Equivalent Sets

Two sets are equivalent (also called equipotent or equinumerous) when they have exactly the same number of elements. Formally, two sets A and B are equivalent if there exists a bijection between them, guaranteeing a one-to-one correspondence where each element of the first set pairs with exactly one unique element of the second set, and vice versa.

In standard mathematical practice, the clearest and most common way to represent this condition is by equating their cardinalities, written as |A| = |B|. It is also common to denote this relationship using the equivalence symbol A ~ B, and some classical texts even employ the triple bar A ≡ B, although the latter notation is less common today.

It is crucial to distinguish between equivalent sets and equal sets to avoid common misconceptions. Two sets are equal when they contain exactly the same elements, whereas they are equivalent when they share the same number of elements, regardless of what those elements actually are.

For this reason, it holds as a general rule that every pair of equal sets is necessarily equivalent. However, the converse is not true: two sets can be equivalent while differing entirely in their individual members.

Examples

To clearly understand how set equivalence works, we examine several practical cases below, covering both finite collections and infinite numerical structures:

1. If we define the set of vowels in the English alphabet as V = {a, e, i, o, u} and the set of the first five prime numbers as P = {2, 3, 5, 7, 11}, we observe that both have the same number of elements. Counting their members shows that |V| = 5 and |P| = 5, which means V ~ P, despite being composed of completely different types of elements.
2. Consider the set of plane geometric figures F = {triangle, square, pentagon, hexagon} and the set of colors C = {red, blue, yellow, green}. Since both contain exactly four elements, equal cardinality is immediately verified: |F| = |C| = 4, making them equivalent sets.
3. Studying natural numbers, let set A = {x ∈ N | x is a factor of 6} and set B = {x ∈ N | x is a multiple of 3 ∧ x ≤ 12}. Writing both collections in roster form yields A = {1, 2, 3, 6} and B = {3, 6, 9, 12}. Each set contains 4 elements (|A| = |B| = 4), demonstrating that A ~ B.
4. Now examine two sets defined under identical algebraic conditions: M = {x ∈ Z | x² = 9} and N = {-3, 3}. Solving the equation for the first set gives the values -3 and 3, so M = {-3, 3}. Here, because M and N share the exact same elements, they are equal sets (M = N) and, consequently, also equivalent sets since |M| = |N| = 2.
5. In the realm of infinite structures, consider the set of natural numbers N = {1, 2, 3, 4,...} and the set of positive even numbers E = {2, 4, 6, 8,...}. While intuition might suggest there are more natural numbers than even numbers, we can establish a direct bijective correspondence by assigning each natural number n to the even number 2n. Because this one-to-one correspondence exists, both sets share the same countably infinite cardinality, formally proving that N ~ E.

Properties of the Equivalence Relation

From a formal algebraic perspective, set equivalence forms an equivalence relation over any given collection of sets. This means it satisfies the reflexive, symmetric, and transitive properties.

1) Reflexive property: every set is equivalent to itself, as it identically shares its own number of elements.

A ~ A

2) Symmetric property: if one set is equivalent to another, the second set is necessarily equivalent to the first.

A ~ B → B ~ A

3) Transitive property: if a first set is equivalent to a second, and that second set is equivalent to a third, then the first set is equivalent to the third.

(A ~ B ∧ B ~ C) → A ~ C

Solved Practice Problems

To determine whether two or more sets are equivalent, the standard procedure is to calculate the cardinality of each set. If counting yields the exact same number of elements, the equivalence relation is confirmed. Below, we walk through four practical exercises step by step.

Problem 1

Given the sets written in roster form M = {m, a, t, h} and N = {3, 6, 9, 12}, determine whether they are equivalent sets, justifying the answer using their respective cardinalities.

Solution

We begin by directly counting the number of distinct elements in each set.

For set M, we count four distinct letters, so its cardinality is |M| = 4. Meanwhile, set N contains four distinct numerical values, giving |N| = 4.

Since both sets satisfy the equality of cardinalities |M| = |N|, we conclude that M ~ N. As this shows, no similarity in the nature of the elements is required for sets to be equivalent.

Problem 2

Consider the sets defined in set-builder notation A = {x ∈ N | x is a factor of 12} and B = {x ∈ Z | x² ≤ 4}. Write both sets in roster form and determine whether they are equivalent.

Solution

We start by listing the elements of each set according to their mathematical conditions.

For set A, we find all natural numbers that evenly divide 12:

A = {1, 2, 3, 4, 6, 12}

For set B, we identify all integers whose square is less than or equal to 4. Solving this inequality yields the integers in the closed interval [-2, 2]:

B = {-2, -1, 0, 1, 2}

Next, we count the elements in each set to compare their sizes. We observe that |A| = 6, while |B| = 5. Because |A| ≠ |B|, they do not have the same number of elements. Therefore, we conclude that A and B are not equivalent sets.

Problem 3

Let sets be given as P = {1, 2, 5}, Q = {x ∈ N | x is a factor of 10 ∧ x ≠ 10}, and R = {a, b, c}. Determine which pairs of sets are equivalent and which pairs are also equal sets.

Solution

First, we express set Q in roster form. The natural factors of 10 are 1, 2, 5, and 10. Excluding 10 according to the given condition leaves Q = {1, 2, 5}.

Comparing P with Q, we see that they contain the exact same elements. Consequently, they are equal sets (P = Q) and, simultaneously, |P| = |Q| = 3, which means they are also equivalent (P ~ Q).

Next, examining set R shows that it contains 3 letters, so |R| = 3. Comparing P with R, we find that |P| = |R|, which means P ~ R; however, clearly P ≠ R because their elements are entirely different.

This confirms that equality between sets always guarantees equivalence, but two sets can be equivalent without being equal.

Problem 4

Consider the sets given in roster form A = {2, 4, 6, 8} and B = {1, 3, 5, 7, 2k + 1}, where k is a positive integer. Find the possible values of k such that A and B are equivalent sets.

Solution

We begin by checking the cardinality of the known set. Counting the elements of A, we see immediately that |A| = 4.

For the sets to be equivalent, the condition |B| = |A| = 4 must hold. However, looking at the initial list for B, there appear to be five elements: four defined numerical values (1, 3, 5, and 7) plus the expression containing the parameter k.

Recall that in set theory, duplicate elements do not increase a set's cardinality. Therefore, for set B to contain exactly four distinct elements, the term 2k + 1 must equal one of the values already listed.

We set up equations for each possible case:

1) If 2k + 1 = 1, solving gives 2k = 0 → k = 0. We reject this value because the problem specifies that k must be a positive integer.

2) If 2k + 1 = 3, subtracting 1 from both sides gives 2k = 2 → k = 1.

3) If 2k + 1 = 5, solving directly gives 2k = 4 → k = 2.

4) If 2k + 1 = 7, we find the final possibility: 2k = 6 → k = 3.

Any of these three values (k = 1, k = 2, or k = 3) makes the fifth term duplicate an existing element in B. As a result, B effectively reduces to four distinct elements (|B| = 4), successfully satisfying the equivalence A ~ B.