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

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The smallest equivalence relation $R_{1}$ containing $(1,2)$ and $(2,1)$ is $\{(1,1)$ $(2,2)$, $(3,3)$, $(1,2)$, $(2,1)\}$. Now we are left with only $4$ pairs namely $(2,3)$, $(3,2)$ $,(1,3)$ and $(3,1) $. If we add any one, say $(2,3)$ to $R_{1},$ then for symmetry we must add $(3,2)$ also and now for transitivity we are forced to add $( 1,3 )$ and $( 3,1)$. Thus, the only equivalence relation bigger than $R_{1}$ is the universal relation. This shows that the total number of equivalence relations containing $(1,2) $ and $(2,1) $ is two.

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Let $A=\{1,2,3\} .$ Then number of relations containing $(1,2)$ and $(1,3)$ which are reflexive and symmetric but not transitive is