PROOF INVOLVING SETS and INDEXED FAMILIES of SETS By: Ahmad Wachidul Kohar, Sri Rejeki (Impome 2012)

PROOF INVOLVING SETS and INDEXED FAMILIES OF SETS By: Ahmad Wachidul Kohar, Sri Rejeki (IMPoME 2012)

2.1 Proof Involving Sets 2.1.1 Terms involving Sets 1. Definition 1 Suppose A and B are sets. means “if , then ”, that is

2. Definition 2 If are sets, we say that and , that is A = B ⇔ ( A ⊆ B ) ∧ ( B ⊆ A) ⇔ (x ∈ A → x ∈ B ) ∧ (x ∈ B → x ∈ A). 3. Definition 3 If A and B are sets, we define the sets A ∪ B = {x : x ∈ A ∨ x ∈ B } A ∩ B = {x : x ∈ A ∧ x ∈ B } That is, x ∈ A ∪ B ⇔ (x ∈ A ∨ x ∈ B ) x ∈ A ∩ B ⇔ (x ∈ A ∧ x ∈ B ) 4. Definition 4 If A ∩ B = ∅, then A and B are said to be disjoint, that is A ∩ B = ∅⇔ ¬(∃x ∈ A ∩ B ) ⇔ ¬(∃x ∈ A)(x ∈ B ) ⇔ (∀x ∈ A)(x B ) ⇔ x ∈ A → x B 5. Definition 5 If A and B are sets, we deﬁne the difference of A and B as A\B = {x : x ∈ A and x B }= A ∩ Bc 6. Definition 6 If A and B are sets, we deﬁne the symmetric difference of A and B as A B = {x : x ∈ A ∩ Bc or x ∈ Ac ∩ B } = ( A ∩ Bc ) ∪ ( Ac ∩ B ) 2.1.2 Direct Proof At the foundational level, the task in writing the proof of a theorem is to show that p → q is a tautology. Theorem 1 If A ⊆ B , then A ∩ B = A Proof by using direct proof method given as follows: Suppose A ⊆ B , and pick x ∈ A ∩ B . Then x ∈ A, so that A ∩ B ⊆ A. Now pick x ∈ A. Then since A ⊆ B , it is also true that x ∈ B , so that x ∈ A ∩ B . Thus A ⊆ A ∩ B , so that A ∩ B = A. Theorem 2 (DeMorgan's Law). If A and B are sets, then

Proof: Proof: We show that ( A ∩ B )c ⊆ Ac ∪ Bc and ( A ∩ B )c ⊇ Ac ∪ Bc . (⊆): Pick x ∈ (A ∩ B)c . Then x A ∩ B , so that either x A or x B . We consider each case. (Case x A): If x A, then x ∈ Ac . Thus x ∈ Ac ∪ Bc . (Case x B ): If x B , then x ∈ Bc . Thus x ∈ Ac ∪ Bc . In either case, we have ( A ∩ B )c ⊆ Ac ∪ Bc

2.1.3 Proof by contrapositive The idea of this method is that p → q is equivalent with , meaning that we can use this latter form to prove a statement. Example If , then A Proof: By contrapositive, suppose A . Then there exist . Thus, so that and which means there exist and or

2.1.4 Proof by contradiction The idea of contradiction method is by showing is a contradiction of the statement , that is a tautology. Another way to write is using its equivalence, which is Example: Given A and B are sets satisfying . Prove that Proof: By contradiction, we obtain Suppose , then

(given)

(Contradict with the given that is ) Exercises 2.1 1. Construct the negation of each statement below. a. x ∈ A ∪ B b. A and B are disjoint c. A = B d. x ∈ A\B e. x ∈ A B f. A ⊆ B ∪ C g. A ∪ B ⊆ C ∩ D h. If A ∪ C ⊆ B ∪ C , then A ⊆ B 2. Prove the following: a. If A ⊆ B , then A ∪ B = B b. If and , then

c. If A ⊆ B , then Ac⊇ Bc d. If , then 3. Suppose A, B, and C are sets. Consider the following statements. …….1) …….2) For each of equation above, one direction of the implication is true and one is false. Prove the direction that is true, and provide a counterexample for the direction that is false. 4. Prove that distributes over and vice versa. 5. If X and Y are disjoint sets, we sometimes write . So, if someone makes a statement like (1) What he is really saying is the compound statement and (2) Prove equations (1) by showing both parts of equation (2) Proof 1. The negation of the forms a) to h) are given as follows (a) x ∈ A ∪ B x ∈ A ∪ B means that x A ∪ B means that (b) A and B are disjoint. A and B are disjoint means that It means there is no

Thus, the negation is (c) A = B

(d) x ∈ A\B

(e) x ∈ A B

(f ) A ⊆ B ∪ C

(g) A ∪ B ⊆ C ∩ D

(h) If A ∪ C ⊆ B ∪ C , then A ⊆ B

2. a. means ………a) means ………b) Using syllogism, from a) and b), we obtain , or b. Since , then . Likewise, also causes

3. For equation 1) For , it will be proven as true by the following proof. (⊆) means or . Since A=B , then . Then, we obtain or . Thus,

means or . Since A=B , then . Then, we obtain or . Thus, For , it will be proven as false by showing a counterexample as follows. Let

, but For equation 2 For , it will be proven as true by the following proof. (⊆) Pick , then and Since and , then . Since and , so . Thus

Pick , then and Since and , then Because of and , so Thus For , it will be proven as false by showing a counterexample as follows. Let

, but 4. a)

5. .. a) …….b) Since a) and b) are proven, it is enough to show that

2.2 Indexed Families of Sets

If we’re working with a few sets at a time, it’s probably sufﬁcient to use A, B , and C to represent them. Yet, if we have many sets, for instance, 10 sets (generally called a family or collection of sets instead of a set of sets), it might be more sensible to put them into a family and address them as A1 , A2 , ..., A10 . In a case like this, we would say that the set {1, 2, 3,..., 10} indexes the family of sets. If we write Nn = {1, 2, 3,..., n}, then we could write a family of n sets as

n A1, A2 , A3 ,..., An Ak : k Nn Ak k 1 and we would say that Nn is an index set for the family of sets. This notation has advantages, for then we could write unions and intersections more succinctly:

A1 A2 A3 ... An  Ak k 1

A1 A2 A3 ... An  Ak k 1 Definition 2.2.3. Let F be a family of sets. Then the union over F is defined by:  A x : A F x A A F

If F is indexed by A, this becomes  A  A x : A x A A F A Definition 2.2.4. Let F be a family of sets. Then the intersection over F is defined by:  A x : A F x A A F If F is indexed by A, this becomes:  A  A x : A x A A F A Example 2.2.5.

Construct the negation of the statement: x ∪F A. Solution: the statement x ∈/ ∪F A means A F x A . If F is indexed by A,, we have

A x A

Example 2.2.5.

Construct the negation of the statement: x ∩F A. Solution: the statement x ∈/ ∩F A means A F x A . If F is indexed by A,, we have

A x A Theorem 2.2.6. (DeMorgan’s Law) Let F be a family of sets. Then,

'  A  A' F F Proof: We prove by mutual subset inclusion

: Pick x ∈ [∪F A]’. Then x ∈/ ∪F A. Therefore, for all A ∈ F, x ∈/ A. But then x ∈

A’ for every A’ ∈ F. Thus, x ∈ ∩F A.

: Pick x ∈ ∩F A’. Then x ∈ A’ for every A ∈ F. Thus, x ∈/ A for every A ∈ F, so that x ∈/ ∪F A. Therefore, x ∈ [∪F A]’. Theorem 2.2.7. (DeMorgan’s Law) Let F be a family of sets. Then,

'  A  A' F F

Proof: We prove by mutual subset inclusion

: Pick x ∈ [∩F A]’. Then x ∈/ ∩F A. Therefore, there exists A ∈ F, x ∈/ A. But then x ∈ A’ for any A’ ∈ F. Thus, x ∈ ∪F A.

: Pick x ∈ ∪F A’. Then x ∈ A’ for any A ∈ F. Thus, x ∈/ A for any A ∈ F, so that x ∈/ ∩F A. Therefore, x ∈ [∩F A]’.

Theorem 2.2.8. Let F be a family of sets indexed by A, and suppose B ⊆ A. Then,

1. ∪β∈B Aβ ⊆ ∪α∈A Aα.

2. ∩β∈B Aβ ⊇ ∩α∈A Aα. Proof:

1. Pick x ∈ ∪β∈B Aβ. Then x ∈ Aβ for any β ∈ B. Since B ⊆ A, it follows that β ∈ A.

Therefore, x ∈ Aα , for any α ∈ A. Since β was chosen arbitrarily, we have shown that x

∈ Aα , for any α ∈ A, so that ∪α∈A Aα.

2. Pick x ∈ ∩α∈A Aα . We must show that x ∈ Aβ for all β ∈ B, so pick β ∈ B. Since

B ⊆ A, it follows that β ∈ A. Therefore, x ∈ Aβ , because x ∈ Aα for any α ∈ A. Since β was chosen arbitrarily, we have shown that x ∈ Aβ for all β ∈ B, so that x ∈ ∩β∈B Aβ.

Notice how we showed x ∈ ∩β ∈B Aβ by picking an arbitrary β ∈ B and showing that x

∈ Aβ . This shows that x ∈ Aβ for all β ∈ B, so that x ∈ ∩β ∈B Aβ . We can write

Theorem 2.2.8 in a slightly different form if the family of sets is not indexed. If F1is a family of sets and F2 ⊆ F1 , we call F2 a subfamily of F1 . Since B ⊆ A in Theorem 2.2.8, { Aβ }β ∈B is a subfamily of { Aα }α∈A . Swapping the notation in Theorem 2.2.8 for an arbitrary family F1 and a subfamily F2 , we have the following.

Theorem 2.2.9.

Suppose F1 is a family of sets, and F2 is a subfamily of F1 . Then,

1. ∪F2 A ⊆ ∪F1 A,

2. ∩ F2 A ⊇ ∩ F1 A.