Sets · Lesson 5 of 6
Complements and Their Properties
“Find what is missing from a set within a stated universe and reason through complement laws and De Morgan’s laws.”
• Find complements relative to a stated universal set. • Connect complement notation, set difference, and shaded regions. • Explain complement laws and double complementation through membership. • Use both De Morgan laws and justify them in words. • Evaluate complement expressions without losing parentheses or domain restrictions.
A complement needs a universe
Suppose A contains the odd numbers from 1 to 9. Asking for the numbers “not in A” is incomplete: do we mean numbers from 1 to 10, all natural numbers, or all real numbers? A complement resolves that ambiguity by looking only within a stated universal set.
If A ⊂ U, the complement A′ consists of all elements of U that do not belong to A. A′ is read “A complement” or “A prime” in this context. The universe U must remain fixed during the calculation.
The condition x ∈ U is essential. A number outside the universe is not part of the complement merely because it is absent from A. This is why A must itself be a subset of U: the complement is defined within the common setting of the discussion.
Problem
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′.
- 1.Start with U, which gives the complete list of possible members.
- 2.Remove 1, 3, 5, 7, and 9 because they belong to A.
- 3.The remaining members are 2, 4, 6, 8, and 10. Thus A′ = {2, 4, 6, 8, 10}.
- 4.Zero and 12 do not enter the answer: although neither belongs to A, neither belongs to this U.
Problem
Keep A = {1, 3}. Compare its complement in U₁ = {1, 2, 3, 4} and U₂ = {1, 2, 3, 4, 5, 6}.
- 1.Within U₁, removing 1 and 3 leaves {2, 4}.
- 2.Within U₂, removing the same two members leaves {2, 4, 5, 6}.
- 3.The set A did not change, but the collection of allowed outsiders did. The two complements differ.
- 4.Always state the universe when a complement could be ambiguous; A′ is not an absolute list attached permanently to A.
Problem
Let U be all prime numbers and A the primes that are not divisors of 42. Find A′.
- 1.Factor 42 = 2 × 3 × 7. Its prime divisors are exactly 2, 3, and 7.
- 2.A excludes these three primes and includes the other primes.
- 3.The members of U not in A are therefore exactly {2, 3, 7}.
- 4.Thus A′ is finite even though U and A are infinite. The prime-domain restriction excludes composite divisors such as 6 and 14.
For a universe of students in a specified class, the complement of a named subgroup means all students in that universe outside the subgroup. In the source’s example, a class divided into girls and boys gives the boys as the complement of the girls. The general rule remains “all class members outside A”; the answer must follow the categories actually specified by the situation.
Seeing a complement in a diagram
The rectangle fixes the universe. If the circle represents A, its complement is everything inside the rectangle outside the circle. Shading the surrounding region makes the two simultaneous conditions visible: in U, and not in A.
If a diagram shows another set B overlapping A, A′ includes the B-only region but excludes the overlap because the overlap is still in A. A complement is determined by the chosen set and U; the presence of extra curves does not change its rule.
The basic complement laws
A and its complement divide the universe into two parts. Every allowed element belongs to exactly one of those parts. From that observation we can derive several useful laws instead of memorising them as disconnected formulas.
The union contains all elements of U because each one passes one of the two tests. It contains nothing outside U, since A and A′ are both subsets of U. Those two facts establish equality with U.
This makes A and A′ disjoint. It does not mean that either set must be empty; both can have many members while sharing none. In the odd/even example, both contain five elements, but their intersection is empty.
A member of U outside A′ is precisely a member that was not removed from A. More directly, for an allowed x, “not outside A” means “in A”. Taking a complement twice returns the starting set. If the universe changes between the two operations, this reasoning no longer describes the same operation.
Removing no elements from U leaves U. Removing all members of U leaves none. These statements also explain why the complement of an empty collection is not automatically an infinite collection: its size is determined by the universe.
De Morgan’s laws
Now consider a complement taken after combining two sets. The words “outside the union” and “outside the intersection” impose different requirements. Translating them carefully shows why the operation switches between union and intersection.
To be outside A ∪ B, an element of U must be outside A and outside B. If it were in even one of them, it would enter the union. Thus it passes both complement tests and belongs to A′ ∩ B′. Conversely, an element outside both sets belongs to neither part of their union, so it is outside the union. This proves both directions of the first law.
To be outside A ∩ B, an allowed element must fail at least one of the two membership tests. It may be outside A, outside B, or outside both. Those possibilities form the union A′ ∪ B′. Conversely, failing either membership test prevents membership in the intersection. This proves the second law.
Notice the difference in the negations. “Not in either set” requires being outside both. “Not in both sets” permits being in one but not the other, as well as being in neither. The first law leaves only the region outside both circles; the second law leaves everything except their overlap.
Problem
Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3}, and B = {3, 4, 5}. Compute both De Morgan results.
- 1.Removing A from U gives A′ = {1, 4, 5, 6}; removing B gives B′ = {1, 2, 6}.
- 2.A ∪ B = {2, 3, 4, 5}, so (A ∪ B)′ = {1, 6}.
- 3.The intersection A′ ∩ B′ is also {1, 6}, verifying the first law for these sets.
- 4.A ∩ B = {3}, so (A ∩ B)′ = {1, 2, 4, 5, 6}.
- 5.The union A′ ∪ B′ is {1, 2, 4, 5, 6}, verifying the second law for these sets.
- 6.These computations are checks on this case. The general membership arguments above establish the laws for arbitrary subsets of one universe.
A complement does not pass through parentheses without changing the operation. In general (A ∪ B)′ is not A′ ∪ B′. Also, (A − B)′ is not A′ − B′. First evaluate the parenthesised set or apply a correctly justified identity.
Complements described by rules
Finite lists are convenient for checking complements, but the definition also works for infinite sets and geometric objects. Keep the universe visible in the description, then negate the membership condition within that universe. Be careful with words such as “all”, “at least one”, and “and”.
Problem
Let U = N and A = {x : x ∈ N and x is prime}. Describe A′.
- 1.The complement contains natural numbers that are not prime.
- 2.It includes the composite numbers and also 1, because 1 is not prime.
- 3.Thus A′ = {x : x ∈ N and x is not prime}. Calling it only the composite numbers would omit 1.
- 4.Similarly, the complement of positive multiples of 3 consists of natural numbers not divisible by 3, not numbers outside N.
Problem
Let U be all triangles in a plane and A the triangles with at least one angle different from 60°. Describe A′.
- 1.An allowed triangle is outside A exactly when it has no angle different from 60°.
- 2.That means every angle equals 60°.
- 3.Such triangles are equiangular, and in ordinary plane geometry they are equilateral. Thus A′ is the set of equilateral triangles in the plane.
- 4.Negating “at least one angle differs” gives “all angles agree with 60°”, not “at least one angle equals 60°”.
Problem
Let U = {1, 2, 3, 4, 5, 6}, B = {2, 3, 5}, and C = {3, 4}. Find (B − C)′.
- 1.Compute the parenthesised difference first. From B remove 3, giving B − C = {2, 5}.
- 2.Now remove {2, 5} from U.
- 3.The result is (B − C)′ = {1, 3, 4, 6}.
- 4.Notice that 3 returns to the answer: it was excluded from the difference, so it belongs to the complement of that difference.
At the fifth exercise boundary, complements of lists, natural-number families, and Venn regions use the same definition. For an inequality in N, determine which natural numbers satisfy it before complementing. For example, 2x + 1 > 10 means x ≥ 5 for natural x, so its complement in N is {1, 2, 3, 4}.
Check your understanding
Each question assumes complements are taken in the stated universe. Translate the condition into words before choosing. When a combined expression appears, identify the set inside the parentheses before taking its complement.
Quiz
For U = {1, 2, 3, 4, 5} and A = {1, 3}, what is A′?
Keep all members of U absent from A. Zero is outside U and 5 must be included.
Which equals (A ∪ B)′?
Outside the union means outside both sets, giving an intersection of complements.
Which equals (A ∩ B)′?
Outside the intersection means failing at least one membership condition, giving a union of complements.
What is (A′)′ when the universe stays fixed?
Taking the complement twice restores the original membership condition.
In N, which number must belong to the complement of the primes?
1 is natural and not prime. The other listed numbers are prime.
Which region represents A ∩ A′?
No element can be both in A and outside A.
In the triangle example, the complement of “at least one angle differs from 60°” means:
To fail “at least one differs”, none can differ, so all three equal 60°.
Key Takeaways
• A complement contains elements of U outside A; its answer depends on U. • A′ = U − A, and both A and A′ lie within the universe. • A ∪ A′ = U, while A ∩ A′ = ∅. • Double complementation returns A when U stays fixed. • The empty set and universal set are complements of each other. • De Morgan’s laws switch union and intersection when complementing both inputs. • Negate the full condition carefully, especially when it includes “at least one” or several restrictions.