Understanding Equivalence Relations and Set Partitions

The question asks for the maximum number of possible equivalence relations that can be defined on the set A = {1, 2, 3, 4}. An equivalence relation on a set A is a relation that satisfies three properties: reflexivity, symmetry, and transitivity.

A key concept is that the number of equivalence relations on a set is equal to the number of ways that set can be partitioned. A partition of a set is a division of the set into non-empty, non-overlapping subsets whose union is the original set.

Calculating the Number of Equivalence Relations using Bell Numbers

The number of partitions of a set with n elements is given by the n-th Bell number, denoted as $B_n$. Our set A has 4 elements, so we need to find $B_4$.

We can calculate Bell numbers using the recurrence relation: $B_{n+1} = \sum_{k=0}^{n} \binom{n}{k} B_k$ where $B_0 = 1$.

Step-by-Step Calculation:

• $B_0$ = 1 (By definition)
• $B_1$ = $\binom{0}{0} B_0 = 1 \times 1 = 1$
• $B_2$ = $\binom{1}{0} B_0 + \binom{1}{1} B_1 = (1 \times 1) + (1 \times 1) = 1 + 1 = 2$
• $B_3$ = $\binom{2}{0} B_0 + \binom{2}{1} B_1 + \binom{2}{2} B_2 = (1 \times 1) + (2 \times 1) + (1 \times 2) = 1 + 2 + 2 = 5$
• $B_4$ = $\binom{3}{0} B_0 + \binom{3}{1} B_1 + \binom{3}{2} B_2 + \binom{3}{3} B_3 = (1 \times 1) + (3 \times 1) + (3 \times 2) + (1 \times 5) = 1 + 3 + 6 + 5 = 15$

Therefore, the 4th Bell number, $B_4$, is 15.

Listing Partitions for Set A = {1, 2, 3, 4}

We can also find the number of equivalence relations by listing all possible partitions of the set A = {1, 2, 3, 4}.

The partitions can be categorized based on the sizes of the subsets:

1. Type 1: One subset containing all elements.
   • {{1, 2, 3, 4}}
   Number of partitions = 1.
2. Type 2: One subset with 3 elements and another with 1 element.
   • {{1, 2, 3}, {4}}
   • {{1, 2, 4}, {3}}
   • {{1, 3, 4}, {2}}
   • {{2, 3, 4}, {1}}
   Number of partitions = $\binom{4}{3} = 4$.
3. Type 3: Two subsets, each with 2 elements.
   • {{1, 2}, {3, 4}}
   • {{1, 3}, {2, 4}}
   • {{1, 4}, {2, 3}}
   Number of partitions = $\frac{\binom{4}{2} \times \binom{2}{2}}{2!} = \frac{6 \times 1}{2} = 3$. (We divide by 2! because the order of the two subsets of size 2 doesn't matter).
4. Type 4: One subset with 2 elements and two subsets with 1 element each.
   • {{1, 2}, {3}, {4}}
   • {{1, 3}, {2}, {4}}
   • {{1, 4}, {2}, {3}}
   • {{2, 3}, {1}, {4}}
   • {{2, 4}, {1}, {3}}
   • {{3, 4}, {1}, {2}}
   Number of partitions = $\binom{4}{2} = 6$. (Choose the 2 elements that go together).
5. Type 5: Four subsets, each with 1 element.
   • {{1}, {2}, {3}, {4}}
   Number of partitions = 1.

Total number of partitions = 1 + 4 + 3 + 6 + 1 = 15.

Conclusion

Both methods confirm that the total number of possible equivalence relations on the set A = {1, 2, 3, 4} is 15.