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.


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    $\begingroup$ Your book is correct. The set of all things in $A$ that are not in $A$ is the set with nothing in it--the empty set. $\endgroup$ – Jonny Dec 18 '14 at 17:44
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    $\begingroup$ You can think of $\setminus$ as being a "take away" operation. So $A\setminus B$ is like taking away everything $A$ and $B$ have in common. With that in mind, $A\setminus A$ would have the interpretation that you're removing every element of $A$ from $A$. What does that leave you with? $\endgroup$ – Cameron Williams Dec 18 '14 at 17:49
  • $\begingroup$ @user92570, use the classes..;) $\endgroup$ – mle Dec 21 '14 at 11:53

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.


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.


There is no element that is in $A$ and not in $A$ at the same time, so the answer is the empty set.


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.


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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