Set difference starts with one set and removes elements that belong to another. A complement uses the same idea after we specify a universal set: the collection of objects under consideration.

From the set of registered students, remove those who have already submitted an assignment. The remaining set contains the registered students who have not submitted it.

For sets A and B, the difference $A\setminus B$ contains exactly the elements of A that do not belong to B. For a complement, first fix a universal set U and assume A is a subset of U. The complement of A within U contains the elements of U that are absent from A.

• $x\in A\setminus B$ means $x\in A$ and $x\notin B$.

$A\setminus B$ is the part of $A$ that $B$ does not reach. Here $x$ belongs to it and $y$ does not, because $$ lies in as well. A complement is the same idea with the frame playing the part of the second set.

Remove 4. The number 8 does not enter the result because it was not in the starting set.

$\{0,1\}\setminus\{1,2\}=\{0\}$, but $\{1,2\}\setminus\{0,1\}=\{2\}$.

The two calculations begin with different sets and leave different elements.

Within $U=\{1,2,3,4\}$, the complement of $\{2,4\}$ is $\{1,3\}$.

The result contains exactly the elements of this U that are not in the chosen set.

Taking complements turns every union into an intersection and every intersection into a union, over families of any size.

• $A\setminus B$ keeps what is in A and not in B; $A^c$ needs a universal set to be fixed first.
• $A\setminus B=A\cap B^c$, so a difference obeys the set laws.

Great work! You now know how to:

• Compute a difference and a complement, naming the universal set first
• Rewrite $A\setminus B$ as an intersection
• Apply De Morgan to a family of any size