Berestovskii V. N.'s A. D. Alexandrovs length manifolds with one-sided bounded PDF

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A set is said to be closed if its complement is open. For example, if E = R with the usual distance, the intervals (a, b), (−∞, b), (a, ∞) are open, the intervals [a, b], (−∞, b], [a, ∞) are closed, and the interval (a, b] is neither open nor closed. 33 PROPOSITION. Every open ball is open. 30 Metric Spaces Chap. 2 PROOF. Fix x and r. To show that B(x, r) is open, we need to show that for every y in B(x, r) there is a number q > 0 such that B(y, q) ⊂ B(x, r). This is accomplished by picking q = r − d(x, y).

23 for this) on E. Show that each subset A is both open and closed. For r ≤ 1, every open ball B(x, r) consists of exactly the point x. Note ¯ 1) = E for every x. ) If E is countable, then it is separable (trivially). If E is uncountable, it is not separable. Show this. 49 Half-spaces. A set H in Rn of the form H = {x : ξ · x ≤ b}, where ξ in Rn and b in R are given, is called a half-space. Show that half-spaces are closed. D. Convergence Let (E, d) be a metric space. Our goal is to discuss the notion of convergence for a sequence of points in E.

This is obvious for y = 1. Let y ∈ [0, 1); note that g(y) is the infimum of the union of all intervals Dq with q > y; that infimum cannot belong to D; so g(y) must belong to C (since it is obvious that g(y) ∈ B). Finally, we show that g : [0, 1] → C is an injection by showing that if y < z, then g(y) < g(z). Fix y < z. Note that there is at least one q in I such that y < q < z, and the corresponding set Dq is contained in {x ∈ D : f (x) > y} but not in {x ∈ D : f (x) > z}. It follows that the number g(y) is to the left of the interval Dq whereas g(z) is to the right.

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A. D. Alexandrovs length manifolds with one-sided bounded curvature by Berestovskii V. N.

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