Sets, subsets and set operations

Which cards belong to either collection?

Put six distinct cards numbered 1, 2, 3, 4, 5 and 6 on a table. Mark cards 1, 2 and 4 with A, and cards 2, 3, 4 and 5 with B. Cards 2 and 4 have both marks; card 6 has neither. A collection may overlap another without becoming two copies of the shared cards.

Write U = {1, 2, 3, 4, 5, 6} for the cards under discussion, A = {1, 2, 4} and B = {2, 3, 4, 5}. The braces list a set’s elements. If we want cards belonging to A or B, the answer is {1, 2, 3, 4, 5}. If we require both marks, the answer is {2, 4}.

The circle diagram places each card according to its two memberships, and the rectangle keeps U visible. Its circle sizes do not represent counts. The precise lists, rather than the areas, say which elements the sets contain.

Read the objects first, then the notation

The lesson Symbols, definitions and mathematical language helps you read ∈ and = with their intended meanings. Moving between words, pictures and symbols supports the change from card lists to diagrams. The lesson Examples, explanations and counterexamples prepares you to distinguish a successful example from a general identity.

The interval example later uses Inequalities, intervals and absolute value. It is optional on a first reading: finish the card operations and empty-set distinction, then return after you can read included and excluded interval endpoints.

For the identity proofs, “and” requires both conditions, “or” allows either or both, and “not” reverses the entire stated condition. The logic and direct-proof lessons develop this language further. Here each proof also explains the membership reasoning in words, so begin with a candidate element before following the symbolic equivalences.

An element belongs; a list describes

The statement 2 ∈ A says that the object 2 is an element of A. The statement 3 ∉ A says it is not. Neither statement compares all the elements of two sets. Sets can contain numbers, letters, points or other sets; decide what the objects are before reading a membership claim.

Order and repeated listings do not change a set: {1, 2, 4} = {4, 1, 2} = {1, 2, 2, 4}. Each has exactly the three distinct elements 1, 2 and 4. A list recording arrivals in order, or a collection recording repeated copies, carries information that this ordinary set does not retain.

A condition can describe a set instead of listing it. The notation {x ∈ U : x is even} means the elements x of the specified U satisfying that condition. For our six-card U it gives {2, 4, 6}. State the domain and a definite condition; “interesting cards” does not determine a set until “interesting” has a precise meaning.

For real numbers, {x ∈ ℝ : −1 ≤ x ≤ 2} is the interval [−1, 2]. It contains every real number between its endpoints, not only the integers −1, 0, 1 and 2. A few dots in a diagram cannot replace this domain-wide definition.

Containment checks every element of the left set

A ⊆ B means that every element of A belongs to B: whenever x ∈ A, x ∈ B. The set {2, 4} is a subset of our B because both of its elements are in B. Our original A is not a subset of B because 1 ∈ A but 1 ∉ B. One such element is enough to refute containment.

Membership and containment ask different questions. In A = {1, 2, 4}, 2 ∈ A and {2} ⊆ A are true, but {2} ∈ A is false: the listed elements are three numbers, not a singleton set. In C = {2, {2}}, both 2 and the set {2} are elements. Braces change the object you are discussing.

Equality is determined by elements: A = B exactly when every object belongs to A if and only if it belongs to B. This is the extensional meaning of set equality. Equivalently, A ⊆ B and B ⊆ A must both hold. Equal counts alone do not suffice: {1, 2} and {1, 3} each have two elements but are different sets.

Containment includes equality, so A ⊆ A always holds. A proper subset A ⊊ B satisfies A ⊆ B and A ≠ B; for example {2, 4} ⊊ {2, 3, 4, 5}. Some sources use ⊂ with different conventions. We use ⊆ for possibly equal containment and ⊊ for strict containment.

Nothing inside is different from an empty set inside

The empty set ∅ = {} has no elements. The set {∅} contains one element, the empty set itself. A box with nothing inside illustrates ∅; an outer box containing one empty inner box illustrates {∅}. These sets are unequal because one has no elements and the other has one.

Thus ∅ ∉ ∅, but ∅ ∈ {∅}. On the other hand, ∅ ⊆ A for every set A, including A = ∅. To refute this subset claim, we would need an element of ∅ absent from A. There is no element of ∅, so no counterexample exists. This containment claim does not say that ∅ is an element of every set.

In particular, ∅ ⊆ {∅} is true, whereas {∅} ⊆ ∅ is false: the element ∅ of the singleton is not an element of ∅. The singleton is a genuine set, not another notation for emptiness. Read each layer of braces before deciding membership or containment.

Inclusive “or” collects; “and” keeps the overlap

The union A ∪ B contains an element x exactly when x ∈ A or x ∈ B, allowing both. For the card example, A ∪ B = {1, 2, 3, 4, 5}. Shared cards 2 and 4 appear once each. Union does not discard the overlap and does not record a second copy of a shared element.

The intersection A ∩ B contains x exactly when x ∈ A and x ∈ B. Both conditions must hold, so A ∩ B = {2, 4}. Sets are disjoint when their intersection is empty; for example {1, 2} ∩ {4, 5} = ∅. An empty intersection does not require either input set to be empty.

These definitions apply to finite and infinite sets. If A is a subset of B, every member of A already belongs to B, so A ∪ B = B and A ∩ B = A. Switching the two input sets changes neither the inclusive-or rule nor the both-memberships rule.

Removing B from A is a directional instruction

The difference A ∖ B contains x exactly when x ∈ A and x ∉ B. Begin in A, then remove shared elements. Our cards give A ∖ B = {1}, whereas B ∖ A = {3, 5}. Reversing the inputs generally changes the answer; difference is not a commutative operation.

The symmetric difference A △ B keeps elements belonging to exactly one of A and B. Thus A △ B = (A ∖ B) ∪ (B ∖ A) = {1, 3, 5}. The two exclusive sides remain, but the overlap {2, 4} does not. An equivalent formula is A △ B = (A ∪ B) ∖ (A ∩ B).

For equal input sets, each element belongs to both or neither, so A ∖ A = ∅ and A △ A = ∅. For disjoint inputs, no element is shared, so symmetric difference equals union. These conclusions follow from the membership rules, rather than from an assumption about circle areas.

“Not in A” still needs a place to look

Choose a universe U containing the sets under discussion. For A ⊆ U, its complement relative to U is Aᶜ = U ∖ A: the elements of U that do not belong to A. Here the superscript c always refers to the same stated U. It does not mean every imaginable object outside A.

With U₁ = {1, 2, 3, 4, 5, 6} and A = {1, 2, 4}, the complement is {3, 5, 6}. Change to U₂ = {1, 2, 3, 4, 5, 6, 7}, retaining the same A, and its complement becomes {3, 5, 6, 7}. The object 7 is outside U₁, so it was never a candidate for the first complement.

A difference A ∖ B is defined from its two input sets without choosing an ambient universe. If both inputs are subsets of one fixed U, we may rewrite it as A ∩ Bᶜ. That rewrite uses the same U for the complement and relies on A ⊆ U. A common universe is also required when combining two complements in a law.

Test one candidate with the exact rule

For a candidate x ∈ U, let p mean x ∈ A and q mean x ∈ B. Its membership in A ∪ B is p or q; membership in A ∩ B is p and q; membership in A ∖ B is p and not q. The four possibilities for p and q cover every element of U.

Keep the kinds of objects separate. A ∪ B is a set, while the statement x ∈ A or x ∈ B has a truth value. Writing A “and” B does not define a logical statement about their elements. Translate through membership before applying a logical rule.

The adjacent explorer can change the overlap, make one set empty, or make the inputs equal while keeping the six-card U fixed. Predict a result, then inspect every row. That verifies the whole declared finite universe, but a general identity still needs a reason that covers an arbitrary candidate.

Outside the whole union means outside both collections

Let A and B be subsets of a fixed U. De Morgan’s first law is (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ. In the card example the left side contains only 6, and so does the right side: card 6 fails membership in both A and B.

For a general proof, take any x ∈ U. It belongs to (A ∪ B)ᶜ exactly when the statement “x ∈ A or x ∈ B” is false. This means x is not in A and is not in B. Equivalently, x ∈ Aᶜ and x ∈ Bᶜ, so x ∈ Aᶜ ∩ Bᶜ. Every step works in both directions, giving the same elements and therefore equal sets.

The whole union is negated, so the inclusive “or” changes to “and” between the two failed memberships. Card 1 is not in B but is in A, so it is not outside their union. This single card exposes the incorrect replacement Aᶜ ∪ Bᶜ.

Failing a shared membership needs only one failure

De Morgan’s other law is (A ∩ B)ᶜ = Aᶜ ∪ Bᶜ, again using one fixed U. For our cards, both sides are {1, 3, 5, 6}. Cards 1, 3 and 5 belong to one input set; they still lie outside the intersection because they do not belong to both.

Take any x ∈ U. It is outside A ∩ B exactly when the two memberships do not both hold. Thus x ∉ A or x ∉ B, allowing both failures. This is exactly x ∈ Aᶜ ∪ Bᶜ. The equivalence holds for every candidate, establishing set equality.

Compare the two laws before memorising their symbols: outside a union requires neither membership, while outside an intersection requires at least one missing membership. Negating the entire condition changes the connector. The four-pattern tables beside the diagrams make these exhaustive possibilities explicit.

Set laws are short versions of element arguments

A set identity states equality for all sets satisfying its conditions. To justify it, compare membership on its two sides for an arbitrary element. For A, B and C ⊆ U, distribution gives A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C): an element must be in A and in at least one of B, C, which is exactly membership in at least one of the two intersections.

The dual distribution law is A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). If x ∈ A, both sides contain it. If x ∉ A, both sides require x ∈ B and x ∈ C. These two cases cover every candidate. Neither law makes difference commutative or allows complements from different universes.

For sets contained in U, complementing twice gives (Aᶜ)ᶜ = A: for x ∈ U, not being outside A is the same as being in A. Also A ∪ Aᶜ = U and A ∩ Aᶜ = ∅, because every element of U has exactly one of the two membership alternatives. These equalities retain the requirement A ⊆ U.

Two containments prove equality; one element can disprove it

Another way to prove X = Y is to show X ⊆ Y and Y ⊆ X. For X = (A ∩ B) ∪ (A ∖ B), every member comes from A, so X ⊆ A. Conversely, take any x ∈ A. Either x ∈ B, placing it in A ∩ B, or x ∉ B, placing it in A ∖ B. In either case x ∈ X, so A ⊆ X and X = A.

To disprove a proposed identity, specify input sets and an element belonging to exactly one side. With A = {1} and B = {2}, A ∖ B = {1} and B ∖ A = {2}; the element 1 separates them. This refutes commutativity of difference, without refuting any union or intersection law.

For a finished calculation, read the operation and the universe again, then check a shared member, an exclusive member and any boundary or empty case that occurs. For a proof, replace those examples with an arbitrary-member argument. A diagram helps you find the reason; its finite drawing is not the whole argument for every possible set.

Worked examples

Try it yourself

1. Let D = {1, {1}, ∅}. Decide 1 ∈ D, {1} ∈ D, {1} ⊆ D, ∅ ∈ D and ∅ ⊆ D. How many elements does D have?

   Show the answer and reasoning

   All five claims are true. D has three distinct elements: the number 1, the set {1}, and the empty set.

   The first, second and fourth claims inspect the list. For {1} ⊆ D, the only required membership is 1 ∈ D. Empty-set containment is true because there is no empty-set element that could fail membership in D; it is a separate reason from ∅ ∈ D.

2. Which of ∅ ∈ ∅, ∅ ∈ {∅}, ∅ ⊆ ∅, {∅} ⊆ ∅ and ∅ = {∅} are true?

   Show the answer and reasoning

   Only ∅ ∈ {∅} and ∅ ⊆ ∅ are true; the other three claims are false.

   An empty set contains no element, while its singleton contains the element ∅. The empty set is a subset of itself. The singleton is not contained in ∅ because its sole element is missing there, and sets with zero and one element cannot be equal.

3. With U = {0,1,2,3,4}, A = {0,2,4} and B = {1,2}, find A ∩ B, A ∖ B, A △ B and (A ∪ B)ᶜ.

   Show the answer and reasoning

   A ∩ B = {2}; A ∖ B = {0,4}; A △ B = {0,1,4}; (A ∪ B)ᶜ = {3}.

   Only 2 is shared. The A-only elements are 0 and 4, and the B-only element is 1. Union is {0,1,2,4}, so the only element of the specified U outside it is 3. The overlap is excluded from symmetric difference.

4. Let A = {0,2}. Find its complement first in U = {0,1,2,3}, then in V = {−1,0,1,2,3}. Why are the answers different?

   Show the answer and reasoning

   U ∖ A = {1,3}, while V ∖ A = {−1,1,3}; the second universe includes the additional eligible element −1.

   Complement is a difference from the chosen universe. A stays fixed, but −1 is a member of V and not of A. It was outside U, so it could not appear in U’s relative complement.

5. An element x ∈ U belongs to A but not to B. Does it belong to (A ∪ B)ᶜ and to (A ∩ B)ᶜ? Explain both decisions.

   Show the answer and reasoning

   It does not belong to (A ∪ B)ᶜ, but it does belong to (A ∩ B)ᶜ.

   Membership in A already puts x in the union, so its union-complement membership fails. It cannot lie in the intersection because B membership fails, so it belongs to the intersection’s complement within U. The two negations concern different entire conditions.

6. Assume A ⊆ B ⊆ U. Simplify A ∪ B, A ∩ B, A ∖ B and A △ B, retaining a reason for the last answer.

   Show the answer and reasoning

   A ∪ B = B; A ∩ B = A; A ∖ B = ∅; A △ B = B ∖ A.

   Every element of A already belongs to B, so the A-only region is empty. Union contains all of B, intersection retains A, and no member survives A ∖ B. Exactly-one membership can therefore occur only on the B-only side, which is B ∖ A.

7. For real sets A = [0,1] and B = (1,2], find A ∩ B and A ∪ B. Is either input empty?

   Show the answer and reasoning

   A ∩ B = ∅ and A ∪ B = [0,2]; neither input set is empty.

   Every point of B is strictly greater than 1, while every point of A is at most 1, so no point is shared. Their union has no gap: A includes 1 and all points down to 0, while B supplies every point above 1 through 2. Disjointness does not mean an input is empty.

8. Prove A ∩ (A ∪ B) = A by showing both containments. Do not argue only from a drawing.

   Show the answer and reasoning

   Every member of A ∩ (A ∪ B) belongs to A. Conversely, every member of A belongs to A ∪ B as well, so it belongs to the intersection. Both containments give equality.

   Choose an arbitrary member for each direction. The first implication is part of the definition of intersection. The second uses the inclusive union rule followed by the intersection rule. No assumption about whether A or B is finite, nonempty, equal or disjoint is needed.

What to carry forward

• Membership asks whether one object is an element. A subset claim checks every element of one set; equality requires the same elements on both sides.
• A set ignores listing order and repetitions. The empty set has no elements, while {∅} has one empty-set element.
• Union uses inclusive “or”, intersection uses “and”, difference uses a directed exclusion, and symmetric difference keeps exactly one membership.
• A complement always refers to a specified universe. Keep that same universe when combining complements or complementing twice.
• De Morgan’s laws negate a whole membership condition and change its connector. Set identities follow by comparing an arbitrary element’s memberships, or by proving both containments.
• An exact finite table verifies its whole stated finite universe. A general claim about arbitrary sets requires the membership argument and its conditions.