Complements and Subsets

Everything not in A

A class has 20 students, the universal set. The 7 who walk to school make the set A. The other 13 do not walk, and they make the complement of A, written A': everything in the universal set that is not in A.

So $n\left(A^{\prime }\right)=n\left(\xi \right)-n\left(A\right)=20-7=13$n(A') = n(ξ) − n(A) = 20 − 7 = 13. On a Venn diagram, A' is the part of the box outside the circle.

A' shaded: the 13 students in the box outside A.

A set and its complement make the whole

Every element of the universal set is either in A or in A', and never in both. In symbols, $A\cup A^{\prime }=\xi$A ∪ A' = ξ and $A\cap A^{\prime }=\varnothing$A ∩ A' = ∅. That is why the two counts add to the whole: 7 + 13 = 20.

The complement of A' is A again, (A')' = A: the students who are not among those who do not walk are the ones who walk.

For a student picked at random, P(A') = 1 − P(A): the chance of not walking is $1-\frac{7}{20}=\frac{13}{20}$1 − 7/20 = 13/20. When the complement is easier to count, count it and subtract from the whole.

The complement depends on the universal set

Let A be the even numbers from 2 to 10. If $\xi$ξ is the whole numbers from 1 to 10, then A' = {1, 3, 5, 7, 9}, 5 elements. If $\xi$ξ is the whole numbers from 1 to 20, then A' also contains 11, 12, 13 and every other number up to 20 except the five in A, 15 elements. A complement is only defined once the universal set is.

Complements of a union and an intersection

Take $\xi$ξ to be the whole numbers from 1 to 10, A = {2, 4, 6, 8} and B = {1, 2, 3, 4, 5}. The numbers in neither set are ${\left(A\cup B\right)}^{\prime }=$(A ∪ B)' = {7, 9, 10}.

Now take the two complements, A' = {1, 3, 5, 7, 9, 10} and B' = {6, 7, 8, 9, 10}. Their intersection, the numbers in both, is $A^{\prime }\cap B^{\prime }=$A' ∩ B' = {7, 9, 10}, the same set. "Not in A or B" means "not in A and not in B": ${\left(A\cup B\right)}^{\prime }=A^{\prime }\cap B^{\prime }$(A ∪ B)' = A' ∩ B'.

In the same way, ${\left(A\cap B\right)}^{\prime }=A^{\prime }\cup B^{\prime }$(A ∩ B)' = A' ∪ B'. Here $A\cap B=$A ∩ B = {2, 4}, so ${\left(A\cap B\right)}^{\prime }$(A ∩ B)' is the other 8 numbers, and $A^{\prime }\cup B^{\prime }=$A' ∪ B' = {1, 3, 5, 6, 7, 8, 9, 10}, 8 numbers.

${\left(A\cup B\right)}^{\prime }$(A ∪ B)' shaded: the numbers outside both circles, 7, 9 and 10. They are exactly the numbers outside A and also outside B.

Subsets

B is a subset of A, written $B\subseteq A$B ⊆ A, when every element of B is also an element of A. Let A = {2, 4, 6, 8, 10}, the even numbers up to 10, and B = {4, 8}, the multiples of 4 up to 10. Both elements of B are in A, so $B\subseteq A$B ⊆ A.

On a Venn diagram the "B only" region is empty: nothing is in B without also being in A. Some diagrams draw B as a circle inside A instead.

B = {4, 8} and A = {2, 4, 6, 8, 10}. The B only region holds nothing, so every element of B is in A.

What a subset does to the union and the intersection

When $B\subseteq A$B ⊆ A, the overlap is the whole of B and the union is the whole of A: $A\cap B=B$A ∩ B = B and $A\cup B=A$A ∪ B = A. Here $A\cap B=$A ∩ B = {4, 8} and $A\cup B=$A ∪ B = {2, 4, 6, 8, 10}.

The direction matters. $B\subseteq A$B ⊆ A does not mean $A\subseteq B$A ⊆ B: 2 is in A and not in B. And the complements go the other way round: $B\subseteq A$B ⊆ A gives $A^{\prime }\subseteq B^{\prime }$A' ⊆ B', since anything outside A is certainly outside B.

Every set, and the empty set

Every set is a subset of itself, $A\subseteq A$A ⊆ A, since every element of A is in A. A subset that is not the whole set is a proper subset, written $B\subset A$B ⊂ A.

The empty set is a subset of every set: it has no elements, so none of them can fail to be in A.

How many subsets a set has

To make a subset of {a, b}, decide for each element whether it is in or out. a can be in or out, and so can b, which gives 2 × 2 = 4 subsets: {a, b}, {a}, {b} and $\varnothing$∅.

For {a, b, c} there are 2 × 2 × 2 = 8: $\varnothing$∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c} and {a, b, c}. A set of n elements has $2^n$2ⁿ subsets, counting the empty set and the set itself.

Each element of {a, b} is in or out. The four paths through the tree are the four subsets, the last of them the empty set.

The usual mistakes

Giving the whole universal set as the complement. With 20 in $\xi$ξ and 7 in A, the complement leaves out the 7: it has 13 members, not 20.

Giving the members of A itself. A' is everyone else, 13 of the 20.

Reading $B\subseteq A$B ⊆ A backward. It says B sits inside A, not the other way round.

Leaving out the empty set, or the set itself, when counting subsets. {a, b, c} has 8 subsets, not 6 or 7.

Eight toppings

In the application below, a pizza is a subset of 8 toppings, so there are $2^8$2⁸ of them, including the plain pizza, which is the empty set. The pizzas with at least two toppings are counted by taking away those with fewer.