Need to prove with empty set

Question

So I've been working on set theory lately, and I have been asking myself: Do I need to do every proof for the empty set?
For example; I need to prove that if $A\cap B=A$ then $A\subseteq B$. You can do that by saying that if $a\in A$ then $a\in A\cap B$, so $a\in B$. Thus, $A\subseteq B$. Does the assumption that $a\in A$ break the proof for the empty set? (As the empty set has no elements) Also, when I say $a\in B$ I'm assuming B has elements. I know that the proof for $A=\emptyset$ or $B=\emptyset$ is trivial (if $A=\emptyset$, for every set B $A\cap B=A$ and $A\subseteq B$; if $B=\emptyset$ then A must be $\emptyset$ so that $A\cap B=A$, and obviously $\emptyset \subseteq \emptyset$), but I don't know if I should show it in an exam.

Answer 1

In the example that you have mentioned, there is not need to consider the case $A=\emptyset$. When you write “if $x\in A$” you are not assuming that $A$ is not empty. It's just the beginning of a sequence of arguments that proves that if $x\in A$, then $x\in B$. In other words, you are proving that every element of $A$ is also an element of $B$, and that's what $A\subseteq B$ means.