Properties of Cartesian product of Sets

For the sets A, B, C and D, we have-
Property 1: $\ A\times(B\cup C)=(A\times B)\cup (A\times C)$
Proof:
Let $\ (x,y)\in A\times (B\cup C)$
$\ \Leftrightarrow x\in A, y\in (B\cup C)$
$\ \Leftrightarrow x\in A, (y\in B$ or $\ y\in C )$
$\ \Leftrightarrow (x\in A, y\in B)$ or $\ (x\in A, y\in C)$
$\ \Leftrightarrow (x,y)\in A\times B$ or $\ (x,y)\in A\times C$
$\ \Leftrightarrow (x,y)\in (A\times B) \cup (A\times C)$
⸫ $\ A\times(B\cup C)=(A\times B)\cup (A\times C)$


Property 2: $\ A\times(B\cap C)=(A\times B)\cap (A\times C)$
Proof:
Let $\ (x,y)\in A\times (B\cap C)$
$\ \Leftrightarrow x\in A, y\in (B\cap C)$
$\ \Leftrightarrow x\in A, (y\in B$ and $\ y\in C )$
$\ \Leftrightarrow (x\in A, y\in B)$ and $\ (x\in A, y\in C)$
$\ \Leftrightarrow (x,y)\in A\times B$ and $\ (x,y)\in A\times C$
$\ \Leftrightarrow (x,y)\in (A\times B) \cap (A\times C)$
⸫ $\ A\times(B\cap C)=(A\times B)\cap (A\times C)$


Property 3: $\ (A\times B)\cup (C\times D)=(A\cup C)\times (B\cup D)$
Proof:
Let $\ (x,y)\in (A\times B)\cup (C\times D)$
$\ \Leftrightarrow (x,y)\in (A\times B)$ or $\ (x,y)\in (C\times D)$
$\ \Leftrightarrow (x\in A, y\in B$) or $\ (x\in C, y\in D)$
$\ \Leftrightarrow (x\in A$ or $\ x\in C$), $\ (y\in B$ or $\ y\in D)$
$\ \Leftrightarrow (x\in A\cup C$), $\ (y\in B\cup D)$
$\ \Leftrightarrow (x,y)\in (A\cup C)\times (B\cup D)$
⸫ $\ (A\times B)\cup (C\times D)=(A\cup C)\times (B\cup D)$


Property 4: $\ (A\times B)\cap (C\times D)=(A\cap C)\times (B\cap D)$
Proof:
Let $\ (x,y)\in (A\times B)\cap (C\times D)$
$\ \Leftrightarrow (x,y)\in (A\times B)$ and $\ (x,y)\in (C\times D)$
$\ \Leftrightarrow (x\in A$ and $\ x\in C$) and $\ (y\in B$ and $\ y\in D)$
$\ \Leftrightarrow (x\in A\cap C$) and $\ (y\in B\cap D)$
$\ \Leftrightarrow (x,y)\in (A\cap C)\times (B\cap D)$
⸫ $\ (A\times B)\cap (C\times D)=(A\cap C)\times (B\cap D)$


Property 5: $\ A\times(B-C)=(A\times B)-(A\times C)$
Proof:
Let $\ (x,y)\in A\times (B-C)$
$\ \Leftrightarrow x\in A$ and $\ y\in (B-C)$
$\ \Leftrightarrow x\in A$ and $\ (y\in B$ and $\ y\notin C)$
$\ \Leftrightarrow ( x\in A$ and $\ y\in B )$ and $\ (x\in A$ and $\ y\notin C)$
$\ \Leftrightarrow ( x,y) \in A\times B$ and $\ (x,y)\notin A\times C$
$\ \Leftrightarrow (x,y)\in (A\times B) -(A\times C)$
⸫ $\ A\times(B-C)=(A\times B)-(A\times C)$


Property 6: $\ (A-B)\times C=(A\times C)-(B\times C)$
Proof:
Let $\ x\in (A-B)\times C$
$\ \Leftrightarrow x\in (A-B)$ and $\ y\in C$
$\ \Leftrightarrow (x\in A$ and $\ x\notin B)$ and $\ y\in C$
$\ \Leftrightarrow ( x\in A$ and $\ y\in C )$ and $\ (x\notin B$ and $\ y\in C)$
$\ \Leftrightarrow ( x,y) \in A\times C$ and $\ (x,y)\notin B\times C$
$\ \Leftrightarrow (x,y)\in (A\times C) -(B\times C)$
⸫ $\ (A-B)\times C=(A\times C)-(B\times C)$


Property 7: $\ A\times B = B\times A$ iff $\ A=\phi$ or $\ B=\phi$ or $\ A=B$
Proof:
Case 1:
Let $\ A=\phi$
Now $\ A\times B = \phi \times B = \phi$   [⸪ $\ A=\phi$ ]
also $\ B\times A = B\times \phi =\phi$   [⸪ $\ A=\phi$ ]
Hence $\ A\times B = B\times A$
Case 2:
Let $\ B=\phi$
Now $\ A\times B = A\times \phi =\phi$   [⸪ $\ B=\phi$]
also $\ B\times A = \phi \times A =\phi$   [⸪ $\ B=\phi$]
Hence $\ A\times B = B\times A$
Case 3:
Let $\ A=B$
Now $\ A\times B = A\times A$   [⸪ $\ A=B$ ]
also $\ B\times A = A\times A$   [⸪ $\ A=B$ ]
Hence $\ A\times B = B\times A$


Q. If $\ A\times B=B\times A$ then show that $\ A=B$
Solution:
Let $\ a\in A \Rightarrow (a,b)\in A\times B ~\forall~ b\in B$
    $\ \Rightarrow (a,b)\in B\times A$   [⸪ $\ A\times B=B\times A$]
    $\ \Rightarrow a\in B$ [definition of Cartesian product of sets]
Conversely,
Let $\ b\in A \Rightarrow (a,b)\in A\times B ~\forall~ a\in A$
    $\ \Rightarrow (a,b)\in B\times A$   [⸪ $\ A\times B=B\times A$]
    $\ \Rightarrow b\in A$ [definition of Cartesian product of sets]
Hence, $\ A=B$


Q. If $\ A\times B=A\times C$ then show that $\ B=C$
Solution:
Let $\ b\in B \Rightarrow (a,b)\in A\times B ~\forall~ a\in A$
    $\ \Rightarrow (a,b)\in A\times C$   [⸪ $\ A\times B=A\times C$]
    $\ \Rightarrow b\in C$ [definition of Cartesian product of sets]
Conversely,
Let $\ c\in C \Rightarrow (a,c)\in A\times C ~\forall~ a\in A$
    $\ \Rightarrow (a,c)\in A\times B$   [⸪ $\ A\times B=A\times C$]
    $\ \Rightarrow c\in B$ [definition of Cartesian product of sets]
Hence, $\ B=C$


Q. If A, B and C be three sets where $\ B\subseteq A$ then show that $\ (B\times C)\subseteq (A\times C)$
Solution:
Let $\ (b,c)\in B\times C$
$\ \Rightarrow b\in B$ and $\ c\in C$
$\ \Rightarrow b\in A$ and $\ c\in C$   [⸪ $\ B\subseteq A$]
$\ \Rightarrow (b,c)\in A\times C$
Hence $\ (B\times C)\subseteq (A\times C)$


Q. If A, B and C be three sets where $\ A\subseteq B$ then show that $\ (A\times C)\subseteq (B\times C)$
Solution:
Let $\ (a,c)\in A\times C$
$\ \Rightarrow a\in A$ and $\ c\in C$
$\ \Rightarrow a\in B$ and $\ c\in C$   [⸪ $\ A\subseteq B$]
$\ \Rightarrow (a,c)\in B\times C$
Hence $\ (A\times C)\subseteq (B\times C)$


Q. If $\ A\subseteq B$ and $\ C\subseteq D$ then show that $\ (A\times C)\subseteq (B\times D)$
Solution:
Let $\ (a,c)\in A\times C$
$\ \Rightarrow a\in A$ and $\ c\in C$
$\ \Rightarrow a\in B$ and $\ c\in D$   [⸪ $\ A\subseteq B$ and $\ C\subseteq D$]
$\ \Rightarrow (a,c)\in B\times D$
Hence $\ (A\times C)\subseteq (B\times D)$