Sets

What a set is

A set is a collection of objects, called its elements or members. Sets are named with capital letters and written in curly brackets, with the elements separated by commas: $A=\{2,4,6,8\}$. Order does not matter, and no element is written twice, so $\{2,4,6,8\}$ and $\{8,6,4,2\}$ are the same set.

A set can be described in three ways:

• By listing its elements: $B=\{a,b,c,d,e\}$. For a long pattern you may write $\{a,b,c,\dots \}$, where the dots mean the pattern carries on.
• In words: $P=\left\{\text{prime numbers less than}12\right\}$, which is $\{2,3,5,7,11\}$.
• With a rule, in set-builder notation: $A=\{x:x\text{is a natural number}\}$ is read as the set of all $x$ such that $x$ is a natural number. The colon means such that. $C=\{x:2⩽x⩽6\}$ is every number from 2 to 6, including both ends. If the universal set says $x$ is an integer, $C$ is $\{2,3,4,5,6\}$.

$\mathrm{n}(A)$ means the number of elements in $A$. For $A=\{2,4,6,8\}$, $\mathrm{n}(A)=4$. Read inequality signs carefully when you count: $\{x:2<x⩽6\}$ leaves 2 out but keeps 6.

The universal set and the complement

The universal set, $ℰ$, contains everything being considered in the question. Every other set in the question is part of it.

The complement of $A$, written $A\prime$, is everything in $ℰ$ that is not in $A$. If $ℰ=\{1,2,3,\dots ,10\}$ and $A=\{2,4,6,8\}$, then $A\prime =\{1,3,5,7,9,10\}$.

A set and its complement together make up the whole of $ℰ$, with no overlap, so $\mathrm{n}(A)+\mathrm{n}(A\prime )=\mathrm{n}(ℰ)$. Here $4+6=10$. This gives a quick way to count a complement without listing it.

Union and intersection

The intersection $A\cap B$ is the set of elements in both $A$ and $B$.

The union $A\cup B$ is the set of elements in $A$ or $B$ or both.

To remember which is which, think of $\cup$ as a cup that holds everything from both sets, and $\cap$ as the narrow part that the two sets share.

With $A=\{1,2,3,4,6,12\}$ and $B=\{2,4,6,8,10,12\}$:

• $A\cap B=\{2,4,6,12\}$
• $A\cup B=\{1,2,3,4,6,8,10,12\}$. Each element is written once, even though 2, 4, 6 and 12 are in both sets.

The symbols combine. $(A\cup B)\prime$ is everything outside both sets. $A\cap B\prime$ is everything in $A$ that is not in $B$: here $\{1,3\}$. Work from the inside of any brackets outwards.

Venn diagrams

A Venn diagram shows the universal set as a rectangle, marked $ℰ$, and each set as a circle inside it. Two overlapping circles make four regions:

• in $A$ only, which is $A\cap B\prime$;
• in both, which is $A\cap B$;
• in $B$ only, which is $A\prime \cap B$;
• in neither, which is $(A\cup B)\prime$.

Every element of $ℰ$ goes in exactly one region. Place the elements of the overlap first, then the rest of each circle, then everything left over outside the circles. Cambridge Core questions use two sets.

A completed Venn diagram

$ℰ=\{1,2,\dots ,10\}$, $A=\left\{\text{factors of}6\right\}=\{1,2,3,6\}$ and $B=\left\{\text{even numbers}\right\}=\{2,4,6,8,10\}$. Reading from the diagram: $A\cap B=\{2,6\}$, $\mathrm{n}(A\cup B)=7$ and $(A\cup B)\prime =\{5,7,9\}$.

Counting with Venn diagrams

Often a Venn diagram shows how many elements are in each region rather than the elements themselves. Then the regions add up to $\mathrm{n}(ℰ)$, and

$$\mathrm{n}(A\cup B)=\mathrm{n}(A)+\mathrm{n}(B)-\mathrm{n}(A\cap B)$$

The overlap is subtracted because adding $\mathrm{n}(A)$ and $\mathrm{n}(B)$ counts the elements in both sets twice.

For a survey question:

1. Put the number in both sets in the overlap first.
2. Subtract it from each set's total to get $A$ only and $B$ only.
3. Subtract everything inside the circles from $\mathrm{n}(ℰ)$ to get the number outside.

If the overlap is not given, find it from the totals: $\mathrm{n}(A\cap B)=\mathrm{n}(A)+\mathrm{n}(B)-\mathrm{n}(A\cup B)$. If a region is unknown, call it $x$, write the other regions in terms of $x$, and set the total equal to $\mathrm{n}(ℰ)$.

The diagram for Worked example 2

The numbers in the four regions add to $\mathrm{n}(ℰ)=40$. The circle for $H$ holds $16+7=23$ students and the circle for $N$ holds $7+12=19$.

Shading regions

Questions ask you to shade a region given in set notation, or to describe a shaded region. Decide region by region: for each of the four regions, ask whether it belongs to the set, and shade it only if the answer is yes.

• $A\cap B$: the overlap.
• $A\cap B\prime$: in $A$ and not in $B$, the part of $A$ outside the overlap.
• $A\prime \cap B$: $B$ only.
• $A\cup B$: both circles completely.
• $(A\cup B)\prime$: outside both circles.
• $(A\cap B)\prime$: everything except the overlap.
• $A\prime$: everything outside $A$, which includes $B$ only and the outside.
• $A\cup B\prime$: in $A$, or outside $B$. That is everything except $B$ only.

A common slip is to shade $A\prime$ as only the outside region. $A\prime$ also contains the part of $B$ that is not in $A$.

Shading for A union B complement

The regions in $A\cup B\prime$ are marked shaded. $B$ only is left blank, because an element there is not in $A$ and is in $B$, so it is in neither $A$ nor $B\prime$.