Chapter  4, Relations
Section  4.1  Ordered Pairs and Cartesian Products
Summary
 Ordered Pair:
 Lets take an example. Suppose means “x is capital city of country y”. The pair is considered as an ordered pair since .
 In an ordered pair , is called first coordinate and is called second coordinate.
 Cartesian Product:
Suppose and are sets. Then the Cartesian product of and , denoted , is the set of all ordered pairs in which the first coordinate is an element of A and the second is an element of B. In other words,
.  Suppose and are sets, then:
 .
 .
 .
 .
 .
 Suppose and are sets. Then = iff either or .
 Truth set of a statement:
Suppose is a statement with two free variables in which ranges over a set and ranges over another set . Then is the set of all assignments to and that make sense in the statement . The truth set of is the subset of consisting of those assignments that make the statement come out true. Thus,
truth set of
Soln1
(a)
(b)
Soln2
(a) .
(b) .
Soln3
It requires plotting points in the plane. So leaving it :)
Soln4
, , , and
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(a)
Thus, .
(b)
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Thus,
(c)
Thus, .
(d)
Thus,
(e) and , thus .
Soln5
Proof for:
()Let be an arbitrary element of . Thus is an ordered pair with in first coordinate and second coordinate in . Thus can be either in or in . Thus we have following cases:

Case 1: :
Thus we can say that . It follows that . 
Case 2: :
Thus we can say that . It follows that .
Thus from both cases, if , then .
()Suppose . Thus we have two cases:

Case 1: :
Thus and . Since , it follows that . Now since and , it follows that . 
Case 2: :
Thus and . Since , it follows that . Now since and , it follows that .
Thus from both cases, if , then .
Thus from both directions .
Proof for:
()Let be an arbitrary element of . Thus and . It follows that , or . Similarly, since , it follows . Thus .
()Let be an arbitrary element of . Thus , it follows that . Similarly, since , it follows that . Thus we have .
Thus from both directions we have .
Soln6
It does not cover all the possible cases. Following cases are missed:
 Case 1: .
 Case 2: .
Soln7 .
Soln8
Theorem is correct.
A×(B\C) = (A×B)(A×C)
()Suppose . Thus and . It follows that , but . Thus .
()Suppose . Thus and . Since and , it follows that . Thus and , or . Thus we have .
Since is arbitrary, we can conclude that .
Soln9
()Suppose . Thus and . Since , it follows that and . Since , we can have following possible cases:

Case 1: :
Since and , it follows that . Thus . 
Case 2: :
Since and , it follows that . Thus .
Thus from both cases, .
()Suppose . Thus we have following cases:

Case 1: .
Here since and , it follows that . Also since , it follows that . Thus we have, and . It follows that . 
Case 2: .
Here since and , it follows that . Also since , it follows that . Thus we have, and . It follows that .
Thus from both cases, .
Thus from both directions, we have .
Soln10
Suppose . Suppose . It follows that . Thus there are following cases:

Case 1: :
Since , it follows that . 
Case 2: : Since , it follows that .
Thus either , or .
Soln11
(a)
Suppose . Thus there atleast exists an such that . Thus and . Since , it follows . Similarly, since , it follows . Thus . Since is arbitrary, it follows that .
(b)
∪p∈PCp =(∪i∈IAi)×(∪i∈IBi)
()Suppose . Thus exist in atleast one of the set, say such that . Thus and . It follows that and . Thus we have . Since is arbitrary, it follows that .
()Suppose . Thus such that and such that . Thus , or where . Since and , it follows that , or . Now, since and , it follows that . Since is arbitrary, it follows that .
Thus, from both directions we have: .
Soln12
The theorem and proof both are not correct. It does not consider if some of the sets or are empty.