Laws of Set Operations

Prom Pig. 2.2, it may be noted that the shaded area in diagram a represents not only A U B but also B U A. Analogously, in diagram b the small shaded area is the visual

FIGURE 2.3

AUSUC

AUSUC

(h)

representation not only of A fi B but also of BPjA When formalized, this result is known as the commutative law (of unions and intersections):

These relations are very similar to the algebraic laws a -f h — h -H a and a x h — h x a.

To take the union of three sets A, B, and C, we first take the union of any two sets and then "union" the resulting set with the third: a similar procedure is applicable to the intersection operation. The results of such operations are illustrated in Fig. 2.3. it is interesting thai the order in which the sets are selected for the operation is immaterial. This fact gives rise to the tfisoe/tffive law (of unions and intersections):

These equations are strongly reminiscent of the algebraic laws a + (b c) — ( a + b) + < and ox(hc) = (tfxii)x c.

There is also a law of operation that applies when unions and intersections arc used in combination. This is the distributive law (of unions and intersections):

These resemble the algebraic law a x (b + c) — (a xb) + {a x c).

Example 7 Verify distributive law, given A =_{4,5), B = {3, 6, 7}, and C = {2, 3}. To verify the first --- part of the law, we find the left- and right-hand expressions separately:

Right: {A u 8) n (A u C) = (3,4, 5, 6, 7} D {2( 4,5} = (3,4, 5J

14 P^rt One Introduction

Since the two sides yield the same result, the law is verified. Repeating the procedure for the second part of the law, we have

Left: A n (8 u C) = {4, 5] n|2, 3, 6, 7| = 0 Right: (AnB)J(AnC) = 0U0 = 0 Thus the law is again verified.

To verify a law means to chcck by a specific example whether the law actually works out. If the law is valid, then any specific example ought indeed to work out. This implies that if the law does not check out in as many as one single example, then ihe law is invalidated. On the other hand, the successful verification by specific examples (however many) does not in itself prove the iaw. To prove a law, it is necessary to demonstrate that the law is valid for all possible cases. The procedure involved in such a demonstration will be illustrated later (see, e.g., Sec. 2.5}.

EXERCISE 2 3

1. Write the following in set notation:

(a) The set of all reai numbers greater than 34.

(b) The set of a(l real numbers greater than 8 but less than 65.

2. Given the sets ^ = ¡2,4,6}, Si = (7,2,61, 5* = (4,2,6}, and S4 = {2,4], which of the following statements are true?

[o) Si = (d) 3 £ S2

(9) Si D

(£>} = a (set of real numbers) (e) 4 £ S3

(h) 0C52

(c) 8 € S2 (0 C R

(>) *3=>P,2|

3. Referring to the four sets given in Prob. 2, find:

(a) Si u S2 (c) S2 r S,

(e) s4 n s2 n s]

(b) Si u s3 (d) S2 r S4

(f) U u 54

4. Which of the following statements are valid?

(a) A u A = A (d) A U U =

U

(g) The complement of

(b) An A = A (e) A n 0 =

0

A is A.

(c) AU& = A (f ) A nU =

A

5. Given A ** (4,5,6}, B = {3,4, 6, 7}, and C = (2, 3,6), verify the distributive law.

6. Verify the distributive iaw by means of Venn diagrams, with different orders of successive shading.

8. Enumerate all the subsets of the set S= [Q,b, c, d). How many subsets are there altogether?

9. Example 6 shows that 0 is the complement of U. But since the null set is a subset of any set 0 must be a subset of ¿A Inasmuch as the term "complement of U" implies the notion of being not in U, whereas the term "subset of U" implies the notion of being in U, it seems paradoxical for 0 to be both of these. How do you resolve this paradox?

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