TITLE: Contradictory definition in set theory book?
QUESTION [10 upvotes]: I'm using a book that defines $A\setminus B$ (apparently this is also written as $A-B$) as $\{x\mid x\in A,x\not\in B\}$, but then there was an exercise that asked to find $A\setminus A$. Wouldn't it be a contradiction according to the definition given? (All $x$ such that $x$ is in $A$ and not in $A$...)
Note: The back of the book gives the answer $\emptyset$ with no explanation.
Note2: The book is Introductory Mathematics by Geoff Smith.
Thanks.

REPLY [3 votes]: By the definition of set builder notation,
$$y \in \{ x \in A \mid x \notin A \} $$
if and only if
$$ y \in A \wedge y \notin A $$
Since the latter is a contradiction, we've proven that, for all $y$,
$$ y \notin \{ x \in A \mid x \notin A \} $$
However, we already know a set that tabulates that membership relation: i.e. we have
$$y \notin \varnothing$$
In particular, for all $y$, we have
$$ y \in \{ x \in A \mid x \notin A \} $$
if and only if
$$ y \in \varnothing $$
and so we conclude
$$ \varnothing = \{ x \in A \mid x \notin A \} $$