Sets

Describing a set

A set is a collection of things, written inside curly brackets: $A = \{2, 4, 6, 8, 10\}$.

• $x \in A$ — "$x$ is an element of $A$"
• $x \notin A$ — "$x$ is not an element of $A$"
• $n(A)$ — the number of elements in $A$. Here $n(A) = 5$.
• $\varnothing$ or $\{\;\}$ — the empty set, with no elements at all

A set can also be described by a rule: $\{x : x \text{ is an even number less than } 12\}$ means the same as the list above. Listing it out is usually safer in an exam — you can see what you are working with.

The universal set and the complement

The universal set $U$ is everything under consideration in that question. The complement $A'$ is everything in $U$ that is not in $A$.

If $U = \{1, 2, 3, \ldots, 10\}$ and $A = \{2,4,6,8,10\}$, then $A' = \{1,3,5,7,9\}$.

Union and intersection

For $A = \{1,2,3,4\}$ and $B = \{3,4,5\}$: $\;A \cup B = \{1,2,3,4,5\}$ and $A \cap B = \{3,4\}$.

An element is written once in the union, however many sets it belongs to.

Subsets

$A \subset B$ means every element of $A$ is also in $B$.

Two consequences worth knowing: if $A \subset B$ then $A \cap B = A$ and $A \cup B = B$ — the overlap is all of the smaller set, and the union is all of the larger.

The empty set is a subset of every set.

Venn diagrams

Draw the rectangle for $U$ first, then overlapping circles inside it.

Fill in the overlap first, then work outwards by subtracting. If 25 students study French and 8 study both languages, then $25 - 8 = 17$ study French only — and 17 is the number that goes in the French-only region of the diagram.

Counting problems

Adding the two sets counts everyone in the overlap twice, so subtract the overlap once.

In a class of 40 where 25 study French, 18 study Spanish and 8 study both:

$$n(F \cup S) = 25 + 18 - 8 = 35$$

so 35 study at least one language, and $40 - 35 = 5$ study neither.

The same equation can be run backwards. If you know the total, both individual sets and how many are in neither, the unknown overlap drops out — which is what most exam questions on this topic actually ask.