Sets, Venn diagrams and counting overlaps

The notation

A set is a collection of distinct objects, its elements. $x \in A$ means $x$ belongs to $A$; $\lvert A \rvert$ (or $n(A)$) is the number of elements of a finite set. The universe $U$ is the set of everything under discussion.

Symbol	Meaning	With $A = \{1, 2, 3, 4\}$, $B = \{3, 4, 5\}$, $U = \{1, \dots, 6\}$
$A \subseteq B$	every element of $A$ is in $B$	false: $1 \in A$ but $1 \notin B$
$A \cup B$	in $A$ or $B$ (or both)	$\{1, 2, 3, 4, 5\}$
$A \cap B$	in both $A$ and $B$	$\{3, 4\}$
$A \setminus B$	in $A$ but not in $B$	$\{1, 2\}$
$\overline{A}$ (also $A^c$)	in $U$ but not in $A$	$\{5, 6\}$
$A \times B$	all ordered pairs $(a, b)$ with $a \in A$, $b \in B$	12 pairs, from $(1, 3)$ to $(4, 5)$

The empty set $\varnothing$ has no elements and is a subset of every set. Every set is a subset of itself. A set with $n$ elements has $2^n$ subsets, because each element is either in or out of a subset, independently; the count includes $\varnothing$ and the set itself.

The Cartesian product has $\lvert A \times B \rvert = \lvert A \rvert \cdot \lvert B \rvert$ elements, and order matters: $(1, 3) \in A \times B$, while $(3, 1)$ belongs to $B \times A$.

The complement depends on the universe. "Not English" among 120 students is a different set from "not English" among all the students in a school. Name $U$ before taking any complement.

Two useful laws

De Morgan's laws say that the complement of a union is the intersection of the complements, and the other way round:

$$\overline{A \cup B} = \overline{A} \cap \overline{B}, \qquad \overline{A \cap B} = \overline{A} \cup \overline{B}.$$

In words: "not (A or B)" is "neither A nor B"; "not (A and B)" is "not A, or not B". These let you count a "neither" group as $\lvert U \rvert - \lvert A \cup B \rvert$.

Also, $A \setminus B = A \cap \overline{B}$, and $A$ splits into two disjoint pieces: $A = (A \setminus B) \cup (A \cap B)$. That split is the idea behind every Venn diagram count.

Intervals are sets too

Sets of real numbers are often intervals, and the same operations apply. With $A = [-2, 3)$ and $B = (1, 5]$: $A \cap B = (1, 3)$, $A \cup B = [-2, 5]$, $A \setminus B = [-2, 1]$. Draw both on one number line and read off the result, paying attention to which endpoints are included.

Counting a union: inclusion and exclusion

Adding $\lvert A \rvert$ and $\lvert B \rvert$ counts the elements of $A \cap B$ twice, so subtract them once:

$$\lvert A \cup B \rvert = \lvert A \rvert + \lvert B \rvert - \lvert A \cap B \rvert.$$

For three sets, the pairwise overlaps are subtracted, but that removes the triple overlap once too often, so it is added back:

$$\lvert A \cup B \cup C \rvert = \lvert A \rvert + \lvert B \rvert + \lvert C \rvert - \lvert A \cap B \rvert - \lvert A \cap C \rvert - \lvert B \cap C \rvert + \lvert A \cap B \cap C \rvert.$$

Filling a Venn diagram from the inside out

When a question asks for "only A", "exactly two" or "neither", draw the diagram and fill it from the innermost region outwards. With two sets there are four regions: only $A$, only $B$, both, and neither. With three sets there are eight: the centre (all three), three "exactly two" regions, three "only one" regions, and the outside.

1. Write the triple overlap in the centre.
2. For each pair, subtract the centre from the pair's total to get "these two only".
3. For each set, subtract its three inner regions from its total to get "this one only".
4. Subtract everything inside the circles from $\lvert U \rvert$ to get "neither".

Then add all eight regions; they must give $\lvert U \rvert$. That final sum is a built-in check.

Reading the wording

"Studies French" includes those who also study German; "studies only French" does not. "At least one" is the union; "exactly one" excludes every overlap; "at most one" includes "neither". Most wrong answers in set questions come from reading one of these as another.