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4. Let 𝐴𝐴 = {1,2}, 𝐡𝐡 = {π‘Žπ‘Ž, 𝑏𝑏, 𝑐𝑐} π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž 𝐢𝐢 = {𝑐𝑐, 𝑑𝑑}. Then show that (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢) =

𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢).

To show that (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢) = 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢), we need to prove that every element in the left-hand side set is also in the right-hand side set, and vice versa.

Let's start by considering an arbitrary element (π‘Žπ‘Ž, 𝑏𝑏) in the set (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢). This means that (π‘Žπ‘Ž, 𝑏𝑏) is both in 𝐴𝐴 Γ— 𝐡𝐡 and in 𝐴𝐴 Γ— 𝐢𝐢. By definition, this implies that π‘Žπ‘Ž is an element of 𝐴𝐴, 𝑏𝑏 is an element of 𝐡𝐡, and π‘Žπ‘Ž is an element of 𝐴𝐴, 𝑏𝑏 is an element of 𝐢𝐢. Since 𝐡𝐡 ∩ 𝐢𝐢 is the set of elements that are common to both 𝐡𝐡 and 𝐢𝐢, it follows that 𝑏𝑏 is an element of 𝐡𝐡 ∩ 𝐢𝐢. Thus, (π‘Žπ‘Ž, 𝑏𝑏) is an element of 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢), which proves that every element in the left-hand side set is in the right-hand side set.

Now, let's consider an arbitrary element (π‘Žπ‘Ž, 𝑏𝑏) in the set 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢). This means that π‘Žπ‘Ž is an element of 𝐴𝐴 and 𝑏𝑏 is an element of 𝐡𝐡 ∩ 𝐢𝐢. Since 𝑏𝑏 is an element of 𝐡𝐡 ∩ 𝐢𝐢, it follows that 𝑏𝑏 is an element of 𝐡𝐡 and 𝑏𝑏 is an element of 𝐢𝐢. Thus, (π‘Žπ‘Ž, 𝑏𝑏) is an element of 𝐴𝐴 Γ— 𝐡𝐡 and (π‘Žπ‘Ž, 𝑏𝑏) is an element of 𝐴𝐴 Γ— 𝐢𝐢, which implies that (π‘Žπ‘Ž, 𝑏𝑏) is an element of (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢). This proves that every element in the right-hand side set is in the left-hand side set.

Since we have shown that every element in the left-hand side set is in the right-hand side set, and vice versa, we can conclude that (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢) = 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢).

To show that (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢) = 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢), we need to demonstrate that any element in the intersection on the left-hand side is also an element in the set on the right-hand side, and vice versa.

First, let's find the elements in (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢):

𝐴𝐴 Γ— 𝐡𝐡 = {(1, π‘Žπ‘Ž), (1, 𝑏𝑏), (1, 𝑐𝑐), (2, π‘Žπ‘Ž), (2, 𝑏𝑏), (2, 𝑐𝑐)}
𝐴𝐴 Γ— 𝐢𝐢 = {(1, 𝑐𝑐), (1, 𝑑𝑑), (2, 𝑐𝑐), (2, 𝑑𝑑)}

The intersection of these two sets is:

(𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢) = {(1, 𝑐𝑐), (2, 𝑐𝑐)}

Now, let's find 𝐡𝐡 ∩ 𝐢𝐢:

𝐡𝐡 ∩ 𝐢𝐢 = {𝑐𝑐}

Finally, we need to determine 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢):

𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢) = {(1, 𝑐𝑐), (2, 𝑐𝑐)}

Comparing the sets, we can see that the elements in the intersection on the left-hand side are the same as the elements in the set on the right-hand side. Therefore, we can conclude that (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢) = 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢).

To show that (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢) = 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢), we need to demonstrate that each set is a subset of the other.

First, let's find (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢):

𝐴𝐴 Γ— 𝐡𝐡 represents the Cartesian product of set 𝐴𝐴 and set 𝐡𝐡. This means forming all possible ordered pairs where the first element comes from 𝐴𝐴 and the second element comes from 𝐡𝐡. In this case, we have:

𝐴𝐴 Γ— 𝐡𝐡 = {(1, π‘Žπ‘Ž), (1, 𝑏𝑏), (1, 𝑐𝑐), (2, π‘Žπ‘Ž), (2, 𝑏𝑏), (2, 𝑐𝑐)}

Similarly, 𝐴𝐴 Γ— 𝐢𝐢 represents the Cartesian product of set 𝐴𝐴 and set 𝐢𝐢, which is given by:

𝐴𝐴 Γ— 𝐢𝐢 = {(1, 𝑐𝑐), (1, 𝑑𝑑), (2, 𝑐𝑐), (2, 𝑑𝑑)}

Next, we need to compute the intersection of these two sets:

(𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢) = {(1, 𝑐𝑐), (2, 𝑐𝑐)}

Now, let's consider 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢):

𝐡𝐡 ∩ 𝐢𝐢 represents the intersection of sets 𝐡𝐡 and 𝐢𝐢, which is given by {𝑐𝑐}. Therefore, 𝐡𝐡 ∩ 𝐢𝐢 = {𝑐𝑐}.

Now, compute 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢):

𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢) = {(1, 𝑐𝑐), (2, 𝑐𝑐)}

By comparing both sets, we can see that (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢) = 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢), as the sets contain the same ordered pairs.

Therefore, we have successfully shown that (𝐴𝐴 Γ— 𝐡𝐡) ∩ (𝐴𝐴 Γ— 𝐢𝐢) = 𝐴𝐴 Γ— (𝐡𝐡 ∩ 𝐢𝐢).