Sets · Lesson 3 of 6
Subsets, Intervals, and Universal Sets
“Understand containment, distinguish it from membership, and describe real-number intervals within a suitable universe.”
• Test subset and proper-subset relationships using the every-element rule. • Distinguish membership from containment, including sets that contain sets. • Explain relationships among natural, integer, rational, irrational, and real numbers. • Translate intervals into inequalities, set-builder form, and number lines. • List subsets systematically and choose a valid universal set.
What it means to be a subset
Every student in a particular class is also a student in that school. The class collection therefore fits entirely inside the school collection. Containment expresses this relationship between two sets: it checks all members of the smaller collection, rather than asking whether one object appears in a list.
A is a subset of B if every element of A is an element of B. In the convention used here, A ⊂ B allows A = B. If even one element of A is absent from B, then A ⊄ B.
The symbol ⇒ means “implies”. In the subset test it says: whenever an object passes the membership test for A, it must also pass the test for B. The condition does not require every member of B to belong to A. For example, {1, 3} ⊂ {1, 3, 5}, although 5 belongs only to the larger set.
Some mathematical writing uses ⊆ for containment allowing equality and reserves ⊂ for strict containment. This chapter follows its source: ⊂ allows equality. A proper subset is stated explicitly by also requiring that the two sets be unequal.
Every set is a subset of itself, because each of its elements is certainly one of its elements. The empty set is a subset of every set too. To fail the every-element test, we would need an element of ∅ missing from the other set. Since ∅ has no elements, such a failure cannot occur. This does not assert that ∅ is a member of every set.
A is a proper subset of B when A ⊂ B and A ≠ B. B is then a superset of A. A singleton is a set containing exactly one element.
If A ⊂ B and B ⊂ A, there can be no member present in just one of the sets, so A = B. Conversely, equal sets satisfy both containments. The symbol ⇔, read “if and only if”, expresses that both directions hold. Equality and mutual containment describe the same relationship.
Problem
Let A = {1, 3}, B = {1, 5, 9}, and C = {1, 3, 5, 7, 9}. Test ∅ ⊂ B, A ⊂ B, A ⊂ C, and B ⊂ C.
- 1.∅ ⊂ B is true because there is no member of ∅ that could violate the test.
- 2.A ⊂ B is false: 3 belongs to A but not to B. This one counterexample is decisive.
- 3.A ⊂ C is true: both 1 and 3 belong to C.
- 4.B ⊂ C is true: 1, 5, and 9 all belong to C.
- 5.Each successful test concerns all elements of the left set; a shared member alone proves nothing about containment.
Problem
Let V = {a, e, i, o, u} and B = {a, b, c, d}. Is either a subset of the other?
- 1.The letter e is in V but not in B, so V ⊄ B.
- 2.The letter b is in B but not in V, so B ⊄ V.
- 3.The shared member a does not change either conclusion. The failure must be checked independently in each direction.
Membership is different from containment
A set can itself be an object collected inside another set. This makes braces especially important. Membership asks whether the whole object is a listed member; containment opens the left-hand set and tests each of its members separately.
Let S = {1, 2, {3, 4}, 5}. Its four members are 1, 2, the set {3, 4}, and 5. The numbers 3 and 4 are inside that nested member, but they are not individually members of S. The outer braces define the level at which the membership test is taking place.
Problem
For S = {1, 2, {3, 4}, 5}, compare {3, 4} ∈ S, {3, 4} ⊂ S, and {{3, 4}} ⊂ S.
- 1.The whole object {3, 4} is listed in S, so {3, 4} ∈ S is true.
- 2.For {3, 4} ⊂ S we would need both 3 ∈ S and 4 ∈ S. Neither is individually listed, so that containment is false.
- 3.{{3, 4}} has just one member: the set {3, 4}. That member is in S, so {{3, 4}} ⊂ S is true.
- 4.The extra pair of braces changes what is being tested. It changes the object, not merely its appearance.
| Statement for S = {1, 2, {3, 4}, 5} | Verdict | Reason |
|---|---|---|
| 1 ∈ S | True | 1 is a listed member. |
| {1, 2, 5} ⊂ S | True | All three numbers occur as members. |
| {1, 2, 5} ∈ S | False | That whole set is not a listed member. |
| ∅ ⊂ S | True | The empty set is a subset of every set. |
| ∅ ∈ S | False | The empty set is not listed. |
| {∅} ⊂ S | False | Its only member ∅ is absent from S. |
Problem
Let A = {1}, B = {{1}, 2}, and C = {{1}, 2, 3}. Does A ∈ B and B ⊂ C imply A ⊂ C?
- 1.A ∈ B is true because {1} is a whole member of B.
- 2.B ⊂ C is true because each of its two members, {1} and 2, is a member of C.
- 3.These statements ensure A ∈ C, but that is a membership conclusion.
- 4.A ⊂ C would require 1 ∈ C. The number 1 is not listed individually in C, so A ⊄ C.
- 5.Membership and containment cannot be substituted for each other in a chain of reasoning.
For numerical elements, write 1 ∈ S rather than 1 ⊂ S: the number 1 is being treated as a number, not a set. Also, from x ∈ A and A ∈ B you cannot in general conclude x ∈ B. Nested membership does not automatically flatten into direct membership.
Subsets of the real numbers
The number sets introduced earlier fit into a containment structure. Understanding that structure lets us select an appropriate domain before solving a problem. It also explains why a solution can belong to one number set and fail to belong to another.
Every natural number is an integer. Every integer k is rational because k = k/1. Every rational number is real. Thus N ⊂ Z ⊂ Q ⊂ R. Each of these containments is proper: 0 is an integer but not natural in our convention, 1/2 is rational but not an integer, and √2 is real but not rational.
For example, −5 = −5/1 is rational, and 3½ = 7/2 is rational. A number need not be written as a fraction to be rational; it must be expressible as such a fraction. Irrational numbers are real numbers that cannot be expressed in that way.
In this source the irrational numbers are denoted by T: T = {x : x ∈ R and x ∉ Q}. Examples include √2, √5, and π.
T ⊂ R, but N ⊄ T: for instance, 1 is rational, so it is not in T. Rational and irrational real numbers have no common member, and together account for every real number. Keep the negative sign and the domain when classifying a number; −√2 is irrational just as √2 is.
Intervals as sets of real numbers
An interval describes a continuous portion of the real number line. Unlike a list of integers between two bounds, it includes every real number allowed by the inequalities. Endpoint notation tells us whether the boundary numbers themselves are included.
Let a and b be real numbers with a < b. The open interval (a, b) contains all real x with a < x < b. The closed interval [a, b] contains all real x with a ≤ x ≤ b. The two half-open intervals include just one endpoint. Parentheses exclude an endpoint; square brackets include it.
| Interval | Condition, with x ∈ R | Included endpoints |
|---|---|---|
| (a, b) | a < x < b | Neither |
| [a, b] | a ≤ x ≤ b | Both |
| [a, b) | a ≤ x < b | a only |
| (a, b] | a < x ≤ b | b only |
An interval with different endpoints contains infinitely many real numbers even when its length is small. Between two different real numbers lies their midpoint; repeating that idea supplies further distinct points. Do not write (1, 2) as an empty set just because it contains no natural numbers. It contains real numbers such as 1.1, 1.5, and 1.99.
Problem
Express {x : x ∈ R and −5 < x ≤ 7} as an interval, and express [−3, 5) in set-builder form.
- 1.For the first set, −5 is excluded and 7 is included. Use a parenthesis on the left and a square bracket on the right: (−5, 7].
- 2.The length is 7 − (−5) = 12.
- 3.For [−3, 5), include −3 and exclude 5. The set-builder description is {x : x ∈ R and −3 ≤ x < 5}.
- 4.Its length is 5 − (−3) = 8. The domain R ensures we are describing an interval, not just selected integers.
Unbounded intervals extend indefinitely. [0, ∞) describes the non-negative real numbers, (−∞, 0) the negative real numbers, and (−∞, ∞) all real numbers. Infinity is not a real-number endpoint to be included, so always use a parenthesis next to ∞ or −∞.
Problem
If A = (−3, 5) and B = [−7, 9], is A ⊂ B?
- 1.Any real number greater than −3 and less than 5 is also at least −7 and at most 9.
- 2.Thus every member of A belongs to B, so A ⊂ B.
- 3.The sets are unequal: −7 belongs to B but not to A. Hence A is a proper subset of B.
Listing subsets and choosing a universe
For a small finite set, listing its subsets makes the every-element rule concrete. A universal set has a different role: it fixes the surrounding collection for a whole discussion. Both ideas require you to keep track of which objects belong directly to which set.
Problem
List all subsets of {a, b}.
- 1.Group the subsets by their number of elements to avoid omissions.
- 2.With no elements there is ∅. With one element there are {a} and {b}. With two elements there is {a, b}.
- 3.The complete list is ∅, {a}, {b}, {a, b}. The empty subset and the whole set must both be included.
- 4.For ∅ itself the only subset is ∅; for {a} the subsets are ∅ and {a}. No unlisted object may be inserted.
A universal set U is the basic collection relevant to a particular discussion. Every set under consideration in that discussion must be a subset of U.
For right triangles or isosceles triangles in a plane, all triangles in that plane form a suitable universe. For a discussion involving integers, Q or R can serve as a universe because each contains every integer. More than one choice can be valid, but the choice must include all relevant members.
Problem
Let A = {1, 3, 5}, B = {2, 4, 6}, and C = {0, 2, 4, 6, 8}. Can U = {0, 1, 2, 3, 4, 5, 6} serve for all three?
- 1.U contains every member of A and B.
- 2.It does not contain 8, which belongs to C. Thus C is not a subset of U.
- 3.This U cannot serve for all three sets. Adding 8 would repair the omission; {0, 1, 2, …, 10} would also work.
- 4.A universal set may contain extra members. It may not omit a member of any set being discussed.
“Universal” means universal for the stated context. It does not mean a set of everything whatsoever. Also, the universal set is not determined uniquely by one subset; state which universe you are using before a later complement calculation.
At the third exercise boundary, the main tasks are containment, nested membership, complete subset lists, interval conversions, and choosing a universe. Use the every-element test for containment and the outermost member list for membership. For intervals, inspect the domain and both endpoints separately.
Check your understanding
These questions distinguish related ideas whose symbols are easy to confuse. Read each expression as a sentence first, then apply the relevant test. For nested sets, identify the whole members before reasoning about their contents.
Quiz
Under this chapter’s convention, which statement holds for every set A?
Containment allows equality, so every set is a subset of itself. Proper containment additionally requires inequality.
For S = {1, {2, 3}}, which statement is true?
The singleton {{2, 3}} has the one member {2, 3}, which is listed in S. The numbers 2 and 3 are not listed individually.
Which interval represents x ∈ R with −2 ≤ x < 4?
Include −2 with a square bracket and exclude 4 with a parenthesis.
Which number proves that Q is a proper subset of R?
√2 belongs to R but not to Q. The other numbers are rational.
Which can be a universal set for {0, 2} and {1, 3}?
It contains every member of both sets. The extra 4 is permitted.
How many subsets does {a, b} have, counting ∅ and the whole set?
They are ∅, {a}, {b}, and {a, b}.
What is the length of (−3, 5]?
Subtract the left endpoint from the right endpoint: 5 − (−3) = 8. Including one endpoint does not add a unit of length.
Key Takeaways
• Containment tests every element of the left-hand set and allows equality here. • A proper subset also requires the two sets to be unequal. • Membership checks one whole object; containment checks the elements of a set. • N ⊂ Z ⊂ Q ⊂ R, while T is the set of irrational real numbers. • Intervals are subsets of R; brackets include endpoints and parentheses exclude them. • List small-set subsets by size, including ∅ and the whole set. • A universal set must contain all members relevant to the discussion.