Some modern mathematics for physicists and other outsiders by Paul Roman

By Paul Roman

Publication through Roman, Paul

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36 Suppose that X and Y are metric spaces and f : X −→ Y is a function. Show that the following two statements are equivalent: (a) f is uniformly continuous; (b) for all sequences {un }n 1 , {xn }n 1 ⊆ X such that dX (un , xn ) −→ 0, we have dY f (un ), f (xn ) −→ 0. 2. 37 (a) Let X = Cb (R) (the space of bounded continuous functions f : R −→ R) be furnished with the supremum metric d∞ (f, g) = sup f (t) − g(t) ∀ f, g ∈ X. t∈R def For f ∈ X and r ∈ R, we set fr (t) = f (t + r). Then fr ∈ X. Show that, if f ∈ X is uniformly continuous, then d∞ (fr , f ) −→ 0 as r → 0+ .

We furnish B(E) with the supremum metric d∞ (f, g) = sup f (s) − g(s) . def s∈E Show that B(E), d∞ is a complete metric space. 26 Suppose that X is a separable metric space and f : X −→ R is a function. Let L be the set of all strict local minimizers of f . Show that L is at most countable. 27 (a) Let (X, dX ) and (Y, dY ) be two metric spaces, let f : X −→ Y be a continuous function and let {Cn }n 1 be a sequence of subsets of X Cn = ∅ and diam Cn −→ 0. Show that diam f (Cn ) −→ 0 such that n 1 as n → +∞.

N , the composite function pk ◦ f : Y −→ Xk is continuous (respectively, uniformly continuous). 30 Chapter 1. 127 If (Xk , dXk ), for k = 1, . . , N are metric spaces and Ek ⊆ Xk , for k = 1, . . , N , then d∞ X N Ek N dX = k=1 Ek k . k=1 We can also consider infinite metric products. So, suppose that we have a sequence (Xn , dXn ) n 1 of metric space such that sup dXn (x, y) : x, y ∈ X, n 1 M for some M > 0. 3(b)). 128 (X, dˆX ) is a metric space. We consider the family B = Brm (x) = m n=1 BrXn (xn ) × Xn : x ∈ X = n m+1 Xn , m ∈ N, r > 0 .

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