Complement of a set

Set operations like union and intersection combine two sets you already have. Complement is different — it needs only one set, but it needs a second, silent ingredient: the universal set $\mathcal{U}$, the full collection of elements under discussion in a given problem (all integers from 1 to 20, all vowels of the alphabet, all pupils in a class, and so on). Without an agreed $\mathcal{U}$, "everything not in $A$" is meaningless — it could run on forever. Once $\mathcal{U}$ is fixed, the complement of $A \subseteq \mathcal{U}$, written $A'$ (also seen as $A^c$ or $\overline{A}$), collects exactly the leftover elements:

where: $x$ is a typical element being tested, $\mathcal{U}$ is the stated universal set, and $A$ is the subset whose complement is being found. Read the definition as a two-part filter — an element earns a place in $A'$ only if it passes both tests: it must belong to $\mathcal{U}$, and it must fail to belong to $A$. On a Venn diagram this is the easiest complement to draw of any set operation. $\mathcal{U}$ is the outer rectangle; $A$ is a circle inside it. The complement $A'$ is every point of the rectangle that is not inside the circle — no shading calculation needed beyond "outside the loop, inside the box."

Four properties follow directly from the definition and are worth knowing by heart, since ICSE questions often ask you to "verify" them on a specific example rather than just quote them:

• $A \cup A' = \mathcal{U}$ — a set and its complement together rebuild the whole universal set.
• $A \cap A' = \varnothing$ — a set and its complement never overlap; nothing can be both "in $A$" and "not in $A$" at once.
• $(A')' = A$ — the complement of the complement returns the original set.
• $\mathcal{U}' = \varnothing$ and $\varnothing' = \mathcal{U}$ — the two extreme cases swap places.

Notice that the complement is not a fixed property of $A$ alone — it depends entirely on which $\mathcal{U}$ was chosen. The same set $A = \{2, 4, 6\}$ has a small complement if $\mathcal{U} = \{1,2,\dots,8\}$ but a much larger one if $\mathcal{U}$ is extended further. Always pin down $\mathcal{U}$ before touching $A'$.

The reasoning is essentially a sorting argument, and it is worth walking through carefully because it is exactly what the counting formula below rests on. Take the universal set $\mathcal{U}$ and imagine sorting every single one of its elements into one of two labelled trays: a tray marked "in $A$" and a tray marked "not in $A$." Every element of $\mathcal{U}$ goes into exactly one tray — none can be left out (it must be tested), and none can go into both (the test "is it in $A$" has only a yes or a no answer for any given element). The "in $A$" tray, by definition, contains exactly the set $A$. The "not in $A$" tray, by definition, contains exactly the set $A'$. Because every element of $\mathcal{U}$ landed in one tray or the other, and no element landed in both, the two trays between them contain all of $\mathcal{U}$, with no double-counting. That is precisely the statement $A \cup A' = \mathcal{U}$ and $A \cap A' = \varnothing$ combined — mathematicians call this a partition of $\mathcal{U}$ into two disjoint pieces. Once you accept that $A$ and $A'$ partition $\mathcal{U}$ with no overlap and nothing left over, counting becomes simple addition: the number of elements in the whole box equals the number in the first tray plus the number in the second tray. For a finite universal set:

where: $n(A)$ is the number of elements in $A$, $n(A')$ is the number of elements in $A'$, and $n(\mathcal{U})$ is the total number of elements in the universal set. This single line is the fastest way to check any complement answer — if the two counts you found do not add up to $n(\mathcal{U})$, something was miscounted.

Let the universal set be $\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and let $A = \{2, 4, 6, 8, 10\}$, the even numbers in $\mathcal{U}$. Find $A'$, and confirm $n(A) + n(A') = n(\mathcal{U})$.

Step 1. Write out every element of $\mathcal{U}$ in order: $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ — ten elements in total, so $n(\mathcal{U}) = 10$.

Step 2. Go through the list and cross out every element that belongs to $A = \{2, 4, 6, 8, 10\}$. What remains, uncrossed, is $A'$:

This is exactly the set of odd numbers in $\mathcal{U}$, which makes sense — within the whole numbers, "not even" means "odd."

Step 3. Count both sets: $n(A) = 5$ and $n(A') = 5$. Add them: $5 + 5 = 10 = n(\mathcal{U})$, matching the counting formula exactly. ✓

Step 4. Verify the two key properties directly. $A \cup A' = \{1,2,3,4,5,6,7,8,9,10\} = \mathcal{U}$ ✓, and $A \cap A' = \varnothing$ since no number in the list is both even and odd ✓.

Final answer: $A' = \{1, 3, 5, 7, 9\}$, with $n(A') = 5$.

Competitive stretch

Now take $\mathcal{U} = \{x : x \text{ is a letter in the word MATHEMATICS}\}$, so the distinct letters give $\mathcal{U} = \{M, A, T, H, E, I, C, S\}$, and $n(\mathcal{U}) = 8$. Let $B = \{M, A, T\}$, so $n(B) = 3$. Removing $M$, $A$, $T$ from $\mathcal{U}$ leaves $B' = \{H, E, I, C, S\}$, so $n(B') = 5$. Check: $3 + 5 = 8 = n(\mathcal{U})$ ✓. This shows the counting rule works just as well for letters as for numbers — it never depends on what kind of objects the elements are, only on how many there are.

• Forgetting to fix $\mathcal{U}$ before finding $A'$ — the same set $A$ gives a completely different complement under a different universal set, so state or confirm $\mathcal{U}$ first, every time.
• Writing elements in $A'$ that are not even in $\mathcal{U}$ — the complement can only ever contain elements drawn from the stated universal set, never anything from "outside" it.
• Believing $A$ and $A'$ can share an element — by the very definition of complement this is impossible; if it seems to happen, the universal set or one of the two lists has been copied wrong.
• Skipping the cardinality check — always verify $n(A) + n(A') = n(\mathcal{U})$ after listing $A'$; a mismatch is the fastest way to catch a missed or duplicated element.

Ask the tutor: Want more practice picking the right universal set or finding complements for sets described in words instead of lists — ask the AI tutor for a few guided problems on this topic.