Figure 3

Here is a proof where we show that the sets are subsets of each other. First consider x  A  (B  C). Then x is in A for sure. Also, x is in B  C. So either x  B, in which case x  A  B, or else x  C, in which case x  A  C. In either case, we have x  (A  B)  (A  C). This shows that A  (B  C)  (A  B)  (A  C).

Now consider y  (A  B)  (A  C). Either y  A  B or y  A  C; we consider the two cases separately. If y  A  B, then y  A and y  B, so y  B  C and hence y  A  (B  C). Similarly, if y  A  C, then y  A and y  C, so y  B  C and thus y  A  (B  C). Since y  A  (B  C) in both cases, we've shown that (A  B)  (A  C)  A  (B  C). We already proved the opposite inclusion, so the two sets are equal. ■

Figure 4

The proofs using Venn diagrams seem much easier than the proofs where we ana­lyze inclusions elementwise. Proofs by pic­ture make many people nervous; on the other hand, the Venn diagram for A, B, C has eight regions [see Figure 4] and these comprise all the logical possibilities, so proofs using Venn diagrams are in fact valid. A much more serious objection to proofs via Venn diagrams is that they hide the thought pro­cess; the logic used to shade the diagrams is not specified. If we had written out the reasoning behind the diagrams in Figure 3, the proof would have been as long as the elementwise proof in Example 4, which relies only on logic. Another reason for avoiding Venn diagrams is that they are hard to draw whenever there are more than three sets. Still, nearly everyone who works with mathematics uses pictures, including Venn diagrams, to help understand mathematical situations.

Table 1 gives a few of the basic relationships in set theory. Many other relationships exist. They can be verified using one of three methods: (1) Venn diagrams, (2) elementwise arguments as in Examples 3 and 4, or (3) applying the laws in Table 1. Sometimes, proofs will combine methods (2) and (3).

EXAMPLE 5 We give three proofs for the relationship

(A  B)  A^c  B

Proof 1. See Figure 5. The picture for (A  B)  A^c is double-hatched and is clearly a subset of B.

Proof 2. We show that x  (A  B)  A^c implies x  B. Consider x in (A  B)  A^c. Then x  A^c, and so x  A. Since x is also in A  B, it is in A or B, so it follows that x must be in B.

Proof 3. Using the laws of algebra in Table 1, we obtain

(A  B)  A^c = A^c  (A  B) commutativity 1b

= (A^c  A)  (A^c  B) distributivity 3b

= (A  A^c)  (A^c  B) commutativity 1b

=   (A^c  B) 7b

= (A^c  B)   commutativity la

= A^c  B. identity law 5a

Figure 5

This identity agrees, of course, with the picture on the left in Figure 5. Now it is clear that A^c  B  B, since if x  (A^c  B), then x must be in B.

The symmetric difference  is also an associative operation:

(A  B)  C = A  (B  C).

We can see this by looking at the Venn diagrams in Figure 6. On the left we have hatched A  B one way and C the other. Then (A  B)  C is the set hatched one way or the other but not both. Doing the same sort of thing with A and B  C gives us the same set, so the sets (A  B)  C and A  (B  C) are equal.

FIGURE 6

Of course, it is also possible to prove this fact without appealing to the pictures. You may want to construct such a proof yourself. Be warned, though, that a detailed argument will be fairly complicated.

Since  is associative, the expression A  B  C is unambiguous. Note that an element belongs to this set provided that it belongs to exactly one or to all three of the sets A, B and C.

Consider two sets S and T. For each element s in S and each element t in T, we form an ordered pair (s, t). Here s is the first element of the ordered pair, t is the second element, and the order is important. Thus (s₁, t₁) == (s₂, t₂) if and only if s₁ = s₂ and t₁ = t₂. The set of all ordered pairs (s, t) is called the product of S and T and written S  T:

S  T = {(s, t) : s  S and t  T}.

If S = T, we sometimes write S² for S x S.

EXAMPLE 6 (a) Let S = {1, 2, 3, 4} and T = {a, b, c}. Then S  T consists of the twelve ordered pairs listed on the left in Figure 7. We could also depict

(1, c) (2, c) (3, c) (4, c) c ○ ○ ○ ○ (1, b) (2, b) (3, b) (4, b) b ○ ○ ○ ○ (1, a) (2, a) (3, a) (4, a) a ○ ○ ○ ○ 1 2 3 4 List of {1, 2, 3, 4}×{a, b, c} Picture of {1, 2, 3, 4}×{a, b, c}