1.8 Cartesian product (exercise)

Note : The results of some of the questions (3 - 7) are of generic nature. As such, they can also be treated as theorems on Cartesian products.

Worked out exercises

Problem 1 : Cartesian product " $A\times B$ " consists of 6 elements. If three of these are (1,2), (2,3) and (3,3), then find Cartesian product set " $B\times A$ ".

Solution : We need to know two sets “A” and “B” in order to evaluate " $B\times A$ ". First elements of ordered pairs of " $A\times B$ " are elements of set “A”. Hence, “1”,”2” and “3” are the elements of set “A”. On the other hand, second elements of ordered pairs of AXB are elements of set “B”. Hence, “2” and “3” are elements of set “B”.

Now, it is given that there are total 6 elements in the Cartesian product, which is equal to the product of numbers of elements in two sets i.e. 3 X 2. It means that we have identified all elements of sets “A” and “B”.

$$A=\left\{\mathrm{1,2,3}\right\}$$

$$B=\left\{\mathrm{2,3}\right\}$$

Following the rule for writing Cartesian product in terms of ordered pairs, we have :

$$\Rightarrow B\times A=\{\left(\mathrm{1,2}\right),\left(\mathrm{1,3}\right),\left(\mathrm{2,2}\right),\left(\mathrm{2,3}\right),\left(\mathrm{3,2}\right),\left(\mathrm{3,3}\right)\}$$

Problem 2 : Two sets are given as : A = {1,2} and B = {3,4}. Find the total numbers of subsets of " $A\times B$ ". Also write power set of AXB in roaster form.

Solution : The total numbers of elements in the Cartesian product " $A\times B$ " is “pq”, where “p” and “q” are the numbers of elements in the individual sets “A” and “B” respectively. The all possible subsets that can be formed including empty set and the product " $A\times B$ " itself is :

$$n=2^{pq}=2^{2X_2}=2^4=16$$

Now, the Cartesian product is :

$$\Rightarrow A\times B=\{\left(\mathrm{1,3}\right),\left(\mathrm{1,4}\right),\left(\mathrm{2,3}\right),\left(\mathrm{2,4}\right)\}$$

The corresponding power set comprises of empty set, 4 sets with elements comprising of one element plus 6 sets with elements comprising of two elements taken at a time plus 4 sets with elements comprising of three elements taken at a time plus set itself. There are total of 16 subsets. The power set is set of all subsets as its elements :

$$\Rightarrow P\left(A\times B\right)=\{\phi ,\{\left(\mathrm{1,3}\right)\},\{\left(\mathrm{1,4}\right)\},\{\left(\mathrm{2,3}\right)\},\{\left(\mathrm{2,4}\right)\},$$

$$\{\left(\mathrm{1,3}\right),\left(\mathrm{1,4}\right)\},\{\left(\mathrm{1,3}\right),\left(\mathrm{2,3}\right)\},\{\left(\mathrm{1,3}\right),\left(\mathrm{2,4}\right)\},\{\left(\mathrm{1,4}\right),\left(\mathrm{2,3}\right)\},\{\left(\mathrm{1,4}\right),\left(\mathrm{2,4}\right)\},\{\left(\mathrm{2,3}\right),\left(\mathrm{2,4}\right)\},$$

$$\{\left(\mathrm{1,3}\right),\left(\mathrm{1,4}\right),\left(\mathrm{2,3}\right)\},\{\left(\mathrm{1,3}\right),\left(\mathrm{1,4}\right),\left(\mathrm{2,4}\right)\},\{\left(\mathrm{1,3}\right),\left(\mathrm{2,3}\right),\left(\mathrm{2,4}\right)\},\{\left(\mathrm{1,4}\right),\left(\mathrm{2,3}\right),\left(\mathrm{2,4}\right)\}$$

$$\{\left(\mathrm{1,3}\right),\left(\mathrm{1,4}\right),\left(\mathrm{2,3}\right),\left(\mathrm{2,4}\right)\}\}$$

It is easy to follow a scheme to write combination in which order is not relevant. We can denote each of the ordered pair with a symbol like :

$$A\times B=\left\{a,b,c,d\right\}$$

As pointed out for generating combination for ordered pair, we can start with the left element and keep changing the last element of the combination till all combinations are exhausted. Here, power set in terms of symbols is :

$$P\left(A\times B\right)=\{\phi ,\{a\},\{b\},\{c\},\{d\},$$

$$\{\left(a,b\right),\{\left(a,c\right)\},\{\left(a,d\right)\},\{\left(b,c\right),\left\{\left(b,d\right)\right\},\left\{\left(c,d\right)\right\},$$

$$\left\{\left(a,b,c\right)\right\},\left\{\left(a,b,d\right)\right\},\left\{\left(a,c,d\right)\right\},\left\{\left(b,c,d\right)\right\},$$

$$\left\{\left(a,b,c,d\right)\right\}\}$$

Problem 3 : If " $A\subset B$ " and "C" is any non-empty set, then prove that :

$$A\times C\subset B\times C$$

Solution : Let us first discuss the logic of the relation here. The elements of set “A” are common to set “B”. Now Cartesian product of set "A" with set “C” will yield ordered pairs, which are common with the ordered pairs of the Cartesian product "B" with "C". However, as set “B” is either larger than or equal to, but not smaller than “A”, it follows that above relation should hold.

Now, we prove the relation analytically. Let an arbitrary ordered pair (x,y) belongs to Cartesian product " $A\times C$ ".

$$\left(x,y\right)\in A\times C$$

According to definition of Cartesian product,

$$\Rightarrow x\in Aandy\in C$$

But “A” is subset of “B”. Hence, $x\in B$ .

$$\Rightarrow x\in Bandy\in C$$

Again, applying definition of Cartesian product,

$$\left(x,y\right)\in B\times C$$

This means that :

$$\Rightarrow A\times C\subset B\times C$$

Problem 4 : If $A\subset BandC\subset D$ , then prove that :

$$A\times C\subset B\times D$$

Solution : Let an arbitrary ordered pair (x,y) belongs to Cartesian product " $A\times C$ ".

$$\left(x,y\right)\subset A\times C$$

According to definition of Cartesian product,

$$\Rightarrow x\in Aandy\in C$$

But “A” is subset of “B”. Hence, $x\in B$ . Also, “C” is subset of “D”. Hence, $y\in D$ .

$$\Rightarrow x\in Bandy\in D$$

Again, applying definition of Cartesian product,

$$\Rightarrow \left(x,y\right)\in CXD$$

This means that :

$$\Rightarrow A\times C\subset B\times D$$

Problem 5 : For any given four sets “A”, “B”, “C” and “D”, prove that :

$$\left(A\times B\right)\cap \left(C\times D\right)=\left(A\cap C\right)\times \left(B\cap D\right)$$

Solution : Let an arbitrary ordered pair (x,y) belongs to intersection set “ $\left(A\times B\right)\cap \left(C\times D\right)$ ”. Then,

$$\left(x,y\right)\in \left(A\times B\right)\cap \left(C\times D\right)$$

Applying definition of intersection,

$$\Rightarrow \left(x,y\right)\in A\times Band\left(x,y\right)\in C\times D$$

Applying definition of Cartesian product,

$$\Rightarrow \left(x\in Aandy\in B\right)and\left(x\in Candy\in D\right)$$

$$\Rightarrow \left(x\in Aandx\in C\right)and\left(y\in Bandy\in D\right)$$

Applying definition of intersection,

$$\Rightarrow \left(x\in A\cap C\right)and\left(y\in B\cap D\right)$$

Again, applying definition of Cartesian product,

$$\Rightarrow \left(x,y\right)\in \left(A\cap C\right)X\left(B\cap D\right)$$

This means that :

$$\Rightarrow \left(A\times B\right)\cap \left(C\times D\right)\subset \left(A\cap C\right)\times \left(B\cap D\right)$$

Similarly, starting from RHS, we can prove that :

$$\Rightarrow \left(A\cap C\right)\times \left(B\cap D\right)\subset \left(A\times B\right)\cap \left(C\times D\right)$$

If sets are subsets of each other, then they are equal. Hence,

$$\Rightarrow \left(A\times B\right)\cap \left(C\times D\right)=\left(A\cap C\right)\times \left(B\cap D\right)$$

Problem 6 : Let “A” and “B” be two non-empty sets. If the numbers of common elements be “n” between sets “A” and “B”, then find the common elements between Cartesian products “ $A\times B$ ” and “ $B\times A$ ”.

Solution : The common elements between sets “A” and “B” is “n”. This means :

$$n\left(A\cap B\right)=n$$

We are required to evaluate the expression,

$$n[\left(A\times B\right)\cap \left(B\times A\right)]$$

We have earlier seen that four given sets,

$$\left(A\times B\right)\cap \left(C\times D\right)=\left(A\cap C\right)\times \left(B\cap D\right)$$

If we put C = B and D = A in this equation, then expression on the left hand side of the equation becomes what is required.

$$\Rightarrow \left(A\times B\right)\cap \left(B\times A\right)=\left(A\cap B\right)\times \left(B\cap A\right)$$

$$\Rightarrow n[\left(A\times B\right)\cap \left(B\times A\right)]=n\left[\left(A\cap B\right)\right]\times n\left[\left(B\cap A\right)\right]$$

$$\Rightarrow n[\left(A\times B\right)\cap \left(B\times A\right)]=nXn=n^2$$

Problem 7 : Let “A”, “B” and “C” be three sets. Then prove that :

$$AX\left(B\prime \cup C\prime \right)\prime =\left(A\times B\right)\cap \left(A\times C\right)$$

Solution : From first De-morgan’s theorem, we know that :

$$\left(B\cup C\right)\prime =B\prime \cap C\prime$$

Applying to the LHS, we have :

$$\Rightarrow A\times \left(B\prime \cup C\prime \right)\prime =A\times [\left(B\prime \right)\prime \cap \left(C\prime \right)\prime ]$$

Now, component of complement set is set itself. Hence,

$$\Rightarrow A\times \left(B\prime \cup C\prime \right)\prime =A\times \left(B\cap C\right)$$

Applying distributive property of product operator over intersection operator, we have :

$$\Rightarrow A\times \left(B\prime \cup C\prime \right)\prime =\left(A\times B\right)\cap \left(A\times C\right)$$

Problem 8 : Let “A”, “B” and “C” be three sets. Then prove that :

$$AX\left(B\prime \cap C\prime \right)\prime =\left(A\times B\right)\cup \left(A\times C\right)$$

Solution : From second De-morgan’s theorem, we know that :

$$\left(B\cap C\right)\prime =B\prime \cup C\prime$$

Applying to the LHS, we have :

$$\Rightarrow A\times \left(B\prime \cap C\prime \right)\prime =A\times [\left(B\prime \right)\prime \cup \left(C\prime \right)\prime ]$$

Now, component of complement set is set itself. Hence,

$$\Rightarrow A\times \left(B\prime \cap C\prime \right)\prime =A\times \left(B\cup C\right)$$

Applying distributive property of product operator over union operator, we have :

$$\Rightarrow A\times \left(B\prime \cap C\prime \right)\prime =\left(A\times B\right)\cup \left(A\times C\right)$$