Sets · Lesson 2 of 6
Types of Sets and Equality
“Classify sets by their distinct members and decide when different descriptions define exactly the same collection.”
• Identify empty sets by testing both the condition and the domain. • Distinguish empty sets from singletons containing zero or another set. • Classify finite and infinite sets and count distinct members of finite sets. • Decide equality by comparing members rather than order or appearance. • Use a missing or extra element to justify that two sets are unequal.
The empty set
A clear rule can select no objects at all. That does not make the rule meaningless: it tells us that there are no qualifying members. Recognising this possibility matters when equations, inequalities, or several conditions are used to define a set.
A set containing no elements is called the empty set, null set, or void set. We write ∅ or { }. The source also uses a phi-shaped symbol for the empty set.
Consider natural numbers strictly between 1 and 2. No natural number meets that condition, so {x : x ∈ N and 1 < x < 2} = ∅. If the domain were R instead, 1.5 would qualify and the set would not be empty. An empty solution set is therefore a conclusion about the whole description, including its domain.
Another example is the set of even prime numbers greater than 2. Every even number greater than 2 has 2 as a divisor as well as 1 and itself, so it cannot be prime. The only even prime is 2, and the extra condition “greater than 2” excludes it. No member remains.
Problem
Determine {x : x ∈ Q and x² − 2 = 0}.
- 1.Solving x² = 2 over the real numbers gives x = √2 or x = −√2.
- 2.Both candidates are irrational, so neither belongs to Q.
- 3.No rational number satisfies the equation. The specified set is ∅, although the real solution set is not empty.
Problem
Describe {x : x is odd and x² = 4}, with x an integer.
- 1.The equation x² = 4 has the integer solutions −2 and 2.
- 2.Neither is odd. The word “and” requires a member to satisfy both conditions, not just one.
- 3.The set is empty. Solving the equation is only the first step; the parity restriction must also be applied.
There is only one empty set, even though many different rules can describe it. A set of natural numbers less than 0 and a set of odd integers divisible by 2 both have no members. Their descriptions differ, but their resulting sets are equal.
∅, {0}, and {∅} are different. The empty set has no members. {0} has one member, the number zero. {∅} has one member, the empty set itself. Braces around an object create a set containing that object; they do not erase it.
Finite and infinite sets
After identifying the members, ask whether the complete collection has a finite number of distinct objects. A very large collection can still be finite. The issue is whether its members can be counted by a finite whole number, rather than whether you personally know that number.
A set is finite if it is empty or has a finite number of distinct elements. For a finite set S, n(S) denotes its number of elements; n(∅) = 0.
A set is infinite if it is not finite. There is no finite count that exhausts all its distinct members.
The days of a week form a finite set with 7 members. The letters of the English alphabet form a finite set with 26 members. The people living in the world at a specified time also form a finite set, even when their exact count is unknown. Unknown does not mean infinite.
Natural numbers form an infinite set. If you claim that a list ending at some natural number m contains them all, m + 1 supplies another natural number beyond that endpoint. The odd natural numbers are also infinite: after any odd number there is another, two greater. Multiples of a nonzero integer continue without a final multiple. Points on a line, and the lines parallel to a fixed axis, also form infinite collections.
The prime numbers are infinite too. This is a mathematical fact about the whole collection, not a consequence of how large the primes we have found happen to be. A restriction such as “primes less than 99” changes the collection: only finitely many natural numbers lie below that bound, so only finitely many of them can be prime.
Problem
Classify the sets of natural-number solutions to (x − 1)(x − 2) = 0, x² = 4, and 2x − 1 = 0.
- 1.The product equation gives x = 1 or x = 2. Both are natural numbers, so the set is {1, 2}, finite with two members.
- 2.The square equation gives real roots −2 and 2. Only 2 is natural, so the specified set is {2}, finite with one member.
- 3.The linear equation gives x = 1/2, which is not natural. The set is ∅, and the empty set is finite.
- 4.The displayed equation does not determine the count on its own. Solve it and apply the stated domain.
Dots must be interpreted with their endpoints. {1, 2, 3, …, 100} is finite, while {1, 2, 3, …}, explicitly described as all natural numbers, is infinite. The integers can be indicated as {…, −2, −1, 0, 1, 2, …}; the pattern continues in both directions.
The real numbers cannot all be captured by an ordinary roster that successively lists every member. They are not simply a longer version of the natural-number list. In this chapter, use a description or membership rule for R rather than suggesting that a few decimal examples followed by dots list all real numbers.
Equal sets
Two descriptions can look completely different and still identify exactly the same members. Conversely, two sets can have the same number of members and still be unequal. Equality concerns the identity of the members, not their count alone.
A = B means every element of A belongs to B and every element of B belongs to A. Equivalently, A and B contain exactly the same elements. If this fails, write A ≠ B.
The sets {1, 2, 3, 4} and {3, 1, 4, 2} are equal. So are the primes less than 6 and the prime factors of 30: both give {2, 3, 5}. Equality can therefore link a roster, a verbal description, and an equation-defined set.
To prove inequality, one correctly identified difference is enough. If 0 belongs to A but not to B, the two sets are unequal, regardless of how many other members they share. To prove equality, you need both directions: no member of A is missing from B, and no extra member occurs in B.
Problem
Compare A = {0}, B = {x ∈ R : x > 15 and x < 5}, C = {x ∈ R : x − 5 = 0}, D = {x ∈ R : x² = 25}, and E, the positive integer roots of x² − 2x − 15 = 0.
- 1.The conditions for B are incompatible, so B = ∅. C = {5}. D = {−5, 5}.
- 2.Factor the equation defining E: x² − 2x − 15 = (x − 5)(x + 3). Its roots are 5 and −3.
- 3.Only 5 is a positive integer, so E = {5}. Hence C = E.
- 4.A differs from the other sets because it contains 0. B is the only empty set. D differs from C and E because −5 belongs only to D among those three.
- 5.The only equal pair is C and E. Counting written symbols would not have established this.
Problem
Compare the sets of letters in ALLOY and LOYAL.
- 1.Remove repeated occurrences of L in both words.
- 2.ALLOY gives {A, L, O, Y}; LOYAL gives {L, O, Y, A}.
- 3.Every member of either set occurs in the other. The sets are equal even though the words and the order of their letters differ.
Problem
Compare A = {n : n ∈ Z and n² ≤ 4} with B = {x : x ∈ R and x² − 3x + 2 = 0}.
- 1.For A, the integer values whose squares are at most 4 are −2, −1, 0, 1, and 2.
- 2.For B, factor (x − 1)(x − 2) = 0, so B = {1, 2}.
- 3.Although both sets contain 1 and 2, 0 belongs to A and not to B.
- 4.Thus A ≠ B. The shared members do not establish equality.
Equal counts do not imply equal sets. For example, {1, 2} and {4, 5} each have two members, but neither 1 nor 2 belongs to the second set. Also, an equation must be solved with the correct signs: x² + 5x + 6 = 0 has roots −2 and −3, not 2 and 3.
The second exercise boundary asks you to identify empty sets, classify finite and infinite collections, and compare equality. The reliable order is to identify the members or the membership rule first, then classify. The same method works for geometric collections, words, and equations.
Check your understanding
Classifying from appearance is unreliable, so these questions ask you to examine the actual membership rule. For equality questions, look for a member present in one set and absent from the other before deciding.
Quiz
Which set is empty?
The roots ±√2 are irrational. The other options contain −2 and 2, 1, and 0 respectively.
How many members does {∅} have?
Its only member is the empty set. The outer set is a singleton, not empty.
Which is infinite?
The integers 101, 102, 103, and further integers never stop. All other collections are bounded or fixed finite lists.
What is n({2, 2, 4, 4, 6})?
The distinct members are 2, 4, and 6. Repeated entries do not increase the count.
Which set equals {−3, 3}?
The integer roots of x² = 9 are −3 and 3. Restricting to N would retain only 3.
A and B each have four distinct members. What follows?
Equality requires the same members. Equal finite counts are compatible with either equal or unequal sets.
Key Takeaways
• An empty set has no members and is finite. • ∅, {0}, and {∅} must be distinguished. • Count distinct elements; a large or unknown finite count is still finite. • Apply the domain after solving an equation or inequality. • Equal sets have exactly the same members, not merely the same number of members. • One missing or extra member proves that two sets are unequal.