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4. Let 𝐴 = {1,2}, 𝐡= {π‘Ž, 𝑏, 𝑐} and 𝐢 = {𝑐, 𝑑}. Then show that (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) =

𝐴 Γ— (𝐡 ∩ 𝐢).

To prove that (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) = 𝐴 Γ— (𝐡 ∩ 𝐢), we need to show that each set is a subset of the other.

First, let's find (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢):
𝐴 Γ— 𝐡 = {(1, π‘Ž), (1, 𝑏), (1, 𝑐), (2, π‘Ž), (2, 𝑏), (2, 𝑐)}
𝐴 Γ— 𝐢 = {(1, 𝑐), (1, 𝑑), (2, 𝑐), (2, 𝑑)}
(𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) = {(1, 𝑐), (2, 𝑐)}

Now, let's find 𝐴 Γ— (𝐡 ∩ 𝐢):
𝐡 ∩ 𝐢 = {𝑐}
𝐴 Γ— (𝐡 ∩ 𝐢) = {(1, 𝑐), (2, 𝑐)}

Since (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) = 𝐴 Γ— (𝐡 ∩ 𝐢) = {(1, 𝑐), (2, 𝑐)}, we can conclude that (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) = 𝐴 Γ— (𝐡 ∩ 𝐢).

To show that (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) = 𝐴 Γ— (𝐡 ∩ 𝐢), we need to prove that they have the same elements.

First, let's find (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢):

𝐴 Γ— 𝐡 = {(1, π‘Ž), (1, 𝑏), (1, 𝑐), (2, π‘Ž), (2, 𝑏), (2, 𝑐)}
𝐴 Γ— 𝐢 = {(1, 𝑐), (1, 𝑑), (2, 𝑐), (2, 𝑑)}

Taking the intersection of these two sets, we get:

(𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) = {(1, 𝑐), (2, 𝑐)}

Now, let's compute 𝐡 ∩ 𝐢:

𝐡 ∩ 𝐢 = {𝑐}

Next, let's compute 𝐴 Γ— (𝐡 ∩ 𝐢):

𝐴 Γ— (𝐡 ∩ 𝐢) = {(1, 𝑐), (2, 𝑐)}

We see that (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) = 𝐴 Γ— (𝐡 ∩ 𝐢), as both sets contain the same elements, which are {(1, 𝑐), (2, 𝑐)}.

Therefore, we have shown that (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) = 𝐴 Γ— (𝐡 ∩ 𝐢).

To show that (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) = 𝐴 Γ— (𝐡 ∩ 𝐢), we need to demonstrate that the two sets are equal. Here is how we can prove it step by step:

1. Start with 𝐴 Γ— 𝐡:
- 𝐴 Γ— 𝐡 represents the Cartesian product of set A and set B.
- 𝐴 Γ— 𝐡 is a set that contains all possible ordered pairs, where the first element belongs to set A and the second element belongs to set B.
- In this case, 𝐴 Γ— 𝐡 will contain the following ordered pairs: {(1,π‘Ž), (1,𝑏), (1,𝑐), (2,π‘Ž), (2,𝑏), (2,𝑐)}.

2. Next, consider 𝐴 Γ— 𝐢:
- 𝐴 Γ— 𝐢 represents the Cartesian product of set A and set C.
- 𝐴 Γ— 𝐢 is a set that contains all the possible ordered pairs where the first element belongs to set A and the second element belongs to set C.
- In this case, 𝐴 Γ— 𝐢 will contain the following ordered pairs: {(1,𝑐), (1,𝑑), (2,𝑐), (2,𝑑)}.

3. Now calculate the intersection of (𝐴 Γ— 𝐡) and (𝐴 Γ— 𝐢):
- (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) represents the intersection between the two sets, which means finding the common elements between them.
- In this case, when we calculate the intersection, we find that the only common elements between (𝐴 Γ— 𝐡) and (𝐴 Γ— 𝐢) are the ordered pairs (1,𝑐) and (2,𝑐).
- Hence, (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) = {(1,𝑐), (2,𝑐)}.

4. Lastly, consider 𝐴 Γ— (𝐡 ∩ 𝐢):
- 𝐡 ∩ 𝐢 represents the intersection between sets B and C, which means finding the common elements between them.
- In this case, 𝐡 ∩ 𝐢 = {𝑐}, as it is the only element that belongs to both sets B and C.
- 𝐴 Γ— (𝐡 ∩ 𝐢) represents the Cartesian product of set A and the intersection of sets B and C.
- Since 𝐡 ∩ 𝐢 = {𝑐}, 𝐴 Γ— (𝐡 ∩ 𝐢) will contain the following ordered pairs: {(1,𝑐), (2,𝑐)}.

5. Comparing the results:
- After calculating (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) and 𝐴 Γ— (𝐡 ∩ 𝐢), we see that they both contain the same ordered pairs: {(1,𝑐), (2,𝑐)}.
- Therefore, (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) = 𝐴 Γ— (𝐡 ∩ 𝐢).

By going through the steps above, we have established the equality between (𝐴 Γ— 𝐡) ∩ (𝐴 Γ— 𝐢) and 𝐴 Γ— (𝐡 ∩ 𝐢).