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METRIC SPACES. MATH 113 - SPRING 2015. PROBLEM SET #1. Problem 1 (Distance to a subset and metric Urysohn's Lemma). Let (E,d) be a metric space. Solution. 1. Observe that A ?. ?. A so d(x, A) ? d(x,. ?. A). For the other inequal- ity, consider ? in. ?. A. There exists a sequence {an} ? AN that converges to ?.
single theorem may be applied to handle seemingly different problems. justification of differentiation under the integral sign, and establishment of a solution 2 Metric Spaces. 2.1 Basic Definitions. First we recall certain fundamental properties of real numbers: for all x, y, z ? R,. (i) |x ? y| ? 0;|x ? y| = 0 if, and only if, x = y;.
MA2223: Metric spaces (2016). Homework 1, Solutions. Sergey Mozgovoy. Problem 1: Which of the following maps d : R ? R > R. 1. d(x, y) = v|x ? y|,. 2. d(x, y) = |x2 ? y2|,. 3. d(x, y) = |x?y|. 1+|x?y|. , for x, y ? R, define a metric on R? Solution: 1. This is a metric (one should verify the axioms). 2. This is not a metric: d(?1,1)
25 Nov 2015 Show that the ?-? definition of continuity of a map f : X > Y between metric spaces is equivalent to the definition of continuity in terms of open subsets. Proof. Suppose f is ?-?-continuous (i.e., continuous in the sense of the ?-? definition of continuity) and that U is an open subset of Y . We need to show that
Topology I - Exercises and Solutions. July 25, 2014. 1 Metric spaces. 1. Let p be a prime number, and d: Z ? Z > [0, +?) be a function defined by dp(x, y) = p? max{m?N : pm|x?y}. . Let (X, d) be a metric space and U = X be an open subset in X. Consider a function dU : U ?U > .. Isolated points can cause a problem.
metric in a function space evolved from classical brachistochrone problem of variational calculus. Solution Since | x – y | ? 0 and | x – y | = 0 iff x = y, (d1) follows. .. is also a metric. The new metric spaces {(X, d m. )/m = 1, 2, } are thus obtained from (X, d). Solution: (i) d m. (x, y) = md(x, y) ? 0 ? x, y ? X. Moreover d m.
MAS331: Metric Spaces 2016-17. Solutions to Problems on Chapter 4. 1. (a) Let (x, y) ? R2. Take a sequence (xn,yn) in R2 tending to (x, y). Then, by the d2-version of Prop.2.9, xn > x and yn > y. By the algebra of limits, xnyn > xy, or, in other words, f(xn,yn) > f(x, y). Therefore f is continuous at (x, y), and therefore (since
18. Consider Q as a metric space with the usual distance function d(x, y) = |x ? y|, and define. S = {x ? Q : 2 < x2 < 3}. Show that S is closed and bounded in Q, but that S is not compact. Solution. Note that Sc = {x ? Q : x2 < 2 or x2 > 3} since ±. v. 2 /? Q, ±. v. 3 /? Q. Let x ? Sc. Then x2 > 3 or x2 < 2. If x2 > 3 and x > 0
Problem Set 4: Solutions. Math 201A: Fall 2016. Problem 1. Let f : X > Y be a one-to-one, onto map between metric spaces X, Y . (a) If f is continuous and X is compact, prove that f is a homeomorphism. Does this result remain true if X is not compact? (b) Suppose that f is uniformly continuous and f?1 is continuous. If Y is.
of compactness of a set in a metric space (e .g . by checking in Rudin). (3) Show that S is not compact by considering the sequence in lp with kth element the sequence which is all zeros except for a 1 in the kth slot. Note that the main problem is not to get yourself confused about sequences of sequences! Solution 5.13
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