Solutions for HW2

1) Let A,B be two sets.

a) I will show that A\B = A\(AÇB) .

Let xÎA\B. Since xÏB, xÏAÇB. Thus xÎA\(AÇB). Hence A\BÍ A\(AÇB).

Let xÎA\(AÇB). Then xÏB. Thus xÎA\B. Hence A\(AÇB)Í A\B.

We conclude that A\B = A\(AÇB).

b) I will show that A = (A\B)È(AÇB).

Let xÎA. There are two cases: xÎB or xÏB. If xÎB, then xÎAÇB. Thus xÎ(A\B)È(AÇB). If xÏB, then xÎA\B. Thus xÎ(A\B)È(AÇB). Hence AÍ (A\B)È(AÇB).

Since (A\B) and (AÇB) are subsets of A, (A\B)È(AÇB)Í A.

c) I will show that (A\B)Ç(B\A)= Æ.

Assume (A\B)Ç(B\A) ¹ Æ, that is (A\B)Ç(B\A) contains an element, say x. By definition, xÎA and xÏA. A contradiction.

d) I will show that (AÈB) \ (AÇB) = (A\B)È(B\A).

Let xÎ(AÈB) \ (AÇB). Then there are two cases: “xÎA and xÏAÇB” or “xÎB and xÏAÇB” (Can you see why ?). In the first case, we get xÎA\B. In the second case, we get xÎB\A. Hence in both cases xÎ(A\B)È(B\A).

Let xÎ(A\B)È(B\A). If xÎ(A\B), then xÎ(AÈB) and xÏ (AÇB). Thus xÎ(AÈB) \ (AÇB). The other case follows similarly.

2) Let d ³ R1 + R2. Assume X1 Ç X2 ¹ Æ. Then there is a point x such that distance between x and O1, denoted d(x,O1), is smaller than R1 and distance between x and O2, denoted d(x,O2), is smaller than R2. Look at the triangle whose vertices are O1, x ,O2. By triangle inequality, we should have d(x,O1)+ d(x,O2) ³d. But this is a contradiction because we get R1+R2 > d(x,O1)+ d(x,O2) ³d ³ R1 + R2.