Suppose that $A_1 \subset M_1$ and $A_2 \subset M_2$.

How to formally prove that $A_1\times A_2 = (A_1 \times M_2 )\cap(M_1\times A_2)$? It's easy to justify informally, I can see why it's true, but I have troubles with proving it rigorously.

Let's say $(x,y)\in A_1 \times A_2$. We need to show it belongs to $(A_1 \times M_2 )\cap(M_1\times A_2)$. And then consider $(x,y) \in (A_1 \times M_2 )\cap(M_1\times A_2)$ to show that $(x,y)\in A_1 \times A_2$.

So the first question that needs to be answered - what is $(A_1 \times M_2 )\cap(M_1\times A_2)$? It's the set of elements that are in both sets. A single element is actually a tuple $(x,y)$.

$(A_1 \times M_2 )=(A_1 \times A_2)\cup (A_1 \times M\setminus A_2)$.

$(M_1 \times A_2 )=(A_1 \times A_2)\cup (M_1\setminus A_1 \times A_2)$.

The intersection of two sets above is $A_1 \times A_2$. Not sure if that's rigorous enough. Maybe you could prove it in a more elegant way?


$A_1\subset M_1$ implies that $A_1\times A_2\subset M_1\times A_2$

$A_2\subset M_2$ implies that $A_1\times A_2\subset A_1\times M_2$

Combining this we find: $$A_1\times A_2\subset (A_1\times M_2)\cap (M_1\times A_2)$$

Conversely if $\langle x,y\rangle\in (A_1\times M_2)\cap (M_1\times A_2)$ then we find that $x\in A_1$ and $y\in A_2$ so that $\langle x,y\rangle\in A_1\times A_2$.

From this we conclude that also: $$(A_1\times M_2)\cap (M_1\times A_2)\subset A_1\times A_2$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.