Prove that $X\times Y=\emptyset$ $\iff$ $X=\emptyset$ or $Y=\emptyset$.

Question

I am looking for verification of an attempt that I have made to prove the following claim:

Prove that $X\times Y=\emptyset$ $\iff$ $X=\emptyset$ or $Y=\emptyset$.

Proof

I will prove the contrapositive.

LHS: Assume $X\neq\emptyset$ and $Y\neq\emptyset$, so there is $a\in X$ and $b\in Y$ such that $(a,b)\in X\times Y$. Thus, $X\times Y\neq\emptyset.$

RHS: Assume $X\times Y\neq\emptyset.$ Then, there is a $(a,b)\in X\times Y$ such that $a\in X$ and $b\in Y$, hence $X\neq \emptyset$ and $Y\neq\emptyset$. $\quad \Box$

Can you check and critique my proof attempt above? I would be grateful for any feedback or other ways in which the problem could be appraoched. Thanks

Answer 1

As others have mentioned in the comments, your proof is good and there aren't any problems with it.

However, you don't have to argue this by contrapositive as you have done. Here is an alternative proof of the claim that is equally valid (this makes use of proof by contradiction).

Prove: $X\times Y=\emptyset$ $\iff$ $X=\emptyset$ or $Y=\emptyset$.

Proof

($\impliedby$) If $X = \emptyset$ or $Y = \emptyset$ or both, then there does not exist any $(x,y) \in X \times Y$ since either $x$ or $y$ or both do not exist by assumption (if you want to formalise this, try arguing by contradiction). Therefore $X \times Y = \emptyset$ (as it has no elements).

($\implies$) If $X \times Y = \emptyset$, then by assumption there exists no pair $(x,y) \in X \times Y.$ Again, we can argue by contradiction that $X$ or $Y$ or both must be empty. Assume that they are both non-empty, then there exists $x_0 \in X$ and $y_0 \in Y.$ However, this means that the pair $(x_0, y_0) \in X \times Y$ which contradicts our assumption that $X \times Y$ is empty. $\quad \Box$