = G′ (x) ∈ L1 [a, b] such that G′ (x) dx = φ(χI ) = χI g I for any interval I ⊂ [a, b]. ) Next we check that φ(f ) = f, g for all f ∈ L∞ [a, b]. Indeed, if |f | ≤ A there are step functions fn → f with fn → f pointwise and |fn | ≤ A.