Figure 2

EXAMPLE 2 (a) If the universe is N, and if A and E are as in Example l(a), then

A ^c= {nN :n  l2} and E^c = {n  N :n is odd}.

(b) If the universe is R, then [0.1]^c = (- , 0) u (1, ), (0, 1)^c = (- , 0]

 [1, ) and {0, 1}^c = (- , 0)  (0,1)  (1, ). For any a  R,

[a, )^c = (- , a) and (a, )^c = (- , a]. ■

Note that the last two Venn diagrams in Figure 2 show that A^c  B^c = (A  B)^c. This set identity and many others are true in general. Table 1 lists some basic identities for sets and set operations. Don't be overwhelmed by them; look at them one at a time. As some of the names of the laws suggest, many of them are analogs of laws from algebra. The idempotent laws are new [certainly, a + a = a fails for most numbers], and there is only one distributive law for numbers. Of course, the laws involving complementation are new. All sets in Table 1 are presumed to

TABLE 1. Laws of Algebra of Sets

1a AB =BA b AB =BA 2a (AB)C =A(BC) b (AB)C =A(BC) 3a A(BC)=(AB)(AC) b A(BC)=(AB)(AC) 4a AA = A b AA = A 5a A=A b AU =U c A= d AU =A 6 (Ac)c =A 7a AAc =U b AAc = 8a Uc = b c =U 9a (AB)c =AcBc b (AB)c =Ac  Bc

be subsets of some universal set U. Because of the associative laws, we can write the sets A  B  C and A  B  C without any parentheses and cause no confusion.

The identities in Table 1 can be verified in either of two ways. One can shade in the corresponding sets of a Venn diagram and observe that they are equal. Alternatively, one can show that sets S and T are equal by showing that S  T and T  S; these inclusions can be verified by showing that x  S implies x  T and by showing that x  T implies x  S. We give examples of both sorts of arguments, leaving most of the veri­fications to the interested reader.

EXAMPLE 3 The DeMorgan law 9a is illustrated by Venn diagrams in Figure 2.

Here is a proof in which we show first that (A  B)^c  A^c  B^c

and then that

A^c  B^c  (A  B)^c.

To show that (A  B)^c  A^c  B^c , we consider an element x in

(A  B)^c. Then x  A  B. In particular, x  A, so we must have

x  A^c. Similarly, x  B, and so x  B^c. Therefore, x  A^c  B^c.

We have shown that x  (A  B)^c implies x  A^c  B^c; hence

(A  B)^c  A^c  B^c.

To show the reverse inclusion, A^c  B^c  (A  B)^c , we

consider x in A^c  B^c. Then x  A^c, so x  A. Also, x  B^c, and so

x  B. Since x  A and x  B, we conclude that x  A  B, i.e.,

x  (A  B)^c. Hence A^c  B^c  (A  B)^c. ■

EXAMPLE 4 The Venn diagrams in Figure 3 show the distributive law 3b. The

picture of the set A  (B  C) is double-hatched in the diagram

on the left; on the right, (A  B)  (A  C) is represented by the

set that is single- or double-hatched.