C1.2; E1.2 - Sets and two-set Venn diagrams | Cambridge IGCSE Mathematics

A person can belong to two clubs without becoming two people. For Core and Extended, C1.2/E1.2 uses sets to organise such overlaps. You will describe sets, read union, intersection and complement notation, and fill two-set Venn diagrams without counting anyone twice.

Defining sets and a universal set

In a class of $30$ , $18$ students play football and $15$ play basketball. Does that mean $33$ different students play a sport? Some could play both. Before counting them, we need a way to say exactly which people each group contains.

A set is a well-defined collection of distinct objects, called its elements. “The prime numbers below ten” defines a set: you can decide whether any proposed number belongs. “The nicest numbers” does not, unless you supply an agreed rule for “nicest”.

Set notation

List a set's elements inside braces, for example $A=\{2,3,5,7\}$ . Order does not matter, and repeating an element does not add another element. The notation $n(A)$ means the number of distinct elements in $A$ , so here $n(A)=4$ .

The universal set, $\mathcal{E}$ , contains all the objects being considered in the problem. Every set in that problem stays within it.

You can describe the same set by a condition. For example,

A=\{x:x\text{ is a prime number less than }10\}.

The colon means “such that”; the letter $x$ stands for a possible element. With $\mathcal{E}=\{1,2,\ldots,10\}$ , only numbers in that universal set are available. The dots continue the stated pattern; they are safe here because every integer from one to ten is intended.

Worked example 1: a condition becomes a list

Question. The universal set is the integers from $1$ to $12$ , inclusive. Set $F$ contains the factors of $12$ . List $F$ and find $n(F)$ .

Plan. Find the distinct positive factors, then count the list.

\begin{aligned} 12&=1\times12=2\times6=3\times4 &&\text{all factor pairs}\\ F&=\{1,2,3,4,6,12\} &&\text{each factor once}\\ n(F)&=6 &&\text{six distinct elements} \end{aligned}

Answer. $F=\{1,2,3,4,6,12\}$ , $n(F)=6$ .

Check. Each listed number divides $12$ exactly; the six available integers $5,7,8,9,10,11$ do not.

Keep the factor rule, but reduce the universal set. This changes which otherwise valid factors are available.

Worked example 2: the universal set matters

Question. Now $\mathcal{E}=\{1,2,3,4,5,6\}$ , and $F$ again contains the numbers in $\mathcal{E}$ that are factors of $12$ . List $F$ and find $n(F)$ .

Plan. Test only the six available elements against the factor rule.

\begin{aligned} F&=\{1,2,3,4,6\} &&\text{available factors only}\\ n(F)&=5 &&\text{count distinct entries} \end{aligned}

Answer. $F=\{1,2,3,4,6\}$ , $n(F)=5$ .

Check. Of the six available numbers, only $5$ fails. Therefore five belong; $12$ cannot be included because it is outside $\mathcal{E}$ .

The list $\{2,3,2,5\}$ describes the same set as $\{2,3,5\}$ . A set records which elements are present, rather than how often a label was written.

Watch out

$n(F)=\{1,2,3,4,6\}$ confuses a count with a set. Write $F=\{1,2,3,4,6\}$ and $n(F)=5$ .

In the next problem, the factor rule and the universal set both restrict the list.

Union, intersection and complement

A Venn diagram draws the universal set as a rectangle and each named set as a closed region, usually a circle. An element inside both circles belongs to both sets. Put it in the overlap once, rather than writing a copy in each circle.

Take $\mathcal{E}=\{1,2,\ldots,10\}$ , $A=\{2,4,6,8,10\}$ and $B=\{2,3,5,7\}$ . The overlap contains $2$ , the only number here that is both even and prime.

[DIAGRAM: asset_name: Elements in two overlapping sets; asset_slug: sets_elements_regions; caption: Each element occupies exactly one of the four regions.; description: A universal rectangle contains overlapping circles A and B. A only contains 4, 6, 8 and 10; the intersection contains 2; B only contains 3, 5 and 7; outside both contains 1 and 9.]

Union, intersection and complement

$A\cup B$ , the union, contains elements in $A$ or $B$ , including those in both. $A\cap B$ , the intersection, contains elements in both $A$ and $B$ .

$A'$ , the complement of $A$ , contains every element of $\mathcal{E}$ that is not in $A$ . The prime mark means “not in this set”, rather than “not in either circle”.

“Or” in a union includes the overlap. Everyday language sometimes uses “or” to demand a choice, but set union means at least one of the sets. Complement always needs a universal set: the complement of the even numbers in this picture is not every odd number in existence.

Worked example 3: finding the union and intersection

Question. For the sets $A$ and $B$ above, list $A\cup B$ and $A\cap B$ .

Plan. Collect all elements inside either circle for the union, and only shared elements for the intersection.

\begin{aligned} A\cup B&=\{2,3,4,5,6,7,8,10\} &&\text{either circle, no repeats}\\ A\cap B&=\{2\} &&\text{both conditions hold} \end{aligned}

Answer. The union contains eight elements; the intersection is $\{2\}$ .

Check. The two universal elements missing from the union are $1$ and $9$ , precisely those outside both circles. Only $2$ appears in both original lists.

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C1.2; E1.2 - Sets and two-set Venn diagrams

Defining sets and a universal set

Union, intersection and complement

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