Talk:Simple theorems in the algebra of sets
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Proof of Proposition 3 is below:
Proof: We shall prove (a) and leave (b) as an exercise. Each side of equation (a) defines a set and we wish to prove that these sets are equal. By Proposition 2, a possible strategy is to show that each side is a subset of the other.
- Pick any element x of the left-hand side (LHS). Then, by definition of ∩, x is in A and x is in B ∪ C; that is, x is in A and either x is in B or x is in C (or both). In the first case, x is both in A and in B, so it's in A ∩ B and a fortiori in (A ∩ B) ∪ (A ∩ C). In the second case, x is both in A and in C and so again it's in (A ∩ B) ∪ (A ∩ C). Thus in either case, x is in (A ∩ B) ∪ (A ∩ C). We have shown that every element of the LHS is automatically in the RHS. But this is precisely what we mean by saying that the LHS is a subset of the RHS.
- Pick any element x of the RHS. Then x is in A ∩ B or x is in A ∩ C (or both). In the first case, x is in A and x is in B; in the second, x is in A and x is in C. In either case, x is in A. Also in the first case x is in B and hence in B ∪ C; in the second case, x is in C and thus again in B ∪ C. We have proved that whatever x is, if it is a member of the RHS, then it is both in A and in B ∪ C and hence by definition is in A ∩ (B ∪ C). We have proved that the RHS is a subset of the LHS.
By Proposition 2, (1) and (2) together prove that LHS = RHS, as required.
[edit] Restructure
Cant this be somehow restructured? It sounds like some textbook entry. And the talk page proof seems like it has been taken from some book. --Soumyasch 06:14, 22 March 2006 (UTC)
[edit] A \ B
According to the set article, "A \ B" and "A - B" mean the same thing (alternate notation). —The preceding unsigned comment was added by 70.189.73.224 (talk • contribs) 12:50, May 21, 2006 (UTC)
- Yes. So? Paul August ☎ 20:35, 21 May 2006 (UTC)
