Theorem 1: For any three sets A, B, C , prove that:

i)  A \times (B \cup C) = (A \times B) \cup (A \times C)

(ii) A \times (B \cap C) = (A \times B) \cap (A \times C) .

Proof:

(i)  Let (a, b) be an arbitrary element of A \times (B \cup C) . Then

(a,b) \cup A \times (B \cup C)

\Rightarrow a \cup A \ and \  b \cup B \cup C      [By Definition]

\Rightarrow a \in A \ and \  (b \in B \ or \  b \in C)      [By Definition of Union]

\Rightarrow (a \in A \ and \  b \in B) \ or \  ( a \in A \ and \  b \in c)

\Rightarrow (a,b) \in A \times B \ or \  (a,b) \in A \times C

\Rightarrow  (a,b) \in (A \times B) \cup (A \times C)

\therefore A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)    … … … … … i)

Now, let (x, y) be an arbitrary element of (A \times B) \cup (A \times C) .

Then, (x,y) \in (A \times B) \cup ( A \times C)

\Rightarrow (x,y) \in (A \times B) \ or \ (x,y) \in (A \times C)

\Rightarrow (x \in A \ and \ y \in B) \ or \ (x \in A \ and \ y \in C)

\Rightarrow x \in A \ and \  ( y \in  B \ or \  y \in C)

\Rightarrow x \in A \ and \  y \in (B \cup C)

\Rightarrow (x, y) \in A \times (B \cup C)

\therefore ( A \times B) \cup ( A \times C) \subseteq A \times ( B \cup C)  … … … … … ii)

Hence from i) and ii) we get A \times (B \cup C) = (A \times B) \cup (A \times C)

 

ii)   Let (a, b) be an arbitrary element of A \times (B \cap C) . Then,

(a,b) \in A \times (B \cap C)

\Rightarrow a \in A \ and \ b \in (B \cap C)

\Rightarrow a \in A \ and \ (b \in B \ and \ b \in C)

\Rightarrow (a \in A \ and \ b \in B) \ and \ (a \in A \ and \ b \in C)

\Rightarrow (a,b) \in A \times B \ and \ (a,b) \in A \times C

\Rightarrow (a,b) \in (A \times B) \cap (A \times C)

\therefore A \times (B \cap C) \subseteq (A \times B) \cap (A \times C)  … … … … … i)

Let (x, y) be an arbitrary element of (A \times B) \cap (A \times C) . Then,

(x, y) \in (A \times B) \cap (A \times C)

\Rightarrow (x,y) \in (A \times B) \ and \ (x, y) \in A \times C

\Rightarrow (x \in A \ and \ y \in B) \ and \  (x \in A \ and \ y \in C)

\Rightarrow x \in A \ and \ (y \in B \ and \ y \in C)

\Rightarrow x \in A \ and \ y \in (B \cap C)

\Rightarrow (x,y) \in A \times ( B \cap C)

\therefore (A \times B) \cap (A \times C) \subseteq A \times (B \cap C)  … … … … … ii)

Hence, from (i) and (ii), we get A \times (B \cap C) = (A \times B) \cap (A \times C) .

 

Theorem 2: For any three sets A, B, C , prove that: A \times (B-C) = (A \times B)-(A \times C) .

Proof:

Let (a,b) be an arbitrary element of A \times (B -C) . Then

(a,b) \in A \times (B-C)

\Rightarrow a \in A \ and \ b \in (B-C)

\Rightarrow a \in A \ and \ (b \in B \ and \ b \notin C)

\Rightarrow (a \in A \ and \ b \in B) \ and \ (a \in A \ and \ b \notin C)

(a,b) \in (A \times B) \ and \  (a,b) \notin (A \times C)

\Rightarrow  (a,b) \in (A \times B) - (A \times C)

\therefore A \times (B-C) \subseteq (A \times B)- (A \times C)  … … … … … i)

Again, let (x,y) be an arbitrary element of (A \times B) -(A \times C) . Then,

(x, y) \in (A \times B) - (A \times C)

(x,y) \in A \times B \ and \  (x,y)  \notin A \times C

\Rightarrow (x \in A \ and \ y \in B) \ and \ (x \in A \ and \ y \notin C)

\Rightarrow x \in  A \ and \  (y \in B \ and \ y \notin C)

\Rightarrow x \in A \ and \  y \in  (B -C)

\Rightarrow (x,y) \in A \times (B -C)

\therefore (A \times B)-(A \times C) \subseteq A \times (B-C)  … … … … … ii)

Hence, from (i) and (ii), we get

A \times (B -C) = (A \times B) -(A \times C)

 

Theorem 3: If A and B are any two non-empty sets, then prove that: A \times B=B \times A \Leftrightarrow A=B .

Proof:

First, let A = B . Then we have to prove that A \times B=B \times A .

A=B \Rightarrow A \times B=A \times A and, B \times A=A \times A \Rightarrow  A \times B = B \times A

Conversely, let A \times B = B \times A . Then we have to Prove that A = B .

Let x be an arbitrary element of A . Then,

x \in A + (x,b) \in A \times B for all b \in B \Rightarrow (x,b) \in B \times A      [ \because A \times B = B \times A ]

\Rightarrow x \in B

\therefore A \subseteq B

Again, let y be an arbitrary element of B . Then,

y \in B \Rightarrow (a,y) \in A \times B for all a \in A \Rightarrow (a,y) \in B \times A     [ \because A \times B = B \times A ]

\Rightarrow y \in A

\therefore B \subseteq A

Hence, A = B .

 

Theorem 4: If A \in B , show that A \times A \subseteq (A \times B ) \cap (B \times A) .

Proof: Let (a, b) be an arbitrary element of A \times A

Then, (a, b) \in A \times A

\Rightarrow a \in A \ and \  b \in A

\Rightarrow (a \in A, b \in A) \ and \  (a \in A, b \in A)

\Rightarrow a \in A, b \in B) \ and \  (a \in B, b \in A)

\Rightarrow  (a,b) \in (A \times B) \ and \  (a,b) \in (B \times A)      [ \because A \subseteq B \therefore  a, b \in A \Rightarrow a, b \in B ]

\Rightarrow (a,b) \in (A \times B) \cap (B \times A)

\therefore A \times A \subseteq (A \times B) \cap (B \times A)

Hence, A \subset B \Rightarrow A \times A \subseteq (A \times B) \cap (B \times A)

 

Theorem 5: If A \subseteq  B , prove that: A \times C \subseteq B \times C for any set C .
Proof:  Let (a, b) be an arbitrary element of A \times C .

Then, (a,b) \in A \times C

\Rightarrow a \in A \ and \  b \in C

\Rightarrow a \in B \ and \  b \in C      [ \because A \subseteq B \therefore a \in A \Rightarrow a \in B ]

\Rightarrow (a,b) \in B \times C

\therefore A \times C \subseteq B \times C

 

Theorem 6:  If A \subseteq B and C \subseteq D , prove that: A \times C \subseteq B \times D

Proof:  Let (a,b) be an arbitrary element of A \times C ,

Then (a,b) \in A \times C

a \in A \ and \  b \in C

a \in B \ and \  b \in D      [\because A \subseteq B \ and \  C \subseteq D ]

(a,b) \in B \times D

\therefore A \times C \subseteq B \times D

 

Theorem 7: For any sets A, B,C, D prove that: (A \times B) \cap (C \times D) =(A \cap C) \times (B \cap D) .

Proof:

Let (a,b) be an arbitrary element of (A \times B) \cap (C \times D) . Then,

(a,b) \in (A \times B) \cap (C \times D)

\Rightarrow (a,b) \in A \times B \ and \  (a,b) \in C \times D

\Rightarrow (a \in A \ and \ b \in B) \ and \ (a \in  C \ and \  b \in D)

\Rightarrow (a \in A \ and \ a \in C) \ and \ (b \in  B \ and \  b \in D)

\Rightarrow a \in (A \cap C) \ and \ b \in B \cap D

\Rightarrow (a,b) \in (A \cap C) \times (B \cap D)

\therefore (A \times B) \cap (C \times D) \subseteq (A \cap C) \times (B \cap D )

Similarly, (A \cap C) \times (B \cap D) \in (A \times B) \cap (C \times D)

Hence, (A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)

 

Theorem 8: For any three sets A, B, C prove that:

(i) A \times (B' \cup C')'= (A \times B) \cap (A \times C)

ii) A \times (B' \cap C')' =  (A \times B) \cup (A \times C)

Proof:

i)  A \times (B' \cup C')' = A \times ((B')' \cap (C')') = A \times (B \cap C) = (A \times B) \cap (A \times C)          [ By De-Morgan’s law ]

ii) A \times (B' \cap C')' = A \times (B')' \cup (C')' = A \times (B \cup C) = ( A \times B) \cup (A \times C)          [ By De-Morgan’s law ]

 

Theorem 9:  Let A and B be two non-empty sets having n elements in common, then prove that A \times B and B \times A have n^2 elements in common.

Proof:

We have, (A \times B ) \cap (C x D) = (A \cap C) \times (B \cap D)

\Rightarrow (A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A)

\Rightarrow (A \times B) \cap (B \times A)  = ( A \cap B) \times (A \cap B)     [On replacing C by B and D by A ]

It is given that A \cap B has latex n elements, so (A \cap B) \times (B \cap A) has n^2 elements.

But (A \times B) \cap (B \times A) = (A \cap B) \times (A \cap B)

(A \times B) \cap (B \times A) has n^2 elements

Hence A \times B \ and \  B \times A have n^2 elements in common

Theorem 10: Let A be a non-empty set such that A \times B=A \times C . Show that B=C .

Proof:

Let b be an arbitrary element of B .

Then, (a,b) \in A \times B for all a \in A

\Rightarrow  (a,b) \in A \times C for all a \in A

\Rightarrow b \in C

Thus, b \in B \Rightarrow b \in C

\therefore B \subset C

Let c be an arbitrary element of C .

Then, (a,c) \in A \times C for all a \in A

\Rightarrow (a,c) \in A \times B for all a \in A

\Rightarrow c \in B

c \in C \Rightarrow c \in B

\therefore C \subset B

From i) and ii) we get B = C .