Fall 2010 Midterm Solutions

# Fall 2010 Midterm Solutions - R then inf E is an element of...

This preview shows pages 1–2. Sign up to view the full content.

Math 140A, Fall 2010, Midterm, 11/8/10 Instructions . Answer all questions. You may use without proof anything which was proved in class. Cite a theorem either by name, if it has one, or by brieFy stating what it says. 1. (20 points) Give an example of an open cover of the interval (0 , 1) R which has no ±nite subcover. Solution . Let U i = (1 /i, 1), i 2. Then { U i } is a cover of (0 , 1) with no ±nite subcover. 2. (10 points each). True or false? ²or each one, give a brief reason (a complete proof is not necessary) if true, and a counterexample or brief reason as appropriate if false. (a) The set of irrational real numbers is uncountable. Solution . True: R is uncountable and Q is countable, hence R \ Q is still uncountable; otherwise R = ( R \ Q ) Q would be countable. (b) Let E R be the set of rational numbers x such that 0 < x < 1. Is 1 / 2 an interior point of E ? Solution . No: any ball centered at 1 / 2 necessarily contains an irrational number, which won’t be in E . (c) If E is a bounded subset of

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: R , then inf E is an element of E . Solution . No, e.g., inf(0 , 1) = 0. (d) Let U n be a sequence of open sets in R and let U = ∞ i n =1 U n . Then every point of U is an interior point of U . Solution . No, e.g., U n = (-1 /n, 1 /n ). Then U = { } , and 0 is not an interior point of U . (e) Any bounded sequence in R converges. Solution . No, e.g., a n = (-1) n . (f) Any Cauchy sequence in a metric space converges. Solution . No, e.g., a n consists of the ±rst n digits of the decimal expansion for √ 2. 1 3. (20 points) Show using the defnition oF convergence that lim n →∞ n 2 n 2 + 1 = 1 . No points will be awarded if the deFnition isn’t used! Solution . Let ǫ > 0. Let N be an integer satisfying N > r (1 /ǫ )-1. Then for n ≥ N , we have v v v v n 2 n 2 + 1-1 v v v v = 1 n 2 + 1 < 1 ( r (1 /ǫ )-1) 2 + 1 = ǫ. This gives the desired convergence. 2...
View Full Document

## This note was uploaded on 03/19/2012 for the course MATH 171a taught by Professor Staff during the Spring '08 term at UCSD.

### Page1 / 2

Fall 2010 Midterm Solutions - R then inf E is an element of...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online