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.
$\endgroup$ 35 Answers
$\begingroup$There is no contradiction. The set of all $x$ such that $x\in A$ and $x\notin A$ is empty. Since it is impossible that $x$ is both in $A$ and not in $A$ simultaneously.
$\endgroup$ 0 $\begingroup$The other answers are, of course, all correct. But here's another way to think about it, that will, perhaps, build on your intuition.
You're right that there is a contradiction. The statements "$x \in A$" and $x \notin A$ are mutually exclusive, or contradictory. Therefore, the statement that "$x \in A \mathrel{\mathrm{and}} x \notin A$" must be false for all values of $x$.
Consequently, the set definition is equivalent to $\{ x \mid \mathrm{False} \}$. Therefore, this set must be empty.
$\endgroup$ $\begingroup$There is no element that is in $A$ and not in $A$ at the same time, so the answer is the empty set.
$\endgroup$ 0 $\begingroup$It is useful to note that $A - B = A \cap B^C$. Taking $B = A$, we see $A- A = A \cap A^C = \emptyset$ by definition of complement.
$\endgroup$ $\begingroup$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
$$ \varnothing = \{ x \in A \mid x \notin A \} $$
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