1. Other
  2. Other
  3. idenities listed below problem 1 prove the following set identity...

Question: idenities listed below problem 1 prove the following set identity...

Question details

Idenities listed below Problem 1. Prove the following set identity for any sequence of sets {An) without using Identity 4: Problem 2. Prove the following set identity for any three sets A, B, C without using identity 3: AU(BnC) (AUB)n (AuC)

Identity 1. For any set A and B, ACAUB. Proof. This identify says that A is always a subset of AU B, no matter what B is. To show this, we need to argue that if x E A, then x E AUB. From the definition of a union, x E AUB means that x E Aorx E B. Thus, if x E A, then obviously x E A or x є B, proving that x E A U B. Identity 2. For any set A, (Ac)c-A Proof. Note that two sets are equal if each one is a subset of another. What we have are two sets (Ac)e, which is the complement of Ac, and the set A itself. The identify claims that if we take the complement of A twice, we get the set back. To prove such an identify, we always follow a two-step approach: first, we prove that (Ac)c C A, and second, we show that A c (AC . Proof of (Ac)°C A: Let x . Proof of A C (A): Let x (Acy. Then, x Ac. But, this im plies that x E A. A. Then, x ¢ A. This implies, x (AC).

Identity 3. For any three sets A, B, C, we have A n (B U C) = (A n B) U (A n C) Proof We again follow the two-step approach: . Proof of An (BUC) c (An B) U (An C): Ifx є An (BUC), then х є A and x є BU C. When we say x є B υ C, this can mean two things: x E B or x E C. If x E B, then we must have x E A B because E A is already assumed to be true. Also, An Bc(AnB) U (AnC) by Identity 1. Thus, we have that if x B, then x E A n B and x E (A n B) U (A n C). Now, assume that x E C. Then, we have . Proof of (AnB) U (Anc) с АП (BUC): If x E(ΑΠΒ) U (Anc), then x E (AnB) or z E (Anc) If r e (An B), then x is in both A and B. Sincex e B means that r e BUC, we have x E A and x B UG. i.e., x E A n (B U C). Similarly, if x (A n C), then x A and x E B U C, again implying that

Identity 4 (De Morgans Law). The following identity on countable unions or intersections is true: Proof. The notation UAn means that there is a sequence of sets An and we are taking the union of the entire sequence of sets. Note that being finite is a special case of being countable (just add countably infinite null sets). Our two-step approach: . (U-1 An)С n_iAS: Let z є (uro: i An). Then, z Ģ (U-1 An). Since x does not belong to the union of all Ans, it means that x Ø An for any n. In other words, x є AS, for all n. This means that x є na 1 AS . n-1AS c (UR-An)·Let x ε n-1AS. Then, z E A. for all n. This is equivalent to saying that x fAn, for all n. Since x does not belong to any of the Ans, it cannot belong to their unions Un1n, or

Solution by an expert tutor
Blurred Solution
This question has been solved
Subscribe to see this solution