If an element is not in the set difference $A\setminus B \: \:?$ Given two sets$$A=\{1,3,5\}\quad,\quad B=\{1,3,8\}.$$ Then I compute the $A\setminus B=\{5\}.$ 
But my book$^\dagger$said: ... . Also, observe that $x\notin A\setminus B$ does not mean that $x\notin A\lor x\in B$. Why?
I don't know how to explain the Why, please give me some example.

$^\dagger$a friendly introduction to analysis, second edition, page $4$.
EDIT: My intuition tells me that it does mean that. I guess there is some special case about null-set or about the restriction of the universal set.

You could also use the identity $A\setminus B =A \cap B^c$, where $B^c$ is the complement of $B$ with respect to the universe, say $U$ Then $$ x \notin A \setminus B \iff x \notin A \cap B^c.$$ Applying DeMorgan’s laws (in the “set form” one gets $$ x \notin A \cap B^c \iff x \in (A \cup B)^c \iff x \in A^c \cap B,$$ this is, $x \notin A$ and $x \in B$.
