MATH HOMEWORK 2 1

Transcription

1 MATH HOMEWORK 2 1 Problem (1.40). For i Z, let A i = {i 1,i+1}. Determine the following: (a) (b) (c) A 2i (A i A i+1 ) (A 2i 1 A 2i+1 ) Solution. (a) Since A 2i = {2i 1,2i+1} and A 2i = A 2 A 4 A 6 A 8 A 10 = {1,3} {3,5} {9,11}, it follows that A 2i = {1,3,5,7,9,11}. (b) Observe that A i A i+1 = for any i Z. Therefore (A i A i+1 ) =. (c) NotethatA 2i 1 = {2i 2,2i}andA 2i+1 = {2i,2i+2}. Since2iiscommon to each set, A 2i 1 A 2i+1 = {2i}. So the union of the intersections is a set of even numbers, namely (A 2i 1 A 2i+1 ) = {2,4,6,8,10}. Remark. Although I did not do this in my writeup of the homework solution, it is always a good idea use actual values of i to (literally) see what the sets are. Common L A TEX problems include spacing (mathematical expressions that should be in display mode and not inline) and missing braces for indices, so that A 2i is incorrectly displayed as A 2 i (notice the misalignment of the 2 and i when no braces are used).

2 MATH HOMEWORK 2 2 Problem (1.45). For n N, let A n = ( 1 n, 1 n ). Determine n N A n and n NA n. Solution. The first few sets are: ( A 1 = ( 1,1), A 2 = 1 2, 1 2 ), A 3 = ( 1 3, 1 ) (, and A 4 = 1 3 4, 1 ). 4 For any n > 1, A n is a subset of A 1. Therefore n NA n = A 1 = ( 1,1). For the intersection, note that as n grows larger, the set A n is an interval that is smaller in width. Therefore only element that is common to all the intervals is 0. Hence n NA n = {0}. Remark. Use\displaystyle if you wish to have the large intersection or union symbol with the index underneath the symbols (as opposed to subscript form). For example, $\displaystyle \bigcap_{n\in \N} A_n$ will produce n NA n. There is a shortcut \ds if you use the custom package.

3 MATH HOMEWORK 2 3 Problem (1.46). Which ofthe followingarepartitions ofa={a,b,c,d,e,f,g}? For each collectin of subsets that is not a partition of A, explain your answer. (a) S 1 = {{a,c,e,g},{b,f},{d}} (b) S 2 = {{a,b,c,d},{e,f}} (c) S 3 = {A} (d) S 4 = {{a},,{b,c,d},{e,f,g}} (e) S 5 = {{a,b,c},{b,g},{b,f}} Solution. Only S 1 and S 3 are partitions of A. This can be verified by checking that the union of each element(of S 1 and of S 3 ) is A itself, no element is empty, and no two elements have a nonempty intersection. This last condition is vacuosly true for S 3 since there is only one element: the set A itself. On the other hand, S 2 is not a partition as the union of the elements of S 2 is not the set A since the union would not contain g. The set S 4 is not a partition of A as the empty set is a member of S 4. Lastly, S 5 is not a partition of A since the elements {b,g} and {b,f} have a nonempty intersection (the element b is common to both). Remark. Note that {b,f} is indeed a subset. However, to remove any ambiguity, we should really state that {b,f} is a subset of A. Be very careful about using the phrase subset of and subset in when referring to partitions. The statement The set {b,f} is a subset of S 5. is incorrect because S 5 is a collection of sets, so that subsets of S 5 would be a sets of sets (or sets of subsets of A). The correct statement is The set {b,f} is a subset in S 5. To summarize, take care with how elements of a set are described when the elements are sets themselves.

4 MATH HOMEWORK 2 4 Problem (1.52). Give an example of three sets A, S 1, and S 2 such that S 1 is a partition of A, S 2 is a partition of S 1 and S 2 < S 1 < A. Solution. Let A = {1,2,3,4,5,6}. Since A = 6 and we require S 1 to be a partition of A with S 1 < 6, one possible choice is S 1 = {{1,2},{3,4},{5,6}}. In this case, S 1 = 3. To obtain a partition S 2 of S 1 with S 2 < 3, we can take S 3 to be the partition S 3 = { { {1,2},{3,4} }, { {5,6} } }. Since S 3 = 2, all conditions are satisfied. Remark. A partition of A is a collection of nonempty subsets of A, and the union of these subsets must be equal to A. So if S is a partition of A, each element of S must be a subset of A containing at least one element of A.

5 MATH HOMEWORK 2 5 Problem (1.66). For A = {a R : a 1} and B = {b R : b = 1}, give a geometric description of the points in the xy-plane belonging to(a B) (B A). Solution. The set A is simply interval ( 1,1), and B = { 1,1}. The product A B is the collection of all the points (x,y) whose x-value satisfies 1 x 1 and whose y-value is either 1 or 1. Geometrically, these are two horizontal line segments: onewithendpoints( 1, 1)and(1, 1)andtheotherwithendpoints ( 1,1) and (1,1). Similarly, B A is the collection of points whose x-value is either 1 or 1, and whose y-value satisfies 1 y 1. These points make two vertical line segments with endpoints ( 1, 1) and ( 1,1), and (1, 1) and (1,1). Lastly, the union (A B) (B A) is the square whose corners are located at (±1, ±1). Remark. The product of two sets A and B is A B. We do not write (A B) unless there is a need for explicit order of operations. For example, if we have the expression A B B A, it is possible to interpret this as (a) A (B B) A, which consists of triplets (x,y,z) with x A, y B, and z A; or as (b) (A B) (B A), which is the union of two separate products. The ambiguity arises in which operation we should carry out first: the union or the product. This is why there are parentheses used in the problem statement. However, if we are referring to only A B, then there is no reason to include the parentheses as there is only one operation:.

6 MATH HOMEWORK 2 6 Problem (1.78). Let I denote the interval [0, ). For each r I, define A r = {(x,y) R R : x 2 +y 2 = r 2 } B r = {(x,y) R R : x 2 +y 2 r 2 } C r = {(x,y) R R : x 2 +y 2 > r 2 }. (a) Determine A r and r. r I r IA (b) Determine r I B r and r IB r. (c) Determine r I C r and r IC r. Solution. First note that the equation x 2 +y 2 = r 2 is the equation of a circle centered at (0,0) with radius r. So A r is simply the set of points on a circle of radius r centered at (0,0). In particular, A 0 is a set consisting of just the origin. Similarly, the inequality x 2 +y 2 r 2 consists of points on or inside the circle of radius r, and centered at (0,0). Lastly, x 2 +y 2 > r 2 consists of points outside of the circle of radius r, and centered at (0,0). With these notions, it is easy to check that (a) r I A r = R R and r IA r = ; (b) r I B r = R R and r IB r = {(0,0)}; and (c) r I C r = R R {(0,0)} and r IC r =. Remark. Many of you failed the grasp what each set looks like, and therefore had difficulty understand what the unions and intersections could be. Yet rather than send me an or ask me about it in my office (or even in class) a large number either left this problem blank (large deduction of points) or got it completely wrong.