# Is it always true that $(A_1\cap A_2)\times(B_1\cap B_2)=(A_1\times B_1)\cap (A_2\times B_2)$?

Is it always true that $$(A_1\cap A_2)\times(B_1\cap B_2)=(A_1\times B_1)\cap (A_2\times B_2)\qquad ?$$ I believe it is, but I wanted to make sure than I am on the right path. I proved containment for both sides by picking arbitrary elements $(a,b)$ to be in the LHS, and I have shown that they are in the RHS, and vice-versa. is this correct to solve a problem like this?

• yes, if that's what you did it is correct. Commented Jan 21, 2013 at 17:51
• You can also prove this by showing that the statement of specification upon each set is logically equivalent to the other.
– 000
Commented Jan 21, 2013 at 18:14

The question was answered already in the comments, so I thought I'd make a community wiki answer. Yes, to show that two sets $x$ and $y$ are equal it is enough to show that $x \subseteq y$ and $y \subseteq x$. This follows from the Axiom of Extensionality: sets are equal if and only if they have the same elements.