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From my textbook

The smallest equivalence relation $R_1$ in the set {$1,2,3$} containing $(1,2)$ and $(2,1)$ is {$(1,1)(2,2)(3,3)(1,2)(2,1)$}

Again,

In mathematics, an equivalence relation is a binary relation that is at the same time a reflexive relation, a symmetric relation and a transitive relation.

Is {$(1,2)(2,1)(1,1)$} the transitive relation here? If not,how is it an equivalence relation?

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    $\begingroup$ There is no the trasitive relation. This relation just has the property of being transitive. And $(1,2),(2,1),(1,1)$ demonstrates that the application of the transitivity rule works for this triple. $\endgroup$ – M. Winter Mar 17 '17 at 8:19
  • $\begingroup$ So, {$(1,1)(2,2)(3,3)(1,2)(2,1)(1,1)(2,2)(3,3)(1,2)(2,1)$} is not an equivalence relation? $\endgroup$ – user237454 Mar 17 '17 at 8:22
  • $\begingroup$ A relation is a set, and a set contains everything at most once. This is not satisfied by your example. $\endgroup$ – M. Winter Mar 17 '17 at 8:22
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    $\begingroup$ A reminder that in the definition of being a transitive relation, we say that if $xRy$ and $yRz$ then we must have $xRz$. When we are calling things $x,y,z$ they are allowed to be referring to the same element. Just because there are no examples of $x,y,z$ with all of them different such that $xRyRz$ this shouldn't bother us. Remember that $=$ is the prototypical example of an equivalence relation. The transitive property there is just saying if $x=y$ and $y=z$ then we could combine these as $x=y=z$ and remove the middle leaving only the ends to say $x=z$. $\endgroup$ – JMoravitz Mar 17 '17 at 8:54
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$\{(1,2), (2,1), (1,1)\}$ is a transitive relation because it satisfies the definition of transitive relations, which is that for every $x,y, z\in\{1,2\}$, we know that if $(x,y)\in R$ and $(y,z)\in R$, then we also have $(x,z)\in R$.

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Since R must be an equivalence relation, it must be reflexive and therefore ${(1,1),(2,2),(3,3)} \subseteq R$. Now, because we need that $\{(1,2),(2,1)\} \subseteq R$, we get $\{(1,1),(2,2),(3,3),(1,2),(2,1)\} \subseteq R$. We've shown why no element of $R$ can be removed , and the previous set is itself a well defined equivalence relation, so this makes $R = \{(1,1),(2,2),(3,3),(1,2),(2,1)\} $ the smallest equivalence relation possible with the imposed condition.

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