# For any sets A and B, prove that

Question:

For any sets A and B, prove that

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

Solution:

Given: A and B two sets are given.

Need to prove: $(A \times B) \cap(B \times A)=(A \cap B) \times(B \cap A)$

Let us consider, $(x, y)^{\in}(A \times B) \cap(B \times A)$

$\Rightarrow(x, y)^{\in}(A \times B)$ and $(x, y)^{\in}(B \times A)$

$\Rightarrow\left(x \in_{A}\right.$ and $\left.y \in B\right)$ and $\left(x \in_{B}\right.$ and $\left.y \in_{A}\right)$

$\Rightarrow\left(x^{\in} A\right.$ and $\left.x^{\in} B\right)$ and $\left(y \in_{B}\right.$ and $\left.y \in_{A}\right)$

$\Rightarrow x^{\in}(A \times B)$ and $y \in_{(B \times A)}$

$\Rightarrow(x, y)^{\in}(A \times B) \cap(B \times A)$

From this, we can conclude that

$\Rightarrow(A \times B) \cap(B \times A) \subseteq(A \cap B) \times(B \cap A) \cdots(1)$

Let us consider again, $(a, b)^{\in}(A \cap B) \times(B \cap A)$

$\Rightarrow \mathrm{a}^{\in}(\mathrm{A} \cap \mathrm{B})$ and $\mathrm{b} \in_{(\mathrm{B} \cap \mathrm{A})}$

$\Rightarrow\left(\mathrm{a}^{\in} \mathrm{A}\right.$ and $\left.\mathrm{a}^{\in} \mathrm{B}\right)$ and $\left(\mathrm{b}^{\in} \mathrm{B}\right.$ and $\left.\mathrm{b} \in \mathrm{A}\right)$

$\Rightarrow\left(a^{E} A\right.$ and $\left.b^{\in} B\right)$ and $\left(a^{\in} B\right.$ and $\left.b^{\in} A\right)$

$\Rightarrow(a, b)^{\in}(A \times B)$ and $(a, b)^{\in}(B \times A)$

$\Rightarrow(\mathrm{a}, \mathrm{b})^{\in}(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})$

From this, we can conclude that,

$\Rightarrow(A \cap B) \times(B \cap A) \subseteq(A \times B) \cap(B \times A) \cdots(2)$

Now by the definition of set we can say that, from (1) and (2),

$(A \times B) \cap(B \times A)=(A \cap B) \times(B \cap A)[$ Proved $]$