Show that the number of equivalence relation in the set $\{1{,}2{,}3\}$ containing $(1{,}2)$ and $(2{,}1)$ is two.

Step 1: Understand Equivalence Relations

An equivalence relation $R$ on a set $A$ must satisfy three properties: reflexivity, symmetry, and transitivity. The given set is $A = \{1{,}2{,}3\}$. We are looking for equivalence relations that contain the ordered pairs $(1{,}2)$ and $(2{,}1)$.

Step 2: Apply Reflexivity

For any equivalence relation $R$ on $A$, it must be reflexive. This means that for every element $a \in A$, the pair $(a,a)$ must be in $R$. So, $(1{,}1)$, $(2{,}2)$, and $(3{,}3)$ must be part of any equivalence relation on $A$.

Step 3: Incorporate Given Pairs and Symmetry

We are given that $(1{,}2)$ and $(2{,}1)$ are in the relation $R$. This already satisfies the symmetry property for these two elements. We combine these with the reflexive pairs to form a base for our relations.

Step 4: Consider Transitivity and Possible Partitions

Now we consider transitivity. Since $(1{,}2) \in R$ and $(2{,}1) \in R$, transitivity implies that if $(1{,}2) \in R$ and $(2,x) \in R$, then $(1,x) \in R$. Similarly for other combinations. This means that 1 and 2 must be in the same equivalence class. We have two possibilities for partitions of $A$ that include 1 and 2 in the same class: \\ 1. The partition is $\{\{1{,}2\}, \{3\}\}$. This corresponds to the equivalence relation $R_1 = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}1)\}$. \\ 2. The partition is $\{\{1{,}2{,}3\}\}$. This means all elements are related to each other, forming the universal relation $R_2 = A \times A = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}1), (1{,}3), (3{,}1), (2{,}3), (3{,}2)\}$. Both $R_1$ and $R_2$ contain $(1{,}2)$ and $(2{,}1)$ and satisfy all properties of an equivalence relation.

Step 5: Conclusion

We have found two distinct equivalence relations on the set $A = \{1{,}2{,}3\}$ that contain the pairs $(1{,}2)$ and $(2{,}1)$. These are $R_1 = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}1)\}$ and $R_2 = \{(1{,}1), (2{,}2), (3{,}3), (1{,}2), (2{,}1), (1{,}3), (3{,}1), (2{,}3), (3{,}2)\}$.

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