Set Theory - Extended Maths

Universal set and subsets

A set is a collection of distinct objects, called its elements, listed inside curly brackets: \(A = \{1, 2, 3, 4\}\). We write \(2 \in A\) to say "2 is an element of \(A\)", and \(5 \notin A\) for "5 is not".

Universal set \(U\): The full collection of everything under discussion. Every other set in the question is drawn from \(U\).

Subset \(A \subseteq B\): \(A\) is a subset of \(B\) when every element of \(A\) is also in \(B\). For example \(\{1, 2\} \subseteq \{1, 2, 3, 4\}\).

So if the universal set is \(U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\), then any set of numbers you build from it, such as the even numbers \(\{2, 4, 6, 8, 10\}\), is a subset of \(U\).

Intersection, union, complement

Three operations combine or contrast sets. Keep the symbols straight and the rest is bookkeeping.

Intersection \(A \cap B\): The elements in both \(A\) and \(B\). Think "and", the overlap.

Union \(A \cup B\): The elements in \(A\) or \(B\) or both. Think "or", everything gathered together, each element listed once.

Complement \(A'\): The elements of the universal set \(U\) that are not in \(A\). Think "everything else".

Cup versus cap

The union symbol \(\cup\) is a cup that holds everything, so it collects both sets. The intersection symbol \(\cap\) is a cap that only covers the shared overlap. Picture the shape and you will not mix them up.

Set-builder and Venn basics

Listing every element gets clumsy for big sets, so set-builder notation describes a set by a rule instead. Read the colon as "such that":

\[ \{\, x : x \text{ is even and } x < 10 \,\} = \{2, 4, 6, 8\} \]

A Venn diagram pictures all this. The rectangle is the universal set \(U\); each set is a circle inside it. Overlapping circles share the region that represents the intersection \(A \cap B\), everything inside either circle is the union \(A \cup B\), and anything left in the rectangle but outside a circle is that set's complement.

A worked example

Worked example

Given \(U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\), \(A = \{2, 4, 6, 8\}\) and \(B = \{1, 2, 3, 4, 5\}\), find \(A \cap B\), \(A \cup B\) and \(A'\).

\(A \cap B\): which elements are in both? Scan \(A\): 2 is in \(B\), 4 is in \(B\), 6 and 8 are not. So \(A \cap B = \{2, 4\}\).

\(A \cup B\): gather every element of either set, once each: \(\{1, 2, 3, 4, 5, 6, 8\}\).

\(A'\): everything in \(U\) that is not in \(A\). Remove \(2, 4, 6, 8\) from \(U\): \(\{1, 3, 5, 7, 9, 10\}\).

\(A \cap B = \{2, 4\}\), \(A \cup B = \{1, 2, 3, 4, 5, 6, 8\}\), \(A' = \{1, 3, 5, 7, 9, 10\}\)

Notice the intersection \(\{2, 4\}\) is smaller than either set, while the union \(\{1, 2, 3, 4, 5, 6, 8\}\) is at least as big as both. That is always the case: intersection narrows down, union builds up.

Where this is assessed

Applying the operations correctly is Criterion A knowledge, and using the symbols \(\cap\), \(\cup\), \(\subseteq\) and \(A'\) precisely with clean set notation is Criterion C communicating. Venn diagrams are a favourite way to show working.

Check yourself

1. \(A = \{1, 2, 3, 4\}\) and \(B = \{3, 4, 5, 6\}\). Find \(A \cap B\). +

The elements in both sets are 3 and 4, so \(A \cap B = \mathbf{\{3, 4\}}\).

2. For the same \(A\) and \(B\), find \(A \cup B\). +

Gather every element once: \(A \cup B = \mathbf{\{1, 2, 3, 4, 5, 6\}}\).

3. If \(U = \{1, 2, 3, 4, 5, 6\}\) and \(A = \{2, 4, 6\}\), find \(A'\). +

The complement is everything in \(U\) not in \(A\), which is \(A' = \mathbf{\{1, 3, 5\}}\).

Part of the Extended Maths library. Spotted an error or want a topic added? That feedback makes the notes better.