I'm trying to prove $A\cap\emptyset=\emptyset$. I've seen several proofs for this which all seemed to essentially go about proving it by noticing that $\emptyset\subset A\cap\emptyset$ by definition and then supposing some x in $A\cap\emptyset$ and then saying that this is a subset of $\emptyset$.I see the logic here and Im certainly not saying it's wrong obviously but I feel like there's certain logical gaps there. So, I went about trying to do it differently and I came up with the following but was wondering if this is correct since I haven't been able to find it elsewhere, and, if it is correct, why this convention isn't more often used
Suppose $x\in \emptyset\cap A$ ,thus $x\in A$ and $x\in\emptyset$. By definition, $x$ cannot belong to $\emptyset$ , thus there is no $x$ such that $x\in A$ and $x\in A\cap\emptyset$ , hence $\emptyset\cap A=\emptyset$.
This is correct. Another way to say it: $A\cap\emptyset\subset\emptyset$ (property of intersection). And the only subset of $\emptyset$ is $\emptyset$. So $A\cap\emptyset=\emptyset$.