Set Notation in IGCSE Maths
Set notation is the language used to describe collections of objects in mathematics. Once you know the symbols, problems involving sets, Venn diagrams, and probability all become much easier to read. This page covers every piece of set notation needed for IGCSE Maths, from listing the members of a set through to set-builder notation and the special symbols for natural numbers, integers, rational numbers, and real numbers.
The Set Notation Symbols
Set notation is a compact language for describing collections of objects. Each symbol below has a precise meaning. Learn them once and they appear throughout sets, probability, and Venn diagrams.
| Symbol | Meaning | Example |
|---|---|---|
| \(\{\;\}\) | The set containing the listed elements. | \(P = \{2, 3, 5, 7\}\) |
| \(\in\) | "Is an element of". The object is a member of the set. | \(3 \in P\) |
| \(\notin\) | "Is not an element of". The object is not in the set. | \(4 \notin P\) |
| \(n(A)\) | The number of elements in set \(A\). | If \(P = \{2,3,5,7\}\) then \(n(P) = 4\). |
| \(\varnothing\) | The empty set. A set with no elements. Also written \(\{\;\}\). | \(\{\text{odd numbers that are multiples of } 10\} = \varnothing\) |
| \(\text{ℰ}\) | The universal set. All elements being considered in the question. | \(\text{ℰ} = \{\text{whole numbers from 11 to 20}\}\) |
| \(A'\) | The complement of \(A\). Every element of \(\text{ℰ}\) that is not in \(A\). | If \(\text{ℰ} = \{4, 5, 6, 7, 8\}\) and \(A = \{4, 7\}\), then \(A' = \{5, 6, 8\}\). |
| \(\subset\) | "Is a subset of". Every member of one set is also in the other. | \(\{2,4\} \subset \{1,2,3,4,5\}\) |
| \(\not\subset\) | "Is not a subset of". At least one element is missing from the other set. | \(\{2, 9\} \not\subset \{1,2,3,4,5\}\) |
| \(\cap\) | "Intersection". Elements in both sets at the same time. | {1, 2, 3} ∩ {2, 3, 4} = {2, 3} |
| \(\cup\) | "Union". Elements in either set, or both. Each element listed once. | {1, 2} ∪ {2, 3} = {1, 2, 3} |
The Universal Set and Complement
Every set problem sits inside a bigger set called the universal set, written \(\text{ℰ}\). The universal set is whatever the question is about: it might be all positive integers less than 20, or all students in a school, or all letters of the alphabet.
The complement of a set \(A\), written \(A'\), is everything in the universal set that is not in \(A\). The two sets together always make up the whole of \(\text{ℰ}\).
Note that \(n(A) + n(A') = n(\text{ℰ})\). In this example: \(4 + 6 = 10\). This is a useful check.
Subsets
A set \(B\) is a subset of \(A\) if every member of \(B\) is also a member of \(A\). It is written \(B \subset A\).
💡 Example
Is \(B = \{4, 8, 12\}\) a subset of \(A = \{2, 4, 6, 8, 10, 12, 14\}\)?
Every element of \(B\) is also in \(A\), so \(B \subset A\).
What's happening?
Every element of the smaller set must also appear in the bigger set. If just one element is missing, it is not a subset.
💡 Example
Is \(T = \{6, 9, 14\}\) a subset of \(A = \{2, 4, 6, 8, 10, 12, 14\}\)?
\(6\) and \(14\) are both in \(A\), but \(9\) is not, so \(T \not\subset A\).
What's happening?
One element missing from \(A\) is enough to break the subset rule. Stop as soon as you find one.
Two useful facts: every set is a subset of itself, and the empty set \(\varnothing\) is a subset of every set.
Intersection and Union
Intersection: \(A \cap B\)
The intersection of two sets is the set of elements that are in both sets at the same time. Think of it as the overlap.
If two sets share no elements, their intersection is the empty set: \(A \cap B = \varnothing\). The two sets are then called disjoint.
Union: \(A \cup B\)
The union of two sets is the set of elements that are in either set, or in both. Each element is listed only once, even if it appears in both.
Counting a union without listing it
If you know how many elements are in each set and how many they share, you can count the union without writing it out. Add the two sizes, then subtract the overlap once, because the shared elements were counted twice.
For example, if \(n(A) = 9\), \(n(B) = 6\) and \(n(A \cap B) = 4\):
\[\begin{array}{rcl} n(A \cup B) &=& 9 + 6 - 4 \\ &=& 11 \end{array}\]Reading them aloud
A useful trick: read \(\cap\) as "and" and \(\cup\) as "or".
\(x \in A \cap B\) means "\(x\) is in \(A\) and \(x\) is in \(B\)".
\(x \in A \cup B\) means "\(x\) is in \(A\) or \(x\) is in \(B\) (or both)".
Set-Builder Notation
Set-builder notation describes a set using a rule rather than a list. It is especially useful when a set is infinite or too large to write out.
The colon \(:\) is read as "such that". Some books use a vertical bar \(|\) instead, but the meaning is the same.
When the rule builds each element from another set
A rule can define every element by a formula applied to the members of a set you already have. Work through the given set one member at a time, put each one into the formula, and collect the results.
If \(A = \{3, 5, 8\}\), then \(\{\, x : x = 2y,\; y \in A \,\}\) means "double each member of \(A\)":
\[\begin{array}{rcl} 2 \times 3 &=& 6 \\ 2 \times 5 &=& 10 \\ 2 \times 8 &=& 16 \end{array}\]So the set is \(\{6, 10, 16\}\). The same reading works for \(x = y^2\) or \(x = 3y + 1\).
When the rule is an inequality to solve
Sometimes the rule needs solving before you can list anything. Solve it exactly as you would any inequality, then apply the number-set condition to decide which solutions count.
For \(\{\, x : 4x + 3 \leq 19,\; x \in \text{ℕ} \,\}\):
\[\begin{array}{rcl} 4x + 3 &\leq& 19 \\ 4x &\leq& 16 \\ x &\leq& 4 \end{array}\]The natural numbers up to 4 are \(\{1, 2, 3, 4\}\). Note that \(x \in \text{ℤ}\) would also allow \(0\) and every negative integer, so the set would be infinite.
💡 Example
Describe \(\{\, x : x > 5 \,\}\) in words.
The set of all \(x\) such that \(x\) is greater than 5.
What's happening?
The colon splits the notation into two parts: the variable on the left, and the rule on the right.
💡 Example
List the elements of \(\{\, x : 1 \leq x \leq 6,\; x \in \text{ℤ} \,\}\).
\(x\) must be an integer (because of \(x \in \text{ℤ}\)).
\(x\) must be between 1 and 6, inclusive.
So the set is \(\{1, 2, 3, 4, 5, 6\}\).
What's happening?
The condition after the comma narrows the type of number allowed. Without it, \(x\) could be anything (decimals, fractions, irrationals).
💡 Example
List the elements of \(\{\, x : x^2 - 2x - 8 = 0,\; x \in \text{ℕ} \,\}\).
Solve the equation first, then apply the number-set condition.
\[\begin{array}{rcl} x^2 - 2x - 8 &=& 0 \\ (x - 4)(x + 2) &=& 0 \end{array}\]So \(x = 4\) or \(x = -2\).
Only \(4\) is a natural number, so the set is \(\{4\}\).
What's happening?
The rule can be an equation, not just an inequality. Solve it, then keep only the solutions that belong to the stated number set. The same equation over \(\text{ℤ}\) would give \(\{-2, 4\}\), and a rule with no solutions of the required type gives the empty set.
The Special Number Sets
Four sets of numbers appear so often that they have their own symbols.
These nest inside each other: \(\text{ℕ} \subset \text{ℤ} \subset \text{ℚ} \subset \text{ℝ}\). Every natural number is also an integer, every integer is also rational, and every rational number is also real.
Worked Examples
💡 Example 1: Listing a set from a description
Write out the set \(A = \{\text{prime numbers less than 20}\}\) as a list. Find \(n(A)\).
A prime number has exactly two factors: 1 and itself.
Check each number below 20: 2, 3, 5, 7, 11, 13, 17, 19 are prime.
\(A = \{2, 3, 5, 7, 11, 13, 17, 19\}\)
\(n(A) = 8\)
What's happening?
The rule defines membership. To turn the rule into a list, work through the candidates and keep the ones that satisfy the rule. \(n(A)\) just counts them.
💡 Example 2: Complement
\(\text{ℰ} = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}\) and \(M = \{\text{multiples of 4 in } \text{ℰ}\}\). List \(M'\).
\(M = \{12, 16, 20\}\)
\(M'\) is everything in \(\text{ℰ}\) not in \(M\):
\(M' = \{11, 13, 14, 15, 17, 18, 19\}\)
Check the sizes add up: ✓
\[\begin{array}{rcl} n(M) + n(M') &=& 3 + 7 \\ &=& 10 \\ &=& n(\text{ℰ}) \end{array}\]What's happening?
The complement always splits the universal set into two pieces. The check that the sizes add to \(n(\text{ℰ})\) catches any missed element.
💡 Example 3: Intersection and union
\(A = \{2, 4, 6, 8, 10\}\) and \(B = \{1, 2, 3, 4, 5\}\). Find \(A \cap B\) and \(A \cup B\).
Intersection (in both):
Union (in either, each listed once):
What's happening?
For \(\cap\), look for shared elements. For \(\cup\), gather them all but never repeat. The pilled numbers appear in both sets but are still listed only once in the union.
💡 Example 4: Set-builder notation
Express the set of natural numbers between 5 and 12 (inclusive) using set-builder notation, then list it.
The variable: \(x\). The rule: \(5 \leq x \leq 12\). The number type: \(x \in \text{ℕ}\).
Set-builder form:
\(\{\, x : 5 \leq x \leq 12,\; x \in \text{ℕ} \,\}\)
Listed: \(\{5, 6, 7, 8, 9, 10, 11, 12\}\)
What's happening?
Two pieces of information together: the range of values, and the type of number. Without \(x \in \text{ℕ}\), the set would also include 5.5, 7.3, and so on.
🔑 Key Points
- A set is a collection of elements, listed in curly brackets or described by a rule.
- \(n(A)\) is the number of elements in set \(A\).
- The empty set \(\varnothing\) has no elements.
- The universal set \(\text{ℰ}\) contains everything in the problem.
- The complement \(A'\) is everything in \(\text{ℰ}\) that is not in \(A\). Always: \(n(A) + n(A') = n(\text{ℰ})\).
- \(\cap\) means "and" (intersection), \(\cup\) means "or" (union).
- \(\text{ℕ} \subset \text{ℤ} \subset \text{ℚ} \subset \text{ℝ}\).
⚠️ Common Pitfalls
- Confusing \(\in\) with \(\subset\). Use \(\in\) for a single element; use \(\subset\) for a whole set.
- Confusing \(\varnothing\) with \(\{0\}\). The empty set has zero elements; the set \(\{0\}\) has one element (which happens to be zero).
- Listing the same element twice in a union. Each element appears only once in any set.
- Forgetting the number-type condition in set-builder notation. \(\{x : x > 0\}\) without \(x \in \text{ℕ}\) (or similar) means every positive real number, not just \(\{1, 2, 3, \ldots\}\).
- Reading \(\cap\) as "or" and \(\cup\) as "and". It is the other way round.