Problem Set 7
  7points

The problems are taken from the text. Keep in mind starred problems have solutions in the back!

NOT GRADED (do not turn in)
Chapter 2.2 1(g), 2(e,d, h,i), 6(d,e)
Chapter 2.3 1 (a,c,g,i,m), 4 (a,b, c), 18 (a,b,c)
Chapter 2.4 2 (all), 3(a), 4(b, f), 8(a,b)

GRADED (please turn in)
Chapter 2.2
10(d), 11(b,d), 13 (d), 15(c)
Chapter 2.3
1(d, j, n), 5(a,b), 6(b), 8(b), 18 (c,d)
Chapter 2.4
5(a,e,f), 6(b, e, f, h),


Partial Solutions

2.2 10(d). Remember to work with elements in the set when you are trying to show inclusions of sets. Let x be in D-A. Then x is in D but x is not in A. Then since D is a subset of B, x is in B. Since C is a subset of A, x is not in C, for otherwise it would be an element of A. Therefore, x is in B-C.

2.2 13(d). Ax B= {((2,4),(4,1)), ((2,4),(2,3)), ((3,1),(4,1)), (3,1),(2,3))}
BxA = {((4,1),(2,4)), ((4,1),(3,1)), ((2,3), (2,4)), ((2,3),(3,1))}

2.3 1(d) Union of B is N-{1}, where N is the natural numbers. Notice that 1 is not in this union because B_n is missing 1 for every n. The intersection of B is the empty set, as there is no natural number in every one of the B_n.


1 (n) The union over D is given by the union over all intervals (-n, 1/n) for all natural numbers n. It consists of real numbers x such that x is in (-n, 1/n) for some natural number n. Notice that -n can be arbitarily small as n can be arbitararily large. Thus x can be any negative number. On the other hand, as n is a natural number, the largest 1/n can be is 1 itself. Therefore, x can be any real number that is either negative, or less than 1.
Union over D is (-infinity, 1).

The intersection over D is given by those elements that are in (-n, 1/n) for every natural number n. Since 1/n can be arbitarily close to 0, there are no positive number that are less than 1/n for every natural number n. Therefore, the intersection includes no positive numbers. On the other hand, if x is negative, then it will be in (-n, 1/n) for all natural nubmer n only if x is greater than -1. Clearly 0 is in (-n,1/n) for all n. Therefore the intersection over D is given by (-1,0].

8(b) FALSE.

Here's a counterexample, thanks to Ryan O'Neal: Let B={a,b,c}, and A_1={a,b}, A_2 = {a}. Then the union of A_1 and A_2 is {a,b} and B minus this union is {c}. On the other hand, B-A_1 = {c} and B-A_2={b,c}, so the union of B-A_1 and B-A_2 is {b,c}. These two sets are not equal.

Many of you thought it was true and gave me a "proof" based on the fact that the left hand side is always included in the right hand side. Here is a proof of that:
 Let x be in B but x not in the union over alpha in Delta of A_alpha. Then x not in A_alpha for any alpha, (since if x is in A_alpha for some alpha, then x is in the union of A_alpha over all alpha). Therefore, x is B-A_alpha for any alpha. Therefore x is in the union of B-A_alpha.

The inclusion of sets does not go the other way, however.

Section 2.4
5 Let S be the set in question.
5(a) i) 5 is in S. ii) If k is in S, then k+5 is in S.
5(e) Let a, r be (fixed) real numbers.
i) a is in S. ii) If x is in S, then xr is in S.
5(f). (i) (Union from i=1 to 1 of A_i)= A_1. (ii) (Union from i=1 to n+1 of A_i) = ) (Union from i=1 to n of A_i) union A_{n+1}.

6(b) We verify this is true for n=1. In particular, 8(1)-5 = 3, and 4(1) -1 =3 as well. Now we assume it's true for n, and show it's true for n+1.

Since it's true for n, we have 3+11+19+....+ (8n-5) = 4n^2 -n.
We note that 4(n+1)^2 -(n+1) = 4(n^2 + 2n+1) - (n+1) = 4n^2 -n + (8n+4-1) = (4n^2-n)+8n+3.
Therefore
3+11+18+....+ 8n-5 + 8(n+1)-5 = 4n^2-n + 8(n+1)-5 = 4n^2-n +8n+3= 4(n+1)^2-(n+1), as desired.

6(h). We first check it's true for n=1. In this case, 1/2! = 1-1/(2)! certainly hods. Now we assume it is true for n, and prove for n+1. The left hand side for n+1 is
1/2! + 2/3! + .... + n/(n+1)! + (n+1)/(n+2)! = [1-1/(n+1)!] + (n+1)/(n+2)! by inductive hypothesis.
We simplify by making common denominators, and find  equals
[1-1/(n+1)!] + (n+1)/(n+2)! = [(n+1)! - 1]/(n+1)! + (n+1)/(n+2)! = (n+2)[(n+1)! - 1]/(n+2)! + (n+1)/(n+2)!
= [(n+2)! - (n+2) + (n+1)]/(n+2)! = [(n+2)! - 1]/ (n+2)! = 1-1/(n+2)!, as desired. This establishes the inductive step so by PMI we have the result.