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 \text{ and } b \cup B \cup C [By Definition]

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

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

\Rightarrow (a,b) \in A \times B \text{ 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) \text{ or } (x,y) \in (A \times C)

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

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

\Rightarrow x \in A \text{ 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)  \text{. Then, } 

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

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

\Rightarrow a \in A \text{ and } (b \in B \text{ and } b \in C)

\Rightarrow (a \in A \text{ and } b \in B) \text{ and } (a \in A \text{ and } b \in C)

\Rightarrow (a,b) \in A \times B \text{ 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)  \text{. Then, } 

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

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

\Rightarrow (x \in A \text{ and } y \in B) \text{ and } (x \in A \text{ and } y \in C)

\Rightarrow x \in A \text{ and } (y \in B \text{ and } y \in C)

\Rightarrow x \in A \text{ 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 \text{ and } b \in (B-C)

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

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

(a,b) \in (A \times B) \text{ 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)  \text{. Then, } 

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

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

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

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

\Rightarrow x \in A \text{ 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 \text{ 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  \text{. 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  \text{. 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 \text{ and } b \in A

\Rightarrow (a \in A, b \in A) \text{ and } (a \in A, b \in A)

\Rightarrow a \in A, b \in B) \text{ and } (a \in B, b \in A)

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

\Rightarrow a \in B \text{ 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 \text{ 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 \text{ and } b \in C

a \in B \text{ and } b \in D [\because A \subseteq B \text{ 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)  \text{. Then, } 

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

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

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

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

\Rightarrow a \in (A \cap C) \text{ 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 \text{ and } B be two non-empty sets having n elements in common, then prove that A \times B \text{ 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 \text{ 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 \text{ 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 .