Solving the Relation: ##n((AXB) \cap (BXA)) = n(A \cap B)^2##

[cr7einstein]

Homework Statement

If I am given $n(A)$ and $n(B)$ for two sets A and B, and also provided with $n(A\cap B)^2$. We are supposed to find $n((AXB) \cap (BXA))$.

Homework Equations

My teacher said that the formula for $n((AXB) \cap (BXA)) = n(A \cap B)^2$. I am not sure how do you get to this result.

The Attempt at a Solution

So my question is, how do you prove (or derive the relation)-
$n((AXB) \cap (BXA)) = n(A \cap B)^2$

 

[SammyS]

Homework Statement

If I am given $n(A)$ and $n(B)$ for two sets A and B, and also provided with $n(A\cap B)^2$. We are supposed to find $n((AXB) \cap (BXA))$.

Homework Equations

My teacher said that the formula for $n((AXB) \cap (BXA)) = n(A \cap B)^2$. I am not sure how do you get to this result.

The Attempt at a Solution

So my question is, how do you prove (or derive the relation)-
$n((AXB) \cap (BXA)) = n(A \cap B)^2$

Clarification

By $\displaystyle\ AXB\$ do you mean the direct product $\displaystyle\ A\times B\$ ?

Also is $\displaystyle\ n(A\cap B)^2\$ actually the square of the number $\displaystyle\ n(A\cap B)\$ ?

.

 

[cr7einstein]

Yes. Sorry, I don't know how to get the thinner cross. And yes, for the second case too.

 

[SammyS]

Yes. Sorry, I don't know how to get the thinner cross. And yes, for the second case too.

\times

$\text{\times gives you}\ \times$

 

[geoffrey159]

You must show that $(A\times B) \cap (B\times A) = (A\cap B)\times (A\cap B)$. The answer is then straightforward if you know how to calculate the cardinal of a cartesian product.
More generally, $(A\times B) \cap (C\times D) = (A\cap C) \times (B\cap D)$ (There is no such rule for a union)
btw: the question makes sense only if A and B are finite sets

 

[cr7einstein]

Thanks!