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For Set A = {1,2,3,4}

Is it possible to generate a relation that is reflexive and symmetric, but not transitive? The textbook says, (1,1),(2,2),(3,3),(4,4),(1,2),(2,1)(2,3)(3,2) but isn't this transitive?

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up vote 2 down vote accepted

No it is not transitive, because it contains $(1,2)$ and $(2,3)$ but not $(1,3)$.

An example on the real numbers: If $x\sim y$ means $|x-y|<1$, then $\sim$ is reflexive, symmetric, and not transitive.

The example you gave could be modified slightly by removing $4$ from $A$ (and $(4,4)$ from the relation), while still being reflexive, symmetric, and not transitive. In that case, it could be defined as $a\sim b$ if and only if $|a-b|\leq 1$.

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It contains (1,2),(2,1) and there is (1,1) – Aaron Nov 13 '12 at 4:37
That is true, but does not say that the relation is transitive. Transitivity requires that for all $a,b,c$, if $(a,b)$ and $(b,c)$ are in the relation, then so is $(a,c)$. To show that the relation is not transitive it suffices to find one instance of $a,b,c$ where $(a,b)$ and $(b,c)$ are in the relation but $(a,c)$ is not. The example above is one such instance. – Jonas Meyer Nov 13 '12 at 4:39
Thank you so much. – Aaron Nov 13 '12 at 4:44

You have $1\sim 2$ and $2\sim 3$; do you also have $1\sim 3$, as required for transitivity? (I’m using $\sim$ to denote the relation.)

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So does it go for all ordered pair in the relation? there is (1,2) and (2,1) and there is (1,1) which is transitive – Aaron Nov 13 '12 at 4:38
@Aaron: You have to read definitions carefully and literally. In order for a relation $\sim$ to be transitive, it has to be the case that whenever $a\sim b$ and $b\sim c$, then $a\sim c$. This has to happen in all possible cases. – Brian M. Scott Nov 13 '12 at 4:42
Thank you so much – Aaron Nov 13 '12 at 4:43

It is not transitive since $(1,2)$ and $(2,3)$ are in $R$ but $(1,3)$ is not.

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