Set theoretic difference $A-B$ contradicting $\emptyset \in B$? Set theoretic difference is defined as follows:

Let $A,B$ be sets. Then the set theoretic difference of $A,B$ (written $A- B$) is the set defined by:
$x \in A-B$ iff $x \in A$  and $x \notin B$

In the notes I'm studying, I came across an example that $\mathbb{N} - \mathbb{Z}=\emptyset$. I realize that the point they're trying to get across is that the natural numbers are contained in the integers, but isn't it true that the empty set is in all sets? If so, does this not mean that $\emptyset \in A$ and $\emptyset \in B$, which contradicts the definition?
Sorry if this is too simple a question.
 A: Indeed, the empty set is a subset of both the naturals and the integers. Indeed, it is the subset of the naturals containing all natural numbers that are not also integers.
It does not show up as an element of either the naturals or the integers, although it is an element of the power sets of both. If we were looking for $\mathcal P(\Bbb N) - \mathcal P(\Bbb Z)$, the empty set would not be an element of that difference, because it is in both. However, the empty set would be that difference, because there is no subset of $\Bbb N$ that is not also a subset of $\Bbb Z$.
Does that help to clarify?
A: The empty set is not an element of either $\mathbb{N}$ or $\mathbb{Z}$.  The empty set is a subset of both (and, indeed, any set).  Symbolically,
$$ \emptyset \not\in \mathbb{N},\quad \emptyset\not\in\mathbb{Z},
\qquad\text{but}\qquad
\emptyset \subseteq \mathbb{N}, \quad \emptyset\subseteq\mathbb{Z}. $$
A: The empty set is a subset of A and B not necessarily an element of A or B. The empty set belongs to A - B if and only if it belongs to A and does not belong to B. In your case of A = N and B = Z the set A-B is the empty set because there is no element of N which does not belong to Z.        
