Math 4317 : Real Analysis I Mid-Term Exam 1 25 September 2012
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1 Instructions: Answer all of the problems. Math 4317 : Real Analysis I Mid-Term Exam 1 25 September 2012 Definitions (2 points each) 1. State the definition of a metric space. A metric space (X, d) is set X with a function d : X X [0, ) such that 0 d(x, y) for all x, y X; d(x, y) = d(y, x) for all x, y X; d(x, y) = 0 if and only if x = y; d(x, y) d(x, z) + d(z, y) for all x, y, z X. 2. State the definition of an open set in a metric space. A set G is a metric space (X, d) is open if for any point x G, there exists a r > 0 such that B r (x) G.
2 3. State the definition of the greatest lower bound of a set of real numbers. THe number α is the greatest lower bound for a set E if α is a lower bound, i.e. α x for all x E and if α is any other lower bound for the set E then we have that α α. 4. State the definition for a set to be countable. A set E is countable if there exists a bijection ϕ : N E. 5. State the Least Upper Bound Property of R. A non-empty subset of real numbers E that is bounded above, has a least upper bound.
3 or (1 point each) 1. If F is a set in a metric space that is not closed, then it is open. 2. If F is closed and K is compact, then F K is compact. 3. For any collection {G α } of open sets, G α is open. 4. For any finite collection {F 1,..., F n } of closed sets, n j=1 F j is closed. 5. Let (X, d) be a metric space. For x X and r > 0 the set is open. 6. If x, y R n, then B r (x) = {y X : d(x, y) < r} ( n j=1 x n ) 1 ( jy j j=1 x2 2 n j j=1 y2 j 7. In a metric space X, the set X is the only set that is both open and closed. ) For any collection of compact sets {K α }, α K α is compact. 9. If E is a nonempty subset of R that is bounded from below, then inf E E. 10. The set of all irrational numbers is countable. 11. A set is open if and only if it is equal to its interior. 12. A set is closed if and only if it contains its boundary. 13. A set in an arbitrary metric space is compact if and only if it is closed and bounded. 14. In a metric space (X, d), one has that r>0 B r(x) =. 15. Every infinite subset of a countable set is countable.
4 Proofs (15 pts each) 1. A set S in a metric space (X, d) is called totally bounded if for any ɛ > 0 there exists a finite subset T of S such that S x T B ɛ (x). Let K X be a compact subset of the metric space (X, d). Prove that K is totally bounded. Solution: Let ɛ > 0 be given, and consider for each x K the set B ɛ (x). This is clearly open since it is an open ball. Also note that {B ɛ (x) : x K} is an open cover of K. Since K is compact, we have that this open cover has a finite subcover, and so there exists a finite set T K such that K x T B ɛ (x). Thus, we have that K is totally bounded.
5 2. Let A be a non-empty set of real numbers which is bounded below. Let A = { x : x A}. Prove that inf A = sup( A). Solution: Note that since A is non-empty and bounded below, we have that inf A exists. Also note that A is non-empty and bounded above, so we have that sup( A) exists. First note, we have that inf A x x A. This is equivalent to Thus, we have that inf A x x A. inf A sup( A) since sup( A) is the least upper bound for the set A and we have that inf A is an upper bound for the set A. So, we have that Conversely, we have that inf A sup( A). (1) sup( A) y y ( A). Since y ( A), we have that y = x for some x A, and so we have that This implies that sup( A) x x A. sup( A) x x A. Thus, we have that sup( A) is a lower bound for the set A, and so we must have that since inf A is the greatest lower bound. Combining (1) and (2) we see that sup( A) inf A (2) inf A = sup( A).
6 3. Let a R with a > 1. Prove that (1 + a) n 1 + na for all n N. Solution: This is known as Bernoulli s inequality. Let a R with a > 1. We proceed by induction. For n = 0, (1 + a) 0 = 1 = 1 + (0)a which is trivially true. Assume that the inequality is true for some k 0. Then (1 + a) k 1 + ka. Consider the case of k+1. Since a > 1, then 1+a > 0. By assumption, (1+a) k 1+ka. Hence, (1+a) k+1 = (1+a)(1+a) k (1+a)(1+ka). Also, (1+a)(1+ka) = 1+a+ka+ka 2 = 1+(k+1)a+ka 2. Since k 0 and a 2 0, then ka 2 0 and 1+(k+1)a+ka 2 1+(k+1)a. Combining inequalities we have (1 + a) k (k + 1)a. Thus, the inequality is true when for k + 1, and by induction is true for all n N.
7 4. Let X = {x = (x 1, x 2, x 3,...) : x i R and sup i x i < }. Show that (X, d) is a metric space when d(x, y) := sup x i y i. i Solution: Let x and y be bounded infinite sequences in R. It is clear that we have since x i y i 0 for all i. 0 d(x, y) If x = y, then we clearly have that d(x, y) = 0. Suppose that d(x, y) = 0, then we have 0 = sup x i y i and so x i y i = 0 for all i, and so x i = y i for all i, or x = y. So we have that d(x, y) = 0 if and only if x = y. We also have d(x, y) = sup x i y i = sup (x i y i ) = sup y i x i = d(y, x). Finally, for the triangle inequality for absolute value, note that x i y i x i z i + z i y i sup x i z i + sup z i y i = d(x, z) + d(z, y). But, this implies that d(x, y) = sup x i y i d(x, z) + d(z, y) so the triangle inequality holds for the metric d.
8 5. Show directly from the definitions that G = {(x, y) R 2 : 4x 2 + y 2 < 1} is open in R 2 with the metric d ((x, y), (u, v)) = (x u) 2 + (y v) 2. Solution: Let (x, y) G, and set r = 1 2 (1 4x 2 + y 2 ). Then we have that B r (x, y) G. Indeed, if (u, v) B r (x, y), then we have (2u)2 + v 2 (2u 2x) 2 + (v y) 2 + (2x) 2 + y 2 2 (u v) 2 + (v y) 2 + (2x) 2 + y 2 < 2r + 4x 2 + y 2 = 1.
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