Intersection of set A and empty set is identical to the empty set I came across this question:

Show that $A \cap \emptyset= \emptyset$

The solution is that since (earlier in the book I am following) $A\cap B\subseteq B$, this proves that $A\cap \emptyset\subseteq\emptyset$.
Also that the empty set is a member of all sets, so that proves $\emptyset\subseteq A\cap\emptyset$, by the axiom of extensionality, this proves that $A\cap\emptyset=\emptyset$.
But I found this weird: for all $x, A\cap B$ iff $x\in A$ and $x\in B$. So surely $A\cap\emptyset\subseteq\emptyset$ means that $x\in\emptyset$, which contradicts with the fact that there is no element in the empty set!
Could anyone help please?
 A: First of all you can see here, and here, difference of $\in$ and $\subseteq$. Then you can say $\emptyset$ has no element, so $ A \cap \emptyset$ has no element which means $A \cap \emptyset= \emptyset$. (Or you can use the other argument).
A: $(\forall x\in\varnothing : x\in A)$ is true because you can find no counterexamples in the empty set.   There are no element in the empty set which is not in $A$.
$$\forall x\in \varnothing : x\in A \quad\iff\quad \neg\exists x\in\varnothing:x\notin A$$
$$\forall x (x\in \varnothing \to x\in A) \quad\iff\quad \neg\exists x(x\in\varnothing\wedge x\notin A)$$
This is what is known as a vacuous truth.   Everything is true about the elements of an empty set; because there are none where it might be false.
A: I'll write the empty set as {}.
As you stated,
$x\in A\cap B \iff x\in A \wedge x\in B$.
If we accept $A\cap$ {} = {},
$x\in$ {} $\iff x\in A \wedge x\in$ {}  
For any x you could substitute in, you have a false statement on both sides.
So the double implication is true for all x.  
