1 Selected Homework Solutions

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1 Selected Homework Solutions Mathematics 4600 A. Bathi Kasturiarachi September Selected Solutions to HW # HW #: (.) 5, 7, 8, 0; (.2):, 2 ; (.4): ; (.5): 3 (.): #0 For each of the following subsets of R R, determine whether it is equal to the Cartesian product of two subsets of R. (a) f(x; y)jx is an integerg = Z R. (b) f(x; y)j0 < y 6 g = R (0; ]: (c) f(x; y)jy > xg. This cannot be written as a Cartesian product of two subsets of R. Suppose X = f(x; y)jy > xg = A B for some A; B subsets of R. The points (0; ) 2 X and (; 2) 2 X. Since 2 A and 2 B, we conclude that (; ) 2 A B. This is not true since (; ) =2 X. (d) f(x; y)jx is not an integer and y is an integerg = Z Z. (e) f(x; y)jx 2 + y 2 < g. This cannot be written as a Cartesian product of two subsets of R. For instance the points (0:8; :) and (0:; 0:8) are both in the set, since 0: : 2 = 0:65 <. However, if the given set could be written as a Cartesian product, then (0:8; 0:8) should also be in the set. But it is not, since 0: :8 2 = :28 >. Noticed we used 0:8 since this is a number greater than p 2. (.5): #3 Let A = A A 2 and B = B B 2. (a) Show that if B i A i for all i, then B A. For purposes of the entire problem, we will de ne the following. Let X = [ [ B j and Y = A j. Then a a!-tuple can be de ned as j2z + j2z + follows. x : Z +! X where x = (b ; b 2 ; ) is an in nite sequence of elements of X, with b i 2 B i for each i and x : Z +! Y where x = (a ; a 2 ; ) is an in nite sequence of elements of Y, with a i 2 A i for each i. Now let x be an element of B, and treated as a!-tuple of elements of X, x i = x(i) 2 B i A i by hypothesis for all i. Therefore, we can treated x as a!-tuple of elements of Y, x i = x(i) 2 A i for all i. This implies x be an element of A. Hence, B A.

2 (a) Show that the converse of (a) holds if B is non empty. If B =? then B A trivially. From this it will not follow that B i A i for all i. For example, if B =?, B 2 = fb ; b 2 g, A = fa g, and A 2 = fb g then B = B B 2 =?, A = A A 2 = f(a ; b )g. In this example, clearly B A, but B 2 * A 2. Therefore, we will assume that B 6=?. Then there exists (b ; b 2 ; ) 2 B such that b i 2 B i for all i. Fix an index j 2 Z +. Let b 0 j 2 B j be an arbitrary element of B j. Then (b ; b 2 ; ; b j ; b 0 j; b j+ ; ) 2 B A (by hypothesis) =) b 0 j 2 A j. Since b 0 j was arbitrary, we have B j A j. Varying j we obtain the desired result. (b) Show that if A is non empty, then each A i is non empty. Does the converse hold? Suppose A is non empty. Then there exists (a ; a 2 ; ) 2 A such that a i 2 A i for all i. Therefore, A i 6=?. Conversely, suppose that for each i, A i 6=?. Choose a i 2 A i and construct an!-tuple (a ; a 2 ; ) which will belong to A by de nition, so A is non empty. (c) What is the relationship between the set A [ B and the Cartesian product of the sets A i [ B i? i.e. What is the relationship between A [ B and (A i [ B i )? What is the relationship between the set A \ B and the Cartesian product of the sets A i \ B i? i.e. What is the relationship between A \ B and (A i \ B i )? Proposition. Suppose A = A A 2 and B = B B 2 are both non empty. Then A [ B (A i [ B i ) with equality if and only if A i = B i for all i. 2

3 0 Proof : First let K = Y [ X [ j2z + A j Then we can de ne an!-tuple x by 0 A [ j2z + B j A = [ j2z + (A j [ B j ). x : Z +! K where x = (x ; x 2 ; ) is an in nite sequence of elements of K, with x i 2 A i [ B i for each i. Now let x be an element of A[B, then x 2 A or x 2 B. That is, x i = x(i) 2 A i for all i or x i = x(i) 2 B i for all i. Therefore, establishing A [ B x i = x(i) 2 A i [ B i for all i 2 Z +, (A i [ B i ). Conversely, if we take an!-tuple x = (x ; x 2 ; ) 2 all i (A i [ B i ), then for x i 2 A i [ B i =) x i 2 A i or x i 2 B i. However, from the above we cannot conclude that x 2 A i or x 2 (unless A i = B i - some component can belong to an A m and not to any B i while another component can belong to a B n and not to any A i ) Suppose, for some i 2 Z +, A i 6= B i :Then there exists a 2 A i such that a =2 B i. Construct the following!-tuple. y= (y ; y 2 ; ) = (b ; b 2 ; ; b i ; a; b i+ ; ) where each y j = b j 2 B j for all j 6= i and y i = a. It is clear that, y =2 A and y =2 B, so y =2 A [ B. This leaves us with only one case to consider, namely, A i = B i for all i 2 Z +. In this case, (A i [ B i ) = A i = A = B = A [ B. Proposition 2. Suppose A i \ B i 6=? for each i 2 Z +. Then B i A \ B = (A i \ B i ). Proof : Suppose A \ B =?. Then all pairings A i \ B j must also be empty. Hence (A i \ B i ) =? and we have equality. Therefore, let us assume that A \ B 6=?. Let x 2 A \ B. Then x 2 A and x 2 B. This means 3

4 that x = (x ; x 2 ; ) with x i 2 A i and x i 2 B i =) x i 2 A i \ B i for each i. This implies that, x 2 (A i \ B i ). Therefore, A \ B (A i \ B i ). To prove the converse, we will assume that A i \ B i 6=? for each i 2 Z +. Then by part (c), (A i \ B i ) is non empty. So select x 2 (A i \ B i ). Then x i 2 A i \ B i =) x i 2 A i and x i 2 B i for all i =) x 2 A i and x 2 B i =) x 2 A \ B. Therefore, (A i \ B i ) A \ B..2 Selected Solutions to HW #2 HW #2: (.6) 2, 5; (.7):, 4 (.7): # Show that Q is countably in nite. First note that Q = Q [ f0g [ Q +, where the unions are disjoint and Q = fx 2 Q j x > 0g, Q + = fx 2 Q j x < 0g. First show that Q + is countable. Let f : Z + Z +! Q + be de ned by, f((m; n)) = m n. Let x 2 Q +. Then x = m n for some (m; n) 2 Z + Z +. Therefore, f((m; n)) = x, so f is surjective. Since Z + Z + is countably in nite, there exists a surjection g : Z +! Z + Z + (by Theorem 7.). The function f g is a surjection since it the composition of two surjections, and is given by, f g : Z +! Q +. Now appeal to Theorem 7. again to conclude that Q + is countable. Next we want to show that Q is countable. Let h : Z + Z +! Q be de ned by, h((m; n)) = m n. Let x 2 Q. Then x = m n for some (m; n) 2 Z + Z +. Therefore, h((m; n)) = x, so h is surjective. Like before the composition h g is a surjection, and is given by, 4

5 h g : Z +! Q. By Theorem 7. we conclude that Q is countable. Finally, Q = Q [ f0g [ Q + is the disjoint union of three countable sets, so is countable..3 Selected Solutions to HW #3 HW #3: (2.3), 4; (2.6): 3, 4, 6 (2.3): # Let X be a topological space; let A be a subset of X. Suppose that for each x 2 A there is an open set U containing x such that U A. Show that A is open in X. (2.3): #4 Let x 2 A. Then there exists an open set U x such that x 2 U x A (by hypothesis). Let U = [ U x. Notice that U is open since X is a topological x2a space. We claim that U = A. For each x 2 A there exists an open set U x such that x 2 U x [ U x, so A U. The converse is easy, because each x2a U x A =) U = [ U x A. Therefore, A is open. x2a (a). If ft g is a family of topologies on X, show that \ T is a topology on X. Is [ \ T : (i)? 2 \ T a topology on X? T, since? 2 T for all. X 2 \ T, since X 2 T for all. (ii) Let fu g be a non empty collection of open sets in \ T. Then fu g 2 T for all. Then [ U 2 T for all, since each T is a topology. Hence [ U 2 \ T. (iii) Let fu i g n be a nite non empty collection of open sets in \ T. Then n\ fu i g 2 T for all. Then U i 2 T for all, since each T is a topology. 5

6 n\ Hence U i 2 \ T. Therefore, \ T is a topology. In general, [ T is not a topology, unless we agree to include all arbitrary unions and nite intersections of open sets into the collection. Consider the following example. Let X = fa; b; cg. Let T = f?; X; fag; fa; cgg and T 2 = f?; X; fbg; fb; cgg. It is easy to show that these are topologies. However, fa; cg \ fb; cg = fcg =2 T [ T 2, thus it is not a topology. (b). If ft g is a family of topologies on X, show that there is a unique smallest topology on X containing all the collections T, and a unique largest topology contained in all T. The unique smallest topology that contains all of the topologies T of X, is [ T together with all arbitrary unions and nite intersections. Call this collection F. One can easily check that F is a topology. Certainly, F contains all the topologies T. Suppose G is another topology that contains all the topologies T and is strictly coarser than F. This means that there is an open set U in G such that U =2 F. It cannot be the case that U 2 T for some. If so, U 2 F. Therefore, U must be an open set that is constructed by using arbitrary unions and (possibly) nite intersection of open sets from di erent T and T topologies. But in this case, U 2 F by the de nition of F. Therefore, there cannot be a smaller topology other than F that contains all of the topologies T. The unique largest topology that is contained in all of the topologies T of X, is \ T. We have already shown that this is a topology. It is easy to show that this topology is the largest unique topology contained in all of the topologies T. This is left as an exercise. (c). For this part check the following. Unique smallest topology containing the to given topologies is: f?; X; fag; fbgfa; bg; fb; cgg. Unique largest topology contained in each of the given topologies is: f?; X; fagg. (2.6): #3 Consider the set Y = [ ; ] as a subspace of R. Which of the following sets are open in Y? Which are open in R? In this example, it will help a lot if you can draw these sets on a number line. (a) A = fx j 2 < jxj < g. The set A is open in R, since it is the union of two open intervals in R. Namely, 6

7 A = ( ; 2 ) [ ( 2 ; ). It is also open in Y since A = ( ; 2 ) [ ( 2 ; ) \ Y, which shows it is the intersection of Y with an open set in R. b. B = fx j 2 < jxj 6 g. The set B is not open in R, since neither interval [ ; 2 ) nor ( 2 ; ] is open in R. However it is open in Y since, B = ( 2; 2 ) [ ( 2 ; 2) \ Y, which shows it is the intersection of Y with an open set in R. c. C = fx j 2 6 jxj < g. The set C is not open in R, since neither interval ( ; 2 ] nor [ 2 ; ) is open in R. It is also not open in Y since we cannot intersect Y with an open set in R, to obtain an interior end point of an interval. d. D = fx j 2 6 jxj 6 g. The set D is not open in R, nor in Y. Same argument as part (c) holds. e. E = fx j 2 < jxj < and x =2 Z +g. The set E is not open in R. We can write E = ( ; 0) [ f(0; ) r Kg, where K = f n jn 2 Z +g. Then E can be expressed as, E = ( ; 0) [ ( [ n= n + ; ), n ( which shows ( that E is open in R. It is also open in Y since E = E \ Y = [ n)) ( ; 0) [ n+ ; \ Y. n= (2.6): #4 Show that the maps : X Y! X and 2 : X Y! Y are open. Let W be a non empty open set of X Y. Consider (W ) X and 2 (W ) Y. We must show that (W ) is open in X and 2 (W ) is open in Y. Choose x 2 (W ). Since is surjective, there exists (x; y) 2 X Y such that (x; y) = x, where (x; y) 2 W. Since W is open in X Y, there exists an open set U V of X Y, such that (x; y) 2 U V X Y, where U is open in X and V is open in Y. By de nition (U V ) = U and it follows that x 2 U (W ), hence (W ) is open in X. To show 2 (W ) is open in Y, we simply observe that 2 (U V ) = V and that y 2 V 2 (W ). 7

8 (2.6): #6 Show that the countable collection is a basis for R 2 : f(a; b) (c; d) j a < b and c < d; a; b; c; d 2 Qg We will rst prove that the collection f(a; b) j a < b and a; b 2 Qg is a basis for R in the standard topology. (i) Let x 2 R. Since rationals are dense in the reals, there exists a 2 Q such that x < a < x and b 2 Q such that x < b < x +. Therefore, x 2 (a; b) where a; b 2 Q. (ii) Now suppose x 2 (a; b) \ (c; d) where a; b; c; d 2 Q. Since the non empty intersection of two open intervals is open, we can easily nd an open interval (p; q) with p; q 2 Q, such that x 2 (p; q) (a; b) \ (c; d). Therefore, by de nition the collection f(a; b) j a < b and a; b 2 Qg is a basis for R. Finally, we appeal to Theorem 5. in the text (page 86) to conclude that the collection f(a; b) (c; d) j a < b and c < d; a; b; c; d 2 Qg is a basis for the topology on R R = R 2. 8