prove or disprove (K ∩ L) × (M ∩ N) = (K × M) ∩ (L × N);

Question

I have a bad understanding of the Cartesian product and having trouble doing this:

We are given $(K ∩ L) × (M ∩ N) = (K × M) ∩ (L × N)$; how to prove it or disprove it?

This is what i am thinking:

• Let $x ∈ (K∩L)×(M∩N)$.
• then $x=(x_1,x_2)$, where $x_1∈(K∩L) ∧ x_2∈(M∩N)$.
• By definition of intersection $x_1∈K ∧ x_1∈L ∧ x_2∈M ∧ x_2∈N$
• as $x_1∈K ∧ x_2∈M$, then $x = (x_1,x_2) ∈K×M$.
• as $x_1∈L ∧ x_2∈N$, then $x = (x_1,x_2) ∈L×N$.
• as $x∈K×M ∧ x∈L×N$, then $x∈(K × M) ∩ (L × N)$

let's prove this the other way

• let $x ∈ (K × M) ∩ (L × N)$
• by definition of intersection $x∈K×M ∧ x∈L×N$.
• $x ∈K×M$ which means that $x_1∈K ∧ x_2∈M$.
• $x ∈L×N$ which means that $x_1∈L ∧ x_2∈N$.
• by definition of intersection $x_1∈(K∩L)$.
• by definition of intersection $x_2∈(M∩N)$.
• which means $x=(x_1, x_2)∈(K∩L)×(M∩N)$.

Is it all correct? or is it all wrong?

Answer 1

Yes, your proof is correct. However, when you want to prove if-and-only-if statements, you should try to prove both directions at once by using $\iff$ instead of $\implies$. It isn't always possible, but when it is (and it is here!) you can halve the length of your proof. Thus a more concise statement of your proof technique is (if you'll forgive my relabelling the $x_i$)$$(a,\,b)\in(K\cap L)\times(M\cap N)\iff (a\in K\land a\in L)\land (b\in M\land b\in N)\\\iff (a\in K\land b\in M)\land (a\in L\land b\in N)\\\iff (a,\,b)\in K\times M\land (a,\,b)\in L\times N\iff (a,\,b)\in (K\times M)\cap (L\times N).$$